Week 1 Notes

Mathematics for
Data Science 2

Vectors, matrices, systems of linear equations and determinants, with every worked example and every graded-assignment question solved where it fits.

7 chapters53 worked examples~66 min read

VectorsMatricesLinear systemsDeterminants

Chapters

01VectorsPart 1
02MatricesPart 2
03Systems of Linear EquationsPart 3
04Determinants — Part 1Part 4
05Determinants — Part 2Part 5
06Extra Practice ProblemsPractice
07Revision Sheet and Exam ChecklistAll parts

Part 1

Vectors

A vector is a list of numbers that we add and scale coordinate by coordinate. In $\mathbb{R}^2$ it is also an arrow.

1.1Vectors

A vector is a list of numbers. A vector with $n$ entries is a vector in $\mathbb{R}^n$ , and it can be written as a row or as a column. Vectors store a row or column of data, and arithmetic on them is done coordinate by coordinate.

1.2Vector addition and scalar multiplication

Addition

(a_1, \dots, a_n) + (b_1, \dots, b_n) = (a_1 + b_1, \dots, a_n + b_n)

Only vectors of the same dimension can be added.

Scalar multiplication

c\,(a_1, \dots, a_n) = (c a_1, \dots, c a_n)

Shopping-demand example: $(3, 2) + (5, 6) = (8, 8)$ . Repeating a demand twice is scaling: $(8, 8, 4, 10, 4) + (8, 8, 4, 10, 4) = 2(8, 8, 4, 10, 4)$ .

1.3Geometric representation in $\mathbb{R}^2$

A point $(a, b)$ is the arrow from the origin $(0, 0)$ to $(a, b)$ . Writing $\hat{\imath} = (1, 0)$ and $\hat{\jmath} = (0, 1)$ for the unit arrows along the axes,

(a, b) = a\hat{\imath} + b\hat{\jmath}, \qquad (1, 2) = 1\hat{\imath} + 2\hat{\jmath}, \qquad (-1, -1) = -\hat{\imath} - \hat{\jmath}

Figure 1. A vector is a combination of î and ĵ

Figure 2. Adding vectors: head-to-tail, or the parallelogram law

Figure 3. Scalar multiplication stretches, shrinks or flips

Check.

(1, 2) + (2, 1) = (3, 3)

: the diagonal of the parallelogram built on

(1, 2)

and

(2, 1)

ends at

(3, 3)

1.4Vectors in a physical context

A vector has magnitude (the length of the arrow) and direction (the arrowhead). It has no fixed position: sliding it parallel to itself does not change it. Velocity, acceleration and force are vectors.

1.5Going deeper: vectors as lists, arrows and combinations

A list and an arrow. $(a, b)$ is a list of two numbers and also the arrow from the origin to the point $(a, b)$ . Every vector in $\mathbb{R}^2$ is a combination $a\hat{\imath} + b\hat{\jmath}$ of the two unit arrows.
Linear combinations. An expression like $cA + dB$ scales each vector and then adds coordinate by coordinate. Most exam questions are about finding a missing scalar from such an expression.
Subtraction. $A - B = A + (-1)B$ . Geometrically, with both arrows starting at the origin, $A - B$ is the arrow from the tip of $B$ to the tip of $A$ .
Solving a vector equation. Two vectors are equal exactly when every coordinate is equal, so one vector equation is really $n$ ordinary equations, one per coordinate. Start with a coordinate that has a known number on the right.
Dimensions must match. $(1, 2) + (1, 2, 3)$ is not defined, because the vectors live in different spaces.
Length (extra). The length of $(a, b)$ is $\sqrt{a^2 + b^2}$ , by Pythagoras. For example the length of $(3, 4)$ is $5$ .

1.6Exam tips and common mistakes

Exam tips

Write a vector equation as one equation per coordinate before solving.
Check dimensions before adding two vectors.
For $A - B$ , subtract each coordinate of $B$ from the same coordinate of $A$ , in that order.

Common mistakes

Adding vectors of different dimensions.
Mixing up the order in $A - B$ , which gives the negative of $B - A$ .
Multiplying only the first coordinate by the scalar.
Treating a list as an arrow without a starting point at the origin (vectors are free to slide, but coordinates assume the origin).

Practice questions (graded-assignment style)

Q1A combination

Numerical Answer Type (NAT).

Let $A = (2, -1, 3)$ and $B = (1, 4, -2)$ . Find the sum of the coordinates of $3A - 2B$ .

Solution

Idea

Compute 3A − 2B coordinate by coordinate.

$3A = (6, -3, 9)$ and $2B = (2, 8, -4)$ .
$3A - 2B = (4, -11, 13)$ .
Sum of coordinates: $4 - 11 + 13 = 6$ .
Answer: 6

Q2Which sum is not defined?

Multiple Choice Question (MCQ): choose the one correct option.

Which of these sums is not defined?

(1, 2) + (3, 4)

(1, 2, 3) + (4, 5)

2(1, 2) + (0, 1)

(0, 0) + (5, 6)

Solution

Idea

Only vectors of the same dimension can be added.

Option 2 adds a vector in $\mathbb{R}^3$ to one in $\mathbb{R}^2$ .
The other three add vectors of the same dimension.
Answer: option 2.

Q3Find the unknown coordinate

Numerical Answer Type (NAT).

$c\,(1, 2) + (3, -1) = (5, j)$ . Find $j$ .

Solution

Idea

The first coordinate has a known number, so find c first.

First coordinate: $c + 3 = 5$ , so $c = 2$ .
Second coordinate: $2c - 1 = 2 \cdot 2 - 1 = 3$ , so $j = 3$ .
Answer: 3

Q4Geometry of vectors

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements.

(a, b) = a\hat{\imath} + b\hat{\jmath}

2(1, 1)

points in the same direction as

(1, 1)

and is twice as long.

(-1)(1, 2) = (-1, -2)

points in the opposite direction to

(1, 2)

A + B

and

B + A

are in general different vectors.

Solution

Idea

Addition of vectors is commutative; scaling by a positive number keeps the direction.

Options 1, 2 and 3 are true.
Option 4 is false: coordinate-wise addition gives the same vector either way.
Answer: options 1, 2 and 3.

Revision overview

VectorList of n numbers
A vector in Rⁿ, row or column

AdditionSame dimension
Coordinate by coordinate

VectorsWeek 1 · Part 1

Scalar multiplec(a₁, …, aₙ) = (ca₁, …, caₙ)

Picture in R²(a, b) = aî + bĵ
Head-to-tail / parallelogram

AnswersQ1 all four · Q4 c = −2 · Q5 j + k = 5

Revision: vectors on one page

Next · Part 2Matrices

Part 2

Matrices

A matrix is a table of numbers with its own arithmetic. The one rule to remember is how the orders must match when multiplying.

2.1Order of a matrix

A matrix is a rectangular array of numbers in rows and columns; the numbers are its entries. An $m \times n$ matrix has $m$ rows and $n$ columns. The $(i, j)$ -th entry sits in row $i$ , column $j$ .

Figure 4. Rows first, then columns: this matrix is 3 × 4

2.2Special matrices

Type	Definition
Square	rows = columns
Diagonal	square, non-diagonal entries are 0
Scalar	diagonal, all diagonal entries equal
Identity $I$	scalar, diagonal entries all 1

Figure 5. Each type is a special case of the previous one

2.3Matrix addition, scalar multiplication and multiplication

Addition and scalar multiple

(A + B)_{ij} = A_{ij} + B_{ij}, \qquad (cA)_{ij} = c\,A_{ij}

Addition needs the same order; it is entry by entry.

Matrix multiplication

A_{m \times n}\, B_{n \times p} = (AB)_{m \times p}, \qquad (AB)_{ij} = \sum_{k=1}^{n} A_{ik} B_{kj}

Entry $(i, j)$ is the dot product of row $i$ of $A$ with column $j$ of $B$ .

Dimension check

$AB$ is defined only when the number of columns of $A$ equals the number of rows of $B$ .

Figure 6. One entry of AB: row of A times column of B

Important properties

Property	Statement
Associativity of addition	$(A + B) + C = A + (B + C)$
Associativity of multiplication	$(AB)C = A(BC)$
Commutativity of addition	$A + B = B + A$
Multiplication is not commutative	$AB \neq BA$ in general
Scalar over addition	$\lambda(A + B) = \lambda A + \lambda B$
Scalar and a product	$\lambda(AB) = (\lambda A)B = A(\lambda B)$
Left distributivity	$A(B + C) = AB + AC$
Right distributivity	$(A + B)C = AC + BC$

Figure 7. Exam trap: both AB and BA exist but differ

Three layers of "not commutative". With a

2 \times 3

and a

3 \times 4

matrix,

AB

exists but

BA

is not even defined. With square matrices both exist but are usually different. Only special pairs happen to commute (for example

A

and

I

2.4Going deeper: when AB exists, and three ways to see it

When is AB defined? Compare the inner numbers

The five-second test. Write the orders under the matrices and compare the two inner numbers:
$(m \times n)\,(n \times p) \;\Longrightarrow\; AB \text{ has order } m \times p$
If the inner numbers ( $n$ and $n$ ) are equal, the product exists and its order comes from the outer numbers. If they differ, stop: the product is not defined.
Entry view. One entry of $AB$ is a dot product:
$(AB)_{ij} = (\text{row } i \text{ of } A) \cdot (\text{column } j \text{ of } B)$
Row view. Row $i$ of $AB$ is row $i$ of $A$ multiplied by the whole matrix $B$ . For example, if row 1 of $A$ is $(a, b, c)$ and $B$ has three rows, row 1 of $AB$ is $a\,(\text{row 1 of } B) + b\,(\text{row 2 of } B) + c\,(\text{row 3 of } B)$ .
Column view. Column $j$ of $AB$ is the whole matrix $A$ multiplied by column $j$ of $B$ .
Zero products. $AB = 0$ does not mean $A = 0$ or $B = 0$ . For example $B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \neq 0$ , but $B^2 = 0$ .
Identity and powers. $AI = IA = A$ . Powers $A^2, A^3, \dots$ exist only for square matrices.
Expanding a square. Always true:
$(A + B)^2 = A^2 + AB + BA + B^2$
It equals $A^2 + 2AB + B^2$ only when $AB = BA$ .
Cost. Computing $AB$ for $(m \times n)(n \times p)$ means $m \cdot p$ entries, each needing $n$ multiplications: $mnp$ multiplications in all.

2.5Transpose, trace and powers: the rules you will be tested on

Matrix questions rarely ask for a long multiplication. They ask which rule is valid. Below are the rules for the transpose, the trace and matrix powers, with the reason for each and the wrong version that looks tempting.

A. The transpose

Rule	Why / remark
$(A^T)^T = A$	Swapping rows and columns twice puts everything back.
$(A + B)^T = A^T + B^T$ , $(cA)^T = cA^T$	Entry by entry.
$(AB)^T = B^TA^T$ (order reverses)	Entry $(i, j)$ of $(AB)^T$ is entry $(j, i)$ of $AB$ , which is (row $j$ of $A$ ) times (column $i$ of $B$ ). That is (row $i$ of $B^T$ ) times (column $j$ of $A^T$ ).
$(A^k)^T = (A^T)^k$	Apply the product rule $k$ times.

Check with sizes. Let $A$ be $2 \times 3$ and $B$ be $3 \times 2$ . Then $AB$ is $2 \times 2$ , so $(AB)^T$ is $2 \times 2$ . $B^TA^T$ is $(2 \times 3)(3 \times 2) = 2 \times 2$ ✓, but $A^TB^T$ is $(3 \times 2)(2 \times 3) = 3 \times 3$ , the wrong shape. So the reversed order is not a convention: it is forced.

Numbers. $A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 0 \\ 2 & 1 \\ 0 & 4 \end{bmatrix}$ give $AB = \begin{bmatrix} 5 & 2 \\ 2 & 13 \end{bmatrix}$ , and $B^TA^T = (AB)^T = \begin{bmatrix} 5 & 2 \\ 2 & 13 \end{bmatrix}$ ✓.

B. Symmetric and skew-symmetric matrices

Symmetric: $A^T = A$ . Skew-symmetric: $A^T = -A$ . Both must be square. A skew-symmetric matrix has zeros on the diagonal ( $a_{ii} = -a_{ii}$ ) and therefore trace $0$ .
For any matrix $A$ (even a rectangular one), $AA^T$ and $A^TA$ are symmetric: $(AA^T)^T = (A^T)^TA^T = AA^T$ .
For a square $A$ : $A + A^T$ is symmetric and $A - A^T$ is skew-symmetric. So every square matrix splits as
$A = \underbrace{\tfrac12(A + A^T)}_{\text{symmetric}} + \underbrace{\tfrac12(A - A^T)}_{\text{skew-symmetric}}$
For $A = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$ : the symmetric part is $\begin{bmatrix} 1 & 3 \\ 3 & 3 \end{bmatrix}$ and the skew part is $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ ; they add back to $A$ ✓.
The product of two symmetric matrices is not always symmetric. $(AB)^T = B^TA^T = BA$ , so $AB$ is symmetric exactly when $AB = BA$ . For $A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ , $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ : $AB = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ , which is not symmetric.
If $A$ is skew-symmetric, then $A^2$ is symmetric ( $(A^2)^T = (A^T)^2 = (-A)^2 = A^2$ ) while $A^3$ is skew-symmetric ( $(A^3)^T = (-A)^3 = -A^3$ ). Odd powers keep the type, even powers become symmetric.

C. The trace

The trace of a square matrix is the sum of its diagonal entries.

$\operatorname{tr}(A + B) = \operatorname{tr}A + \operatorname{tr}B$ , $\operatorname{tr}(cA) = c\operatorname{tr}A$ , $\operatorname{tr}(A^T) = \operatorname{tr}A$ .
$\operatorname{tr}(AB) = \operatorname{tr}(BA)$ , even when $AB \neq BA$ , and even when $A$ and $B$ are rectangular (so $AB$ and $BA$ have different sizes). The reason: $\operatorname{tr}(AB) = \sum_i\sum_k a_{ik}b_{ki}$ , and swapping the order of the sums gives $\operatorname{tr}(BA)$ . In the example above, $\operatorname{tr}(AB) = 5 + 13 = 18$ and $\operatorname{tr}(BA) = 1 + 5 + 12 = 18$ .
$\operatorname{tr}(AB) \neq \operatorname{tr}A \cdot \operatorname{tr}B$ in general. (For $A = B = I_2$ , $\operatorname{tr}(AB) = 2$ but the product of traces is $4$ .)

D. Powers, and the identities that fail without commutativity

Because $AB \neq BA$ in general, the school identities need a condition:

School identity	For matrices
$(A + B)^2 = A^2 + 2AB + B^2$	$(A + B)^2 = A^2 + AB + BA + B^2$ . The school form holds only when $AB = BA$ .
$(A - B)(A + B) = A^2 - B^2$	$(A - B)(A + B) = A^2 + AB - BA - B^2$ . Holds only when $AB = BA$ .
$(AB)^2 = A^2B^2$	$(AB)^2 = ABAB$ . Holds only when $AB = BA$ .
$AB = 0 \Rightarrow A = 0$ or $B = 0$	False: $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \mathbf{0}$ .

Matrices that commute with everything built from themselves are safe: $A$ commutes with $I$ , with $A^k$ and with $cA$ , so $(A + I)^2 = A^2 + 2A + I$ is always valid.

Diagonal matrices multiply and power entry by entry: $\operatorname{diag}(d_1, d_2, d_3)^k = \operatorname{diag}(d_1^k, d_2^k, d_3^k)$ .
Idempotent means $A^2 = A$ ; then every power $A^k = A$ . Involutory means $A^2 = I$ . Nilpotent means $A^k = 0$ for some $k$ .
If $A^2 = cA$ (like the all-ones matrix with $c = 3$ ), then $A^n = c^{n-1}A$ .

2.6Exam tips and common mistakes

Exam tips

Write the orders next to the matrices before multiplying.
Compute one entry at a time: go along the row of the left matrix and down the column of the right matrix.
For "find the order of A" questions, use the inner and outer numbers backwards (as in GA 3).

Common mistakes

Multiplying entry by entry (that is not matrix multiplication).
Assuming $AB = BA$ , or expanding $(A + B)^2$ as $A^2 + 2AB + B^2$ .
Squaring a matrix by squaring each entry.
Trying to compute $AB$ without checking that the inner numbers match.

Practice questions (graded-assignment style)

Q1Which products exist?

Multiple Select Question (MSQ): select all that apply.

$A$ is $2 \times 3$ , $B$ is $3 \times 4$ , $C$ is $4 \times 2$ . Which products are defined?

AB

BA

BC

CA

AC

Solution

Idea

Check that the inner numbers match.

$AB$ : $(2 \times 3)(3 \times 4)$ , inner 3 = 3, result $2 \times 4$ . Defined.
$BA$ : $(3 \times 4)(2 \times 3)$ , 4 ≠ 2. Not defined.
$BC$ : $(3 \times 4)(4 \times 2)$ , 4 = 4, result $3 \times 2$ . Defined.
$CA$ : $(4 \times 2)(2 \times 3)$ , 2 = 2, result $4 \times 3$ . Defined.
$AC$ : $(2 \times 3)(4 \times 2)$ , 3 ≠ 4. Not defined.
Answer: options 1, 3 and 4.

Q2Count the multiplications

Numerical Answer Type (NAT).

$A$ is $2 \times 3$ and $B$ is $3 \times 4$ . How many multiplications of numbers are needed to compute $AB$ ?

Solution

Idea

Count the entries of AB, then the work per entry.

$AB$ is $2 \times 4$ , so it has 8 entries.
Each entry is a dot product of 3 pairs, which needs 3 multiplications.
Total: $8 \times 3 = 24$ .
Answer: 24

Q3One entry

Numerical Answer Type (NAT).

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ , $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ . Find $(AB)_{12}$ .

Solution

Idea

Row 1 of A times column 2 of B.

Row 1 of $A$ is $(1, 2)$ and column 2 of $B$ is $(1, 0)$ .
$(AB)_{12} = 1 \cdot 1 + 2 \cdot 0 = 1$ .
Answer: 1

Q4Always true?

Multiple Select Question (MSQ): select all that apply.

For square matrices $A, B, C$ of the same order, which statements are always true?

(A + B)C = AC + BC

AB = BA

(AB)C = A(BC)

(A + B)^2 = A^2 + 2AB + B^2

Solution

Idea

Distributivity and associativity hold; commutativity does not.

Options 1 and 3 are standard properties.
Option 2 is false in general.
Option 4 would need $AB = BA$ , since $(A + B)^2 = A^2 + AB + BA + B^2$ .
Answer: options 1 and 3.

Q5A power

Numerical Answer Type (NAT).

$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ . Find the top-right entry of $A^5$ .

Solution

Idea

Compute A² and spot the pattern.

$A^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $A^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$ .
In general $A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$ , so the top-right entry of $A^5$ is 5.
Answer: 5

T1The transpose of a triple product

Multiple Choice Question (MCQ): choose the one correct option.

$A$ , $B$ , $C$ are matrices such that $ABC$ exists. Then $(ABC)^T$ equals

A^TB^TC^T

C^TB^TA^T

B^TA^TC^T

C^TA^TB^T

Solution

Idea

Apply $(XY)^T = Y^TX^T$ twice: the order reverses completely.

$(ABC)^T = ((AB)C)^T = C^T(AB)^T = C^T(B^TA^T) = C^TB^TA^T$ .
Size check with $A$ : $2 \times 3$ , $B$ : $3 \times 4$ , $C$ : $4 \times 5$ . $ABC$ is $2 \times 5$ , so its transpose is $5 \times 2$ . $C^TB^TA^T$ is $(5 \times 4)(4 \times 3)(3 \times 2) = 5 \times 2$ ✓. Option 1 would be $(3 \times 2)\cdot(4 \times 3)$ , which is not even defined.
Answer: option 2.

T2The trace of a product in the other order

Numerical Answer Type (NAT).

Let $A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 2 & 1 \\ 0 & 4 \end{bmatrix}$ . Find $\operatorname{tr}(BA)$ .

Solution

Idea

$\operatorname{tr}(BA) = \operatorname{tr}(AB)$ , and $AB$ is only $2 \times 2$ , much smaller than $BA$ which is $3 \times 3$ .

$AB = \begin{bmatrix} 1\cdot 1 + 2 \cdot 2 + 0 & 1\cdot 0 + 2\cdot 1 + 0 \\ 0 + 1\cdot 2 + 3\cdot 0 & 0 + 1\cdot 1 + 3\cdot 4 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 2 & 13 \end{bmatrix}$ , so $\operatorname{tr}(AB) = 18$ .
Direct check: $BA = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 5 & 3 \\ 0 & 4 & 12 \end{bmatrix}$ , whose diagonal is $1 + 5 + 12 = 18$ ✓.
Answer: 18

T3Which statements about transposes hold?

Multiple Select Question (MSQ): select all that apply.

Let $A$ and $B$ be real matrices of suitable sizes. Choose the correct statements.

AA^T

is always symmetric.

(AB)^T = A^TB^T

A

is square,

A - A^T

is skew-symmetric.

The product of two symmetric matrices is always symmetric.

\operatorname{tr}(AB) = \operatorname{tr}(A)\operatorname{tr}(B)

for square

A

B

Solution

Idea

Check each with the rules of the section, and use a small counterexample when a statement is false.

1 ✓ $(AA^T)^T = (A^T)^TA^T = AA^T$ , for rectangular $A$ too.
2 ✗ The correct rule is $(AB)^T = B^TA^T$ ; $A^TB^T$ is generally a different matrix (or a different size, or undefined).
3 ✓ $(A - A^T)^T = A^T - A = -(A - A^T)$ .
4 ✗ $A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ are symmetric, but $AB = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ is not.
5 ✗ For $A = B = I_2$ , $\operatorname{tr}(AB) = 2$ while $\operatorname{tr}A \cdot \operatorname{tr}B = 4$ .
Answer: options 1 and 3.

T4The skew-symmetric part of a matrix

Numerical Answer Type (NAT).

Write $A = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$ as $S + K$ , with $S$ symmetric and $K$ skew-symmetric. Find the $(1, 2)$ entry of $K$ .

Solution

Idea

The skew part is $K = \tfrac12(A - A^T)$ , so $k_{ij} = \tfrac12(a_{ij} - a_{ji})$ .

$k_{12} = \tfrac12(a_{12} - a_{21}) = \tfrac12(2 - 4) = -1$ .
Full pieces: $S = \begin{bmatrix} 1 & 3 \\ 3 & 3 \end{bmatrix}$ and $K = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ . Check: $S + K = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} = A$ ✓.
Answer: -1

T5When does the school formula survive?

Multiple Choice Question (MCQ): choose the one correct option.

$A$ and $B$ are $n \times n$ matrices. For which condition is $(A + B)^2 = A^2 + 2AB + B^2$ guaranteed?

Always

AB = BA

AB = 0

A

and

B

both symmetric

Solution

Idea

Expand $(A + B)(A + B)$ without assuming commutativity.

$(A + B)^2 = A^2 + AB + BA + B^2$ . This equals $A^2 + 2AB + B^2$ exactly when $BA = AB$ .
Option 3: $AB = 0$ does not make $BA$ equal to $AB$ in general. Option 4: symmetric matrices need not commute (see T3).
Answer: option 2.

T6An idempotent matrix

Numerical Answer Type (NAT).

$A$ is a square matrix with $A^2 = A$ . If $(A + I)^2 = I + kA$ , find $k$ .

Solution

Idea

$A$ commutes with $I$ , so the school expansion is valid; then replace $A^2$ by $A$ .

$(A + I)^2 = A^2 + 2A + I$ .
Using $A^2 = A$ : $A + 2A + I = I + 3A$ . So $k = 3$ .
Answer: 3

T7The trace of a power of a diagonal matrix

Numerical Answer Type (NAT).

$D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ . Find $\operatorname{tr}(D^4)$ .

Solution

Idea

A power of a diagonal matrix is the diagonal matrix of the powers.

$D^4 = \operatorname{diag}(2^4, (-1)^4, 3^4) = \operatorname{diag}(16, 1, 81)$ .
$\operatorname{tr}(D^4) = 16 + 1 + 81 = 98$ .
Answer: 98

T8Powers of a skew-symmetric matrix

Multiple Select Question (MSQ): select all that apply.

$A$ is a real $3 \times 3$ skew-symmetric matrix. Choose the correct statements.

Every diagonal entry of

A

0

A^2

is symmetric.

A^3

is symmetric.

A + A^T

is the zero matrix.

Solution

Idea

Use $A^T = -A$ and $(A^k)^T = (A^T)^k$ .

1 ✓ $a_{ii} = -a_{ii}$ , so $a_{ii} = 0$ .
2 ✓ $(A^2)^T = (A^T)^2 = (-A)^2 = A^2$ .
3 ✗ $(A^3)^T = (-A)^3 = -A^3$ : $A^3$ is skew-symmetric. It would be symmetric only if it were the zero matrix.
4 ✓ $A + A^T = A - A = 0$ .
Answer: options 1, 2 and 4.

Revision overview

Orderm × n: m rows, n columns
(i, j) = row i, column j

Special typesSquare ⊃ diagonal ⊃ scalar ⊃ identity

Add / scaleSame order, entry by entry
(cA)ᵢⱼ = c·Aᵢⱼ

MatricesWeek 1 · Part 2

Product(m×n)(n×p) = m×p
(AB)ᵢⱼ = Σ Aᵢₖ Bₖⱼ
Columns of A = rows of B

PropertiesAssociative, distributive
AB ≠ BA in general

AnswersQ6: 3×4, a₂₃ = 9 · Q9: A² = I, B² = 0 · Q10: 16

Revision: matrices on one page

Next · Part 3Systems of Linear Equations

Part 3

Systems of Linear Equations

Matrices let us write a whole system as one equation, $Ax = b$ , and the number of solutions has exactly three possibilities.

3.1Linear equation and system of linear equations

A linear equation has the form

a_1 x_1 + a_2 x_2 + \cdots + a_n x_n + b = 0

where $x_1, \dots, x_n$ are the variables and $a_1, \dots, a_n, b$ are coefficients (usually real). A system is several linear equations in the same variables.

3.2Matrix representation

A system of $m$ equations in $n$ unknowns

\begin{aligned} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\ &\;\;\vdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n &= b_m \end{aligned}

A x = b, \quad A = \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{m1} & \cdots & a_{mn} \end{bmatrix}, \quad x = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}, \quad b = \begin{bmatrix} b_1 \\ \vdots \\ b_m \end{bmatrix}

So $A$ is $m \times n$ , $x$ has $n$ entries and $b$ has $m$ entries. Equation $i$ is row $i$ of $A$ , and its right-hand side goes to $b_i$ .

Figure 8. Coefficients row by row into A, constants into b

3.3Possible types of solutions

Case	Meaning	Two equations, two variables
Unique solution	exactly one assignment satisfies all equations	lines meet at one point
Infinitely many	more than one assignment works	the same line
No solution	no assignment satisfies all at once	distinct parallel lines

Figure 9. The three possible outcomes

Verification

A proposed solution is valid only if it satisfies all the equations simultaneously.

Not in the source, but useful. For a square system

Ax = b

(as many equations as unknowns),

\det A \neq 0

guarantees a unique solution. If

\det A = 0

, the system has either no solution or infinitely many; the determinant alone does not say which, so you check the equations.

3.4Going deeper: how to read a system and its solutions

What to do with a square system Ax = b

Row picture. Each equation in two unknowns is a line. The solutions are the points that lie on all the lines: one point, a whole line, or nothing.
Column picture. $Ax = b$ asks: which combination $x_1 \cdot (\text{column 1}) + x_2 \cdot (\text{column 2}) + \cdots$ gives $b$ ? Solving a system means finding that combination.
How to spot the type. If one equation is a multiple of another, you get infinitely many solutions (same line). If the left sides are multiples but the right sides are not, there is no solution (parallel lines), like $x + y = 2$ and $x + y = 3$ .
Fewer equations than unknowns. If $m < n$ , the system can never have a unique solution: it has either none or infinitely many.
Elimination keeps the solutions. Adding a multiple of one equation to another gives a system with the same solutions.
Always substitute back. A solution is valid only if it satisfies every equation.

3.5Exam tips and common mistakes

Exam tips

Write the orders: $A$ is $m \times n$ , $x$ has $n$ entries, $b$ has $m$ entries.
Use the same variable order in every row of $A$ , and put 0 for a missing variable.
For a square system use $\det A$ first, and check the equations only when $\det A = 0$ .

Common mistakes

Writing the transpose of $A$ (coefficients column by column instead of row by row).
Putting constants into $A$ instead of $b$ .
Saying " $\det A = 0$ means no solution": it means no solution or infinitely many.
Checking a candidate solution in only one equation.

Practice questions (graded-assignment style)

Q1Same line

Multiple Choice Question (MCQ): choose the one correct option.

How many solutions does $x + y = 2,\; 2x + 2y = 4$ have?

None

Exactly one

Infinitely many

Exactly two

Solution

Idea

Compare the two equations.

The second equation is twice the first, so both describe the same line.
Every point of that line is a solution.
Answer: option 3.

Q2Parallel lines

Multiple Choice Question (MCQ): choose the one correct option.

How many solutions does $x + y = 2,\; x + y = 3$ have?

None

Exactly one

Infinitely many

Exactly two

Solution

Idea

The same left side cannot equal two different numbers.

A point with $x + y = 2$ cannot also have $x + y = 3$ .
The lines are parallel and never meet.
Answer: option 1.

Q3Find the bad value

Numerical Answer Type (NAT).

$x - 2y = 1$ and $3x + ky = 5$ . For which $k$ does the system not have a unique solution?

Solution

Idea

Set the determinant of the coefficient matrix to 0.

$\det\begin{bmatrix} 1 & -2 \\ 3 & k \end{bmatrix} = k + 6$ .
No unique solution when $k + 6 = 0$ , that is $k = -6$ .
Answer: -6

Q4Two equations, three unknowns

Multiple Select Question (MSQ): select all that apply.

A system has 2 equations in 3 unknowns. Choose the correct statements.

A

2 \times 3

x

has 3 entries.

b

has 3 entries.

It can have a unique solution.

Solution

Idea

Equations are rows, unknowns are columns.

Options 1 and 2 are true.
Option 3 is false: b has 2 entries, one per equation.
Option 4 is false: fewer equations than unknowns gives none or infinitely many solutions.
Answer: options 1 and 2.

Q5Solve it

Numerical Answer Type (NAT).

Solve $x + y = 5,\; x - y = 1$ . Find $x$ .

Solution

Idea

Add the two equations.

Adding gives $2x = 6$ , so $x = 3$ .
Then $y = 2$ ; check: $3 + 2 = 5$ and $3 - 2 = 1$ .
Answer: 3

Revision overview

Linear equationa₁x₁ + … + aₙxₙ + b = 0

SystemSeveral equations, same variables

Matrix formAx = b
A is m×n, x has n, b has m
Row i of A = equation i

Linear SystemsWeek 1 · Part 3

Solution typesUnique · infinitely many · none
Point · same line · parallel lines

Square systemdet ≠ 0 → unique
det = 0 → none or infinite

AnswersQ12: option 4 · Q13: unique · Q19: k ≠ −9

Revision: linear systems on one page

Next · Part 4Determinants — Part 1

Part 4

Determinants — Part 1

The determinant turns a square matrix into one number. This part computes it for small matrices and for triangular ones.

4.1First-, second- and third-order determinant

Every square matrix $A$ has a number called its determinant, written $\det(A)$ or $|A|$ . It is used to solve systems, to find inverses, in calculus and more.

Order 1 and 2

\det[a] = a, \qquad \det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc

Example: $\det\begin{bmatrix} 2 & 3 \\ 6 & 10 \end{bmatrix} = 2(10) - 3(6) = 2$ .

Figure 11. Geometric meaning of a 2 × 2 determinant (extra intuition)

Order 3: expand along the first row

\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})

Source example: $A = \begin{bmatrix} 2 & 4 & 1 \\ 3 & 8 & 7 \\ 5 & 6 & 9 \end{bmatrix}$ : $\det(A) = 2(72 - 42) - 4(27 - 35) + (18 - 40) = 60 + 32 - 22 = 70$ .

Figure 12. The three terms of the first-row expansion

Upper triangular matrices

If all entries below the diagonal are 0, the determinant is the product of the diagonal entries.

Figure 13. det = 2 · 8 · 9 = 144

4.2Going deeper: what a determinant tells you, and how to compute it fast

One number from a square matrix. $\det A \neq 0$ means $A$ is invertible and $Ax = b$ has a unique solution. $\det A = 0$ means $A$ is singular. Geometrically, the $2 \times 2$ determinant is the signed area of the parallelogram built on the rows.
Sarrus rule (extra, only for $3 \times 3$ ). Add the three products along the diagonals and subtract the three along the anti-diagonals. For $\begin{bmatrix} 2 & 4 & 1 \\ 3 & 8 & 7 \\ 5 & 6 & 9 \end{bmatrix}$ : $(144 + 140 + 18) - (40 + 84 + 108) = 302 - 232 = 70$ , matching the row expansion.
Triangular and diagonal matrices. The determinant is the product of the diagonal entries. So $\det I = 1$ and $\det\,\text{diag}(a, b, c) = abc$ .
Scaling. $\det(cA) = c^n \det A$ for an $n \times n$ matrix: one factor $c$ for each row.
What is not true. $\det(A + B) \neq \det A + \det B$ in general, and $\det(cA) \neq c\det A$ . Example: $A = B = I$ ( $2 \times 2$ ) has $\det(A + B) = \det(2I) = 4$ , not $1 + 1 = 2$ .
What is true. $\det(AB) = \det A \cdot \det B$ and $\det A^T = \det A$ .

4.3Exam tips and common mistakes

Exam tips

For a $2 \times 2$ use $ad - bc$ (main diagonal minus the other diagonal).
For a $3 \times 3$ expand along the first row with signs $+, -, +$ , or use Sarrus as a check.
Look at the structure first: triangular, a zero row, or two equal rows save a lot of work.

Common mistakes

$\det(2A) = 2\det A$ for a $3 \times 3$ matrix. The correct factor is $2^3 = 8$ .
Using $\det(A + B) = \det A + \det B$ .
Forgetting the minus sign on the middle term in a $3 \times 3$ expansion.
Computing $bc - ad$ instead of $ad - bc$ .

Practice questions (graded-assignment style)

Q1A 2 by 2 determinant

Numerical Answer Type (NAT).

Find $\det\begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix}$ .

Solution

Idea

ad − bc.

$3 \cdot 5 - 1 \cdot 2 = 15 - 2 = 13$ .
Answer: 13

Q2Triangular

Numerical Answer Type (NAT).

Find $\det\begin{bmatrix} 2 & 7 & 1 \\ 0 & -3 & 4 \\ 0 & 0 & 5 \end{bmatrix}$ .

Solution

Idea

Upper triangular: multiply the diagonal.

$2 \cdot (-3) \cdot 5 = -30$ .
Answer: -30

Q3Scaling

Numerical Answer Type (NAT).

$A$ is $4 \times 4$ with $\det A = 3$ . Find $\det(2A)$ .

Solution

Idea

One factor 2 per row.

$\det(2A) = 2^4 \cdot 3 = 16 \cdot 3 = 48$ .
Answer: 48

Q4Determinant rules

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements for square matrices of the same order.

\det(A + B) = \det A + \det B

\det(AB) = \det A \cdot \det B

\det(A^T) = \det A

For a

3 \times 3

identity

I

\det(3I) = 27

Solution

Idea

Products and transposes behave well; sums do not.

Option 1 is false in general.
Options 2 and 3 are standard properties.
Option 4: $\det(3I) = 3^3 \det I = 27$ .
Answer: options 2, 3 and 4.

Q5Sarrus check

Numerical Answer Type (NAT).

Find $\det\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}$ .

Solution

Idea

Expand along the first row (or use Sarrus).

$1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1$ .
Sarrus: $(0 + 40 + 0) - (15 + 24 + 0) = 1$ .
Answer: 1

Revision overview

Orders 1 and 2det[a] = a
det [[a,b],[c,d]] = ad − bc

Order 3Expand along row 1:
+a₁₁M₁₁ − a₁₂M₁₂ + a₁₃M₁₃
Example: 70

Determinants IWeek 1 · Part 4

Triangulardet = product of diagonal
2 · 8 · 9 = 144

Scalingdet(cA) = cⁿ det(A)
3×3: det(2A) = 8 det(A)

AnswersQ20: 8 · Q21: options 3, 5 · Q22: B and C

Revision: determinants, part one, on one page

Next · Part 5Determinants — Part 2

Part 5

Determinants — Part 2

Properties and row operations let us avoid heavy arithmetic, and cofactor expansion handles any size.

5.1Important determinant properties

\det(I) = 1, \qquad \det(A^T) = \det(A), \qquad \det(AB) = \det(A)\det(B)

Inverse. If $A^{-1}$ exists then $AA^{-1} = I$ . Taking determinants, $\det(A)\det(A^{-1}) = 1$ , so

\det(A^{-1}) = \frac{1}{\det(A)}

(so $A^{-1}$ can exist only when $\det A \neq 0$ ).

In the graded assignment, determinant questions are almost never "expand this big matrix". They hand you a few facts ( $\det A = 2$ , " $A$ is skew-symmetric", " $A^2 = 9I$ ") and ask for the determinant of something else. Each such question is solved by one or two of the properties below. Learn every property together with why it is true, because the exam changes the numbers but not the reason.

A. Products, powers, inverses and scaling

Property	Why it holds	Where it shows up
$\det(AB) = \det A \cdot \det B$ (same order)	The basic multiplicative rule of determinants.	Any product of matrices
$\det(A^k) = (\det A)^k$	Apply the product rule $k$ times. It also works for negative $k$ when $A$ is invertible.	$\det(A^2B^3)$ , $\det(A^6)$
$\det(AB) = \det(BA)$	Both equal $\det A \cdot \det B$ , which are just two numbers, so their order does not matter. ( $AB \neq BA$ as matrices, but their determinants agree.)	MSQ options like " $\det(BA) = 27$ "
$\det(A^{-1}) = \dfrac{1}{\det A}$	$AA^{-1} = I$ , so $\det A \cdot \det A^{-1} = \det I = 1$ .	$\det(A^2B^{-1})$ ; also shows $\det A \neq 0$ is needed for $A^{-1}$
$\det(A^T) = \det A$	Rows and columns play the same role.	Every row rule is also a column rule
$\det(cA) = c^n \det A$ for $n \times n$	$cA$ multiplies each of the $n$ rows by $c$ , and each row gives one factor $c$ .	$\det(2A)$ , $\det(8M)$ , $\det(-A)$
$\det(-A) = (-1)^n \det A$	The case $c = -1$ of the rule above.	Sign trap: for odd $n$ , $\det(-A) = -\det A$
$\det(P^{-1}AP) = \det A$	$\det P^{-1} \cdot \det A \cdot \det P = \dfrac{1}{\det P}\det A \det P = \det A$ .	"Similar" matrices have the same determinant
$\det I = 1$ ; triangular or diagonal: product of the diagonal	All terms of the expansion except the diagonal one contain a zero.	Row-reduce first, then multiply the diagonal

Recipe for "find det of an expression"

Peel the expression from the outside in: (1) a scalar in front gives $c^n$ ; (2) a product splits into separate determinants; (3) a power becomes the same power of the number; (4) an inverse becomes $1/\det$ ; (5) a transpose disappears. Example for $3 \times 3$ matrices with $\det A = 3$ , $\det B = 6$ : $\det(2A^2B^{-1}A^T) = 2^3 \cdot 3^2 \cdot \tfrac{1}{6} \cdot 3 = 8 \cdot 4.5 = 36$ .

B. The determinant is linear in one row at a time

If you fix every row except one, the determinant behaves like a linear function of that one row. Two consequences:

Factor out. A common factor $t$ of one row can be pulled out: the determinant is multiplied by $t$ (this is the "scale a row" rule from 5.2).
Split. If one row is a sum $u + v$ , the determinant is the sum of two determinants, one with $u$ in that row and one with $v$ , all other rows unchanged.

Example: split row 1 of the matrix below as $(3, 3, 5) = (2, 3, 1) + (1, 0, 4)$ :

\det\begin{bmatrix} 3 & 3 & 5 \\ 0 & 1 & 2 \\ 4 & 1 & 0 \end{bmatrix} = \det\begin{bmatrix} 2 & 3 & 1 \\ 0 & 1 & 2 \\ 4 & 1 & 0 \end{bmatrix} + \det\begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 2 \\ 4 & 1 & 0 \end{bmatrix} = 16 + (-18) = -2

Do not stretch this. The split works for one row while the others stay fixed. Doing it to a whole matrix is

\det(A + B)

, and that has no simple rule: for

A = B = I

(

2 \times 2

\det(A + B) = \det(2I) = 4

but

\det A + \det B = 2

. The cross terms are the problem.

C. When is $\det = 0$ ? A checklist

A row or column of zeros.
Two equal rows (or two equal columns). Subtract one from the other to get a zero row.
One row a multiple of another (proportional rows), for example $(2, 4, 6)$ and $(1, 2, 3)$ .
One row a combination of the other rows (Q23). Rows in an arithmetic progression are the classic disguise: $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ has $R_1 - 2R_2 + R_3 = 0$ , so $\det = 0$ .
Any fact that forces $A$ to be non-invertible (see D and E below).

D. Determinant of a special matrix: take $\det$ of both sides

When a question gives an equation that $A$ satisfies, take the determinant of both sides and use the rules above. The table lists the standard cases.

Given	Take det of both sides	Conclusion
$A^2 = A$	$(\det A)^2 = \det A$	$\det A \in \{0, 1\}$
$A^2 = I$	$(\det A)^2 = 1$	$\det A = \pm 1$
$A^2 = 9I$ , $n \times n$	$(\det A)^2 = 9^n$	$\det A = \pm 3^n$ ; an extra fact such as $\det A > 0$ picks the sign (GA 13)
$A^3 = 8I$ , $A$ real, $3 \times 3$	$(\det A)^3 = 512$	$\det A = 8$ only: an odd power has just one real root, so there is no $\pm$
$A^k = 0$ (nilpotent)	$(\det A)^k = 0$	$\det A = 0$
$A^TA = I$ (orthogonal)	$(\det A)^2 = 1$	$\det A = \pm 1$
$A^T = -A$ (skew-symmetric)	$\det A = \det(-A) = (-1)^n \det A$	$n$ odd: $\det A = -\det A$ , so $\det A = 0$ . $n$ even: nothing is forced.
$AB = 0$	$\det A \cdot \det B = 0$	At least one of $A, B$ is singular. Neither has to be the zero matrix.

Skew-symmetric, even order. The rule "skew-symmetric implies

\det = 0

" is only for odd

n

. The

4 \times 4

matrix with blocks

\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}

on the diagonal and zeros elsewhere is skew-symmetric and has

\det = 1

. In

\det A = (-1)^n\det A

, when

n

is even the equation just says

\det A = \det A

E. Determinant, invertibility and solutions: one idea

For a square matrix $A$ these statements are all the same statement:

$\det A \neq 0$ $\iff$ $A^{-1}$ exists $\iff$ $Ax = b$ has exactly one solution for every $b$ $\iff$ the rows are independent.
$\det A = 0$ $\iff$ $A$ is singular $\iff$ $Ax = 0$ has a non-zero solution $\iff$ $Ax = b$ has no solution or infinitely many, depending on $b$ .
$AB$ is invertible exactly when both $A$ and $B$ are, because $\det A \cdot \det B \neq 0$ needs both factors non-zero.

Parameter questions. "For which $k$ is $A$ singular?" or "for which $k$ is the solution not unique?" means: compute $\det A$ as an expression in $k$ and set it to $0$ (Q11). If the matrix also has to be a certain kind of matrix, check the structure at the value you find.

F. Cofactors: the "alien cofactor" fact

Expanding along row $i$ with the cofactors of the same row gives $\det A$ . Pair the entries of row $i$ with the cofactors of a different row $k$ and the sum is $0$ :

\sum_{j} a_{ij}C_{kj} = \begin{cases} \det A & i = k \\ 0 & i \neq k \end{cases}

Why. The sum $\sum_j a_{ij}C_{kj}$ is the determinant of the matrix obtained from $A$ by replacing row $k$ with a copy of row $i$ . That matrix has two equal rows, so its determinant is $0$ . The same holds for columns.

Check with $A = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & -1 \\ 5 & 2 & 1 \end{bmatrix}$ . The cofactors of row 1 are $C_{11} = 6$ , $C_{12} = -5$ , $C_{13} = -20$ . Then

2(6) + 1(-5) + 3(-20) = -53 = \det A, \qquad 0(6) + 4(-5) + (-1)(-20) = 0, \qquad 5(6) + 2(-5) + 1(-20) = 0

G. A shortcut when all rows have the same sum

If every row of a square matrix adds up to the same number $s$ , add all the columns into column 1. Column 1 becomes $(s, s, \ldots, s)$ , so $s$ factors out, and what is left reduces quickly. For $\begin{bmatrix} 5 & 2 & 2 \\ 2 & 5 & 2 \\ 2 & 2 & 5 \end{bmatrix}$ every row sums to $9$ :

\det = 9\det\begin{bmatrix} 1 & 2 & 2 \\ 1 & 5 & 2 \\ 1 & 2 & 5 \end{bmatrix} \xrightarrow{R_2 - R_1,\; R_3 - R_1} 9\det\begin{bmatrix} 1 & 2 & 2 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} = 9 \cdot 1 \cdot 3 \cdot 3 = 81

H. Statements that look true but are false

Tempting claim	Counterexample or correct version
$\det(A + B) = \det A + \det B$	$A = B = I_2$ : $4 \neq 2$
$\det(cA) = c\det A$	It is $c^n \det A$ . For $3 \times 3$ , $\det(2A) = 8\det A$ .
$\det(-A) = -\det A$	True only for odd $n$ . For $2 \times 2$ , $\det(-A) = \det A$ .
$\det A = 0$ means $A = 0$	$\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ has $\det = 0$ and is not the zero matrix.
$AB = 0$ means $A = 0$ or $B = 0$	$\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ . Both are singular, though.
Same determinant means same matrix	$\det I = 1$ and $\det\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = 1$ .
A skew-symmetric matrix always has $\det = 0$	Only for odd order. A $4 \times 4$ one can have $\det = 1$ .
Swapping two rows keeps $\det$	It changes the sign.

Which property for which question?

The question says	Reach for
"find $\det$ of an expression in $A$ , $B$ , $A^{-1}$ , $A^T$ , $cA$ "	A, the peeling recipe
" $B$ is obtained from $A$ by row operations"	5.2 table: swap $\to -1$ , add a multiple $\to \times 1$ , scale a row $\to \times t$
"one row is $u + v$ " or "two matrices differ in one row"	B, split or factor one row
" $A$ satisfies $A^2 = \ldots$ or $A^T = \ldots$ "	D, take $\det$ of both sides
"for which $k$ is it singular / not unique"	E, set $\det = 0$
"sum of $a_{ij}$ times cofactors"	F, own row gives $\det A$ , another row gives $0$
A messy $3 \times 3$ or $4 \times 4$	C (look for dependence), G (equal row sums), or row-reduce

5.2Effect of elementary row and column operations

Operation	Effect on the determinant
Switch two rows (or columns)	changes the sign
Add a multiple of one row/column to another	unchanged
Multiply a row by $t$	multiplied by $t$
A row or column of all zeros	determinant is 0

Figure 15. Four rules that make determinants quick to track

5.3Minors and cofactors

Minor and cofactor

The $(i, j)$ -th minor $M_{ij}$ is the determinant of the submatrix left after deleting row $i$ and column $j$ . The cofactor adds a sign:

C_{ij} = (-1)^{i+j} M_{ij}

For a $3 \times 3$ matrix: $M_{11} = a_{22}a_{33} - a_{23}a_{32}$ and $M_{23} = a_{11}a_{32} - a_{12}a_{31}$ . The signs form a checkerboard starting with $+$ .

Figure 16. Deleting a row and column gives a minor; the cofactor adds the checkerboard sign

5.4Expansion using minors and cofactors

\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} = (-1)^{1+1}a_{11}M_{11} + (-1)^{1+2}a_{12}M_{12} + (-1)^{1+3}a_{13}M_{13}

In general, for an $n \times n$ matrix, expand along any row $i$ or any column $j$ :

\det(A) = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} M_{ij} \;\;(\text{fixed row } i), \qquad \det(A) = \sum_{i=1}^{n} (-1)^{i+j} a_{ij} M_{ij} \;\;(\text{fixed column } j)

Exam strategy

Choose the row or column with the most zeros: every zero entry kills a whole term.

5.5Going deeper: row operations, zero determinants and smart expansion

Only "add a multiple of a row" is used, so the determinant never changes

Reduce to triangular. Use only "add a multiple of one row to another" to make the entries below the diagonal 0. The determinant never changes, and the final answer is the product of the diagonal. The figure gives $1 \cdot 1 \cdot 7 = 7$ .
Keep a ledger. If you also swap two rows, multiply the answer by $-1$ each time. If you multiply a row by $t$ , the determinant is multiplied by $t$ (so dividing a row by $t$ to simplify means multiplying the final answer by $t$ ).
When is $\det = 0$ ? A zero row or column, two equal rows, two proportional rows, or any row that is a combination of the others (as in Q23).
Invertibility. $A^{-1}$ exists exactly when $\det A \neq 0$ , and then $\det A^{-1} = \frac{1}{\det A}$ .
Cofactor expansion with zeros. Expand along the row or column with the most zeros. Each zero entry removes a whole term, so the work shrinks.

Choose the column with the most zeros

5.6Exam tips and common mistakes

Exam tips

Reduce with row operations when the numbers are awkward, and expand when there are many zeros.
Write the sign $(-1)^{i+j}$ next to every cofactor: a checkerboard starting with $+$ .
Do a quick structure check (zero row, equal rows, dependent rows) before any arithmetic.

Common mistakes

Forgetting the sign change after a row swap.
Treating "multiply a row by $t$ " as if it left the determinant unchanged.
Using $C_{ij} = M_{ij}$ without the sign $(-1)^{i+j}$ .
Expanding along a row full of non-zero numbers when a row with zeros is available.

Practice questions (graded-assignment style)

Q1Track the changes

Numerical Answer Type (NAT).

$A$ is $3 \times 3$ with $\det A = 4$ . $B$ is made by swapping two rows of $A$ and then multiplying one row by 3. Find $\det B$ .

Solution

Idea

A swap gives −1; scaling one row by 3 gives 3.

Swap: $-4$ .
Multiply one row by 3: $-4 \times 3 = -12$ .
Answer: -12

Q2Row reduction

Numerical Answer Type (NAT).

Find $\det\begin{bmatrix} 2 & 1 & 1 \\ 4 & 3 & 3 \\ 2 & 2 & 5 \end{bmatrix}$ using row operations.

Solution

Idea

Subtract multiples of row 1 to clear the first column.

$R_2 - 2R_1 = (0, 1, 1)$ and $R_3 - R_1 = (0, 1, 4)$ .
$R_3 - R_2 = (0, 0, 3)$ , giving a triangular matrix with diagonal $2, 1, 3$ .
Only "add a multiple" steps were used, so $\det = 2 \cdot 1 \cdot 3 = 6$ .
Answer: 6

Q3A zero determinant

Multiple Choice Question (MCQ): choose the one correct option.

Which matrix certainly has determinant 0?

One with two equal rows

One with all diagonal entries 1

One with all entries positive

A symmetric matrix

Solution

Idea

Subtract one equal row from the other.

Subtracting a row from its copy gives a zero row without changing the determinant, so $\det = 0$ .
The other kinds can have any determinant.
Answer: option 1.

Q4An inverse

Numerical Answer Type (NAT).

$A$ is $2 \times 2$ and $\det A = 5$ . Find $\det(A^{-1})$ .

Solution

Idea

$\det A \cdot \det A^{-1} = 1$ .

$\det(A^{-1}) = \frac{1}{5} = 0.2$ .
Answer: 0.2

Q5Minors and cofactors

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements.

A minor of a

3 \times 3

matrix is a

2 \times 2

determinant.

C_{23} = M_{23}

C_{12} = -M_{12}

Expanding along different rows gives different values of the determinant.

Solution

Idea

$C_{ij} = (-1)^{i+j} M_{ij}$ .

Option 1 is true.
Option 2 is false: $(-1)^{2+3} = -1$ , so $C_{23} = -M_{23}$ .
Option 3 is true: $(-1)^{1+2} = -1$ .
Option 4 is false: every row or column gives the same value.
Answer: options 1 and 3.

5.7Practice: determinant properties

Practice questions: determinant properties (graded-assignment style)

Q6A negative scalar

Numerical Answer Type (NAT).

$A$ is a real $3 \times 3$ matrix with $\det A = -2$ . Find $\det(-3A)$ .

Solution

Idea

$\det(cA) = c^n \det A$ , and $c$ is allowed to be negative.

$n = 3$ and $c = -3$ , so $\det(-3A) = (-3)^3 \det A = -27 \cdot (-2) = 54$ .
Common slips: using $c$ instead of $c^3$ (gives $6$ ), or dropping the sign of $c$ so that $3^3$ is used (gives $-54$ ).
Answer: 54

Q7Peeling an expression

Numerical Answer Type (NAT).

$A$ and $B$ are $3 \times 3$ matrices with $\det A = 2$ and $\det B = -4$ . Find $\det(A^3B^{-1}A^T)$ .

Solution

Idea

Split the product into three determinants and convert each piece.

$\det(A^3) = (\det A)^3 = 8$ .
$\det(B^{-1}) = \dfrac{1}{\det B} = -\dfrac{1}{4}$ .
$\det(A^T) = \det A = 2$ .
Multiply: $8 \cdot \left(-\dfrac14\right) \cdot 2 = -4$ . Answer: -4

Q8True or false: six statements about determinants

Multiple Select Question (MSQ): select all that apply.

Let $A$ and $B$ be real square matrices of the same order. Choose the correct statements.

\det(AB) = \det(BA)

A

3 \times 3

and

A^T = -A

, then

\det A = 0

A

4 \times 4

and

A^T = -A

, then

\det A = 0

A^2 = A

, then

\det A

is either

0

1

AB = 0

and

A

is not the zero matrix, then

\det B = 0

\det A = 0

, then

A

is the zero matrix.

Solution

Idea

For each option take the determinant of the given equation and see what it forces.

Option 1 ✓ $\det(AB) = \det A\det B = \det B\det A = \det(BA)$ , even though $AB \neq BA$ in general.
Option 2 ✓ $\det A = \det A^T = \det(-A) = (-1)^3\det A = -\det A$ , so $2\det A = 0$ and $\det A = 0$ .
Option 3 ✗ For $n = 4$ , $(-1)^4 = 1$ and the argument gives nothing. Counterexample: blocks $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ twice on the diagonal; it is skew-symmetric with $\det = 1$ .
Option 4 ✓ $(\det A)^2 = \det A$ , so $\det A(\det A - 1) = 0$ .
Option 5 ✓ Suppose $\det B \neq 0$ . Then $B^{-1}$ exists and $A = (AB)B^{-1} = 0$ , contradicting $A \neq 0$ . So $\det B = 0$ .
Option 6 ✗ $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ has $\det = 0$ and is not zero.
Answer: options 1, 2, 4 and 5.

Q9A similarity transformation

Numerical Answer Type (NAT).

$A$ and $P$ are $3 \times 3$ matrices, $P$ is invertible and $\det A = -3$ . Find $\det(2P^{-1}A^2P)$ .

Solution

Idea

$P^{-1}$ and $P$ cancel in the determinant, because $\det P^{-1} \cdot \det P = 1$ .

$\det(P^{-1}A^2P) = \dfrac{1}{\det P}\cdot(\det A)^2\cdot\det P = (\det A)^2 = 9$ .
The scalar $2$ sits in front of a $3 \times 3$ matrix: $\det(2M) = 2^3\det M = 8 \cdot 9 = 72$ .
Answer: 72

Q10Two matrices that differ in one row

Numerical Answer Type (NAT).

Let $A$ have rows $r_1, r_2, r_3$ with $\det A = 5$ , and let $B$ have rows $s_1, r_2, r_3$ (only row 1 is different) with $\det B = -2$ . Let $C$ be the $3 \times 3$ matrix with rows $2r_1 - 3s_1,\; r_2,\; r_3$ . Find $\det C$ .

Solution

Idea

Rows 2 and 3 are the same everywhere, so the determinant is linear in row 1: constants come out and sums split.

Split row 1: $\det C = \det[2r_1;\, r_2;\, r_3] + \det[-3s_1;\, r_2;\, r_3]$ .
Pull out the constants: $= 2\det[r_1;\, r_2;\, r_3] - 3\det[s_1;\, r_2;\, r_3] = 2\det A - 3\det B$ .
$= 2(5) - 3(-2) = 10 + 6 = 16$ . Answer: 16
Do not try $2\det A - 3\det B$ for matrices that differ in more than one row: the rule needs the other rows to be identical.

Q11For which $k$ is the matrix singular?

Numerical Answer Type (NAT).

For $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & k & 6 \\ 1 & 1 & 1 \end{bmatrix}$ , the system $Ax = b$ does not have a unique solution for every $b$ exactly when $k$ takes one particular value. Find it.

Solution

Idea

No unique solution means $\det A = 0$ . Compute $\det A$ in terms of $k$ and solve $\det A = 0$ .

Expand along row 1:
$\det A = 1(k - 6) - 2(2 - 6) + 3(2 - k) = k - 6 + 8 + 6 - 3k = 8 - 2k$
$8 - 2k = 0 \Rightarrow k = 4$ .
Check: at $k = 4$ , row 2 is $(2, 4, 6) = 2 \cdot (1, 2, 3)$ , a multiple of row 1, so $\det A = 0$ ✓.
Answer: 4

Q12Entries of one row with cofactors of another

Multiple Choice Question (MCQ): choose the one correct option.

Let $A = \begin{bmatrix} 2 & 1 & 3 \\ 0 & 4 & -1 \\ 5 & 2 & 1 \end{bmatrix}$ with $\det A = -53$ , and let $C_{ij}$ be the cofactors of $A$ . What is the value of $a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13}$ ?

-53

0

53

6

Solution

Idea

The entries come from row 2 but the cofactors from row 1. That is an alien cofactor sum, so it is $0$ , not $\det A$ .

The sum is the determinant of $A$ with row 1 replaced by a copy of row 2:
$\det\begin{bmatrix} 0 & 4 & -1 \\ 0 & 4 & -1 \\ 5 & 2 & 1 \end{bmatrix} = 0$
since two rows are equal.
Direct check with $C_{11} = 6$ , $C_{12} = -5$ , $C_{13} = -20$ : $0(6) + 4(-5) + (-1)(-20) = -20 + 20 = 0$ ✓.
Option 1 would be the answer if the entries were from row 1 ( $2(6) + 1(-5) + 3(-20) = -53$ ).
Answer: option 2.

Q13Equal row sums

Numerical Answer Type (NAT).

Find $\det\begin{bmatrix} 5 & 2 & 2 \\ 2 & 5 & 2 \\ 2 & 2 & 5 \end{bmatrix}$ .

Solution

Idea

Every row adds up to $9$ . Add all columns into column 1, factor out $9$ , then clear the first column.

$C_1 \to C_1 + C_2 + C_3$ (adding multiples of columns leaves $\det$ unchanged) turns column 1 into $(9, 9, 9)^T$ . Factor $9$ out of that column.
$R_2 - R_1$ and $R_3 - R_1$ give the triangular matrix $\begin{bmatrix} 1 & 2 & 2 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ , with $\det = 1 \cdot 3 \cdot 3 = 9$ .
$\det = 9 \cdot 9 = 81$ . (In general $\begin{bmatrix} a & b & b \\ b & a & b \\ b & b & a \end{bmatrix}$ has $\det = (a + 2b)(a - b)^2$ ; here $(9)(3)^2 = 81$ ✓.)
Answer: 81

Q14A cube equal to $8I$

Numerical Answer Type (NAT).

$A$ is a real $3 \times 3$ matrix with $A^3 = 8I$ . Find $\det A$ .

Solution

Idea

Take determinants on both sides. The left side is $(\det A)^3$ . Because the power is odd, there is only one real cube root, so no sign choice is needed.

$\det(A^3) = (\det A)^3$ and $\det(8I) = 8^3 \cdot 1 = 512$ (the identity has 3 rows, each scaled by 8).
$(\det A)^3 = 512$ . $\det A$ is a real number, and the only real cube root of $512$ is $8$ .
Compare with GA 13, where $A^2 = 9I$ gave $\det A = \pm 81$ and a sign condition was needed. An even power leaves a $\pm$ ; an odd power does not.
Answer: 8

Q15Which matrix is singular?

Multiple Choice Question (MCQ): choose the one correct option.

Which of the following matrices is singular?

\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}

\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}

\begin{bmatrix} 2 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 4 \end{bmatrix}

Solution

Idea

Singular means $\det = 0$ . Look for dependence first, and compute only if you do not find any.

Option 1. The rows go up by the same step: $R_2 - R_1 = (3, 3, 3) = R_3 - R_2$ , so $R_1 - 2R_2 + R_3 = (1 - 8 + 7,\; 2 - 10 + 8,\; 3 - 12 + 9) = (0, 0, 0)$ . The rows are dependent, so $\det = 0$ . ✓ singular.
Option 2. $1 \cdot 4 - 2 \cdot 3 = -2 \neq 0$ . Not singular.
Option 3. Expand along row 2 (which has two zeros): $\det = 1\cdot(1 \cdot 2 - 1 \cdot 1) = 1 \neq 0$ . Not singular.
Option 4. $2(12 - 1) - 1(4 - 1) + 1(1 - 3) = 22 - 3 - 2 = 17 \neq 0$ . Not singular.
Answer: option 1.

Revision overview

Propertiesdet I = 1 · det Aᵀ = det A
det AB = det A · det B
det A⁻¹ = 1 / det A

Row operationsSwap: sign flips
Add a multiple: unchanged
Scale a row by t: ×t
Zero row: 0

Determinants IIWeek 1 · Part 5

Minor, cofactorMᵢⱼ: delete row i, column j
Cᵢⱼ = (−1)^(i+j) Mᵢⱼ

ExpansionAny row or column
Pick the one with most zeros

AnswersQ23: 0 · Q24: α = 1 · Q26: options 1, 3, 4

Revision: determinants, part two, on one page

Next · PracticeExtra Practice Problems

Practice

Extra Practice Problems

These are the important questions from the Week 1 graded assignment (Questions 3 to 13), written exactly as asked. Each has a step-by-step solution, and an MSQ solution checks every option so you can see why it is right or wrong. Try the question first, then open the solution.

GA 3Does such a matrix A exist?

Multiple Select Question (MSQ): select all that apply.

Let $B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ -1 & 0 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & 0 \\ -1 & 2 \end{bmatrix}$ . Which of the following options are true for a matrix $A$ , such that $AB = C$ ?

Such a matrix does not exist.

There is a unique matrix

A

satisfying this property.

There are infinitely many such matrices.

A

should be a

2 \times 3

matrix.

A

should be a

3 \times 2

matrix.

Solution

Idea

First fix the order of A from the orders of B and C, then compare entries. Count how many entries are forced and how many are free.

Check the order first: the inner numbers must match, the outer numbers give the answer's order

Match entries with C: some entries are forced, some stay free

Order of $A$ . $B$ is $3 \times 2$ and $AB = C$ is $2 \times 2$ . For $AB$ to exist, the columns of $A$ must equal the rows of $B$ , which is $3$ ; and $AB$ has as many rows as $A$ , which is $2$ . So $A$ is $2 \times 3$ . D true, E false.
Write
$A = \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}$
Row 1 of $AB$ is $(a\cdot 1 + b\cdot 0 + c\cdot(-1),\; a\cdot 0 + b\cdot 1 + c\cdot 0) = (a - c,\; b)$ . Row 2 is $(d - f,\; e)$ .
Match with $C$ : $a - c = 0$ , $b = 0$ , $d - f = -1$ , $e = 2$ .
So $b = 0$ and $e = 2$ are forced, but $a$ can be anything (then $c = a$ ) and $f$ can be anything (then $d = f - 1$ ). Two free choices, so infinitely many $A$ . Example:
$A = \begin{bmatrix} 0 & 0 & 0 \\ -1 & 2 & 0 \end{bmatrix} \quad\text{or}\quad \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 1 \end{bmatrix}$
A false (it exists), B false (not unique), C true.
Answer: C and D.

GA 4The characteristic polynomial of a 2 × 2

Multiple Select Question (MSQ): select all that apply.

Let $A$ be a $2 \times 2$ real matrix and let $\text{trace}(A)$ denote the sum of the elements in the diagonal of $A$ . Which of the following are true?

\det(A - cI)

is a polynomial in

c

of degree 1.

\det(A - cI)

is a polynomial in

c

of degree 2.

\det(A - cI) = c^2 - \text{trace}(A)c + \det(A)

\det(A - cI) = c^2 + \text{trace}(A)c - \det(A)

\det(A - cI) = \text{trace}(A)c - \det(A)

\det(A - cI) = -\text{trace}(A)c + \det(A)

Solution

Idea

Expand the 2 × 2 determinant of A − cI directly; the constant term is det(A) and the middle coefficient is minus the trace.

Let $A = \begin{bmatrix} p & q \\ r & s \end{bmatrix}$ . Then $A - cI = \begin{bmatrix} p - c & q \\ r & s - c \end{bmatrix}$ .
$\det(A - cI) = (p - c)(s - c) - qr = ps - pc - sc + c^2 - qr = c^2 - (p + s)c + (ps - qr)$ .
Here $p + s = \text{trace}(A)$ and $ps - qr = \det(A)$ , so $\det(A - cI) = c^2 - \text{trace}(A)\,c + \det(A)$ .
The highest power of $c$ is $c^2$ , so it has degree 2: B true, A false. Option C is exactly the formula: true. D has the wrong signs, and E and F have no $c^2$ term at all: false.
Quick check with $A = I$ : $\det(I - cI) = (1 - c)^2 = c^2 - 2c + 1$ , and trace $= 2$ , $\det = 1$ ✓.
Answer: B and C.

GA 5Oranges and bananas

Multiple Select Question (MSQ): select all that apply.

Suppose there are two types of oranges and two types of bananas available in the market. Suppose 1 kg of each type of orange costs ₹50 and 1 kg of each type of banana costs ₹40. Gargi bought $x$ kg of the first type of each fruit, orange and banana, and $y$ kg of the second type of each fruit, orange and banana. She paid ₹250 for oranges and ₹200 for bananas. Which of the following options are correct with respect to the given information?

The matrix representation to find

x

and

y

can be

\begin{bmatrix} 50 & 50 \\ 40 & 40 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 250 \\ 200 \end{bmatrix}

The matrix representation to find

x

and

y

can be

\begin{bmatrix} 50 & 40 \\ 50 & 40 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 250 \\ 200 \end{bmatrix}

The matrix representation to find

x

and

y

can be

\begin{bmatrix} 40 & 40 \\ 50 & 50 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 200 \\ 250 \end{bmatrix}

x

can be 2 and

y

can be 3.

There are infinitely many real values possible for

x

and

y

There are only finitely many real values possible for

x

and

y

There are only finitely many natural numbers possible for

x

and

y

Solution

Idea

Write one equation per fruit. Here both equations reduce to x + y = 5, so the system has a whole line of real solutions but only a few whole-number ones.

x + y = 5, drawn once

Build the equations. Oranges: both types cost ₹50/kg, so $50x + 50y = 250$ . Bananas: both types cost ₹40/kg, so $40x + 40y = 200$ .
A: row 1 is the orange equation, row 2 the banana equation: exactly the system above. True.
B: the first row would mean $50x + 40y = 250$ , as if the second type of orange cost ₹40/kg. But both types of orange cost ₹50/kg. False.
C: row 1 is $40x + 40y = 200$ (bananas) and row 2 is $50x + 50y = 250$ (oranges). Same two equations, just listed in the other order, and the order of equations does not change the system. True.
Solve. Dividing, $x + y = 5$ from the first and $x + y = 5$ from the second: they are the same equation, so the two lines coincide (see the figure). Check: $\det\begin{bmatrix} 50 & 50 \\ 40 & 40 \end{bmatrix} = 2000 - 2000 = 0$ .
D: $2 + 3 = 5$ ✓, and $50\cdot 5 = 250$ , $40 \cdot 5 = 200$ . True.
E: every $(x,\; 5 - x)$ works, so infinitely many real pairs. True. F: False, for the same reason.
G: whole numbers with $x + y = 5$ are $(1,4), (2,3), (3,2), (4,1)$ (and $(0,5), (5,0)$ if 0 is counted as natural): a finite list either way. True.
Answer: A, C, D, E and G.

GA 6Scaling a 5 × 5 matrix

Numerical Answer Type (NAT).

Suppose $\det(4A) = n \times \det(A)$ for any $5 \times 5$ real matrix $A$ . What is the value of $n$ ?

Solution

Idea

The exponent is the number of rows, not the scalar: det(cA) = c^m det(A).

The exponent counts rows

For an $m \times m$ matrix, $\det(cA) = c^m \det(A)$ , because multiplying the whole matrix by $c$ multiplies each of the $m$ rows by $c$ , and each row scaling multiplies the determinant by $c$ .
Here $m = 5$ and $c = 4$ : $n = 4^5 = 1024$ . Answer: 1024

GA 7Determinant of a product of powers

Numerical Answer Type (NAT).

$A, B$ are two square matrices of the same order. If $\det(A) = 2$ and $\det(B) = 3$ , what is $\det(A^2 B^3)$ ?

Solution

Idea

det(XY) = det(X) det(Y), so every power of a matrix becomes the same power of its determinant.

Powers of a matrix become powers of its determinant

The determinant of a product is the product of determinants: $\det(XY) = \det(X)\det(Y)$ . So $\det(A^n) = (\det A)^n$ .
$\det(A^2 B^3) = \det(A)^2 \cdot \det(B)^3 = 2^2 \cdot 3^3 = 4 \cdot 27 = 108$ . Answer: 108

GA 8Row operations and a scaled inverse

Numerical Answer Type (NAT).

Let $A$ be a square matrix of order 3 and $B$ be a matrix that is obtained by adding 8 times the first row of $A$ to the third row of $A$ and adding 9 times the second row of $A$ to the first row of $A$ . If $\det(A) = 5$ , then find out the value of $\det(8A^2B^{-1})$ .

Solution

Idea

Adding a multiple of one row to another keeps the determinant. Then use det(XY) = det X det Y, det(X⁻¹) = 1/det X and det(cX) = c^m det X.

Track the determinant at every step

Find $\det B$ . Both changes add a multiple of one row to another row: $R_3 \to R_3 + 8R_1$ and $R_1 \to R_1 + 9R_2$ . Adding a multiple of a row to another row does not change the determinant, so $\det B = \det A = 5$ .
Break up the expression. $A^2B^{-1}$ is a $3 \times 3$ matrix with $\det(A^2B^{-1}) = \det(A)^2 \cdot \frac{1}{\det B} = \frac{25}{5} = 5$ .
The factor $8$ scales all $3$ rows, so $\det(8M) = 8^3 \det M$ . Hence $\det(8A^2B^{-1}) = 512 \cdot 5 = 2560$ . Answer: 2560

GA 9Trace of A⁶ for the all-ones matrix

Numerical Answer Type (NAT).

If $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ , then what will be the value of the sum of the diagonal elements of $A^6$ ?

Solution

Idea

Do not multiply six times. Find A² in terms of A, spot the pattern A² = 3A, then every higher power is a number times A.

The pattern A² = 3A makes every power a multiple of A

Every entry of $A^2$ is (row of 1s) times (column of 1s) $= 1 + 1 + 1 = 3$ , so $A^2 = 3A$ .
Then $A^3 = A \cdot A^2 = 3A^2 = 9A$ , and in general $A^n = 3^{n-1} A$ . So $A^6 = 3^5 A = 243A$ . (Check: $A^6 = (A^3)^2 = 81A^2 = 81 \cdot 3A = 243A$ ✓.)
Every diagonal entry of $243A$ is $243$ , and there are 3 of them: $243 \times 3 = 729$ . Answer: 729

Comprehension for Questions 10 to 12

The scores of a student in mathematics, physics and chemistry in her class-12 board exams are $m, p$ and $c$ respectively, where each score is out of $100$ . She has applied for three engineering streams in a college. Each stream assigns different weights to these three subjects to calculate her final score, which is again out of $100$ .

For example, the weight given to mathematics, physics and chemistry could be $0.2, 0.7$ and $0.1$ respectively by a stream. These weights are then multiplied with the corresponding scores to get the final score. Concretely, if the student has scored $85, 78, 40$ in the three subjects, her final score for this stream is:

0.2 \times 85 + 0.7 \times 78 + 0.1 \times 40 = 75.6

This is called a weighted average. Note that the weights always sum to $1$ . Now, the weights assigned by the three streams for mathematics, physics and chemistry, in this order, are given below:

Stream-1: $0.2, 0.7, 0.1$
Stream-2: $0.5, 0.3, 0.2$
Stream-3: $0.1, 0.4, 0.5$

The final score of the student in stream-1 is $81$ . It is $83$ in stream-2 and $76$ in stream-3. We wish to find the student's marks in the three subjects.

GA 10Reading the coefficient matrix

Multiple Select Question (MSQ): select all that apply.

This is framed as a system of linear equations. Select all true options concerning the coefficient matrix if the vector of unknowns is given as $\begin{bmatrix} m \\ p \\ c \end{bmatrix}$ . Assume that the first equation corresponds to stream-1, second to stream-2 and last to stream-3.

The first row is

0.2, 0.7, 0.1

The last row is

0.1, 0.4, 0.5

The first column is

0.2, 0.7, 0.1

The middle column is

0.5, 0.3, 0.2

Solution

Idea

Rows are equations (one per stream) and columns are unknowns (one per subject). Do not read a row as a column.

Rows are streams, columns are subjects

Each stream gives one equation, with its weights as the coefficients of $m, p, c$ : $0.2m + 0.7p + 0.1c = 81$ , $0.5m + 0.3p + 0.2c = 83$ , $0.1m + 0.4p + 0.5c = 76$ .
In matrix form the system is
$\begin{bmatrix} 0.2 & 0.7 & 0.1 \\ 0.5 & 0.3 & 0.2 \\ 0.1 & 0.4 & 0.5 \end{bmatrix}\begin{bmatrix} m \\ p \\ c \end{bmatrix} = \begin{bmatrix} 81 \\ 83 \\ 76 \end{bmatrix}$
and the coefficient matrix is the left matrix.
A: first row $= (0.2, 0.7, 0.1)$ (stream-1). True. B: last row $= (0.1, 0.4, 0.5)$ (stream-3). True.
C: the first column is the maths weights of the three streams, $(0.2, 0.5, 0.1)$ , not $(0.2, 0.7, 0.1)$ . False. D: the middle column is the physics weights, $(0.7, 0.3, 0.4)$ , not $(0.5, 0.3, 0.2)$ (that is the second row). False.
Answer: A and B.

GA 11Find the average mark

Numerical Answer Type (NAT).

Find $\dfrac{m + p + c}{3}$ .

Solution

Idea

Solve the 3 × 3 system by eliminating one unknown at a time, then always substitute back into all three original equations.

Clear the decimals by multiplying each equation by $10$ :
(1) $2m + 7p + c = 810$
(2) $5m + 3p + 2c = 830$
(3) $m + 4p + 5c = 760$
From (1): $c = 810 - 2m - 7p$ . Put it in (2): $5m + 3p + 1620 - 4m - 14p = 830$ , so $m - 11p = -790$ , i.e. $m = 11p - 790$ .
Put $c$ in (3): $m + 4p + 4050 - 10m - 35p = 760$ , so $9m + 31p = 3290$ .
Substitute $m = 11p - 790$ : $99p - 7110 + 31p = 3290$ , so $130p = 10400$ and $p = 80$ . Then $m = 880 - 790 = 90$ and $c = 810 - 180 - 560 = 70$ .
Check all three: $0.2(90) + 0.7(80) + 0.1(70) = 18 + 56 + 7 = 81$ ✓; $0.5(90) + 0.3(80) + 0.2(70) = 45 + 24 + 14 = 83$ ✓; $0.1(90) + 0.4(80) + 0.5(70) = 9 + 32 + 35 = 76$ ✓.
$\dfrac{m + p + c}{3} = \dfrac{90 + 80 + 70}{3} = \dfrac{240}{3} = 80$ . Answer: 80

GA 12Determinant of the coefficient matrix

Numerical Answer Type (NAT).

Find $\det(A)$ , where $A$ is the coefficient matrix.

Solution

Idea

Expand a 3 × 3 determinant along a row. A quick sanity check: multiply by 10 to remove decimals and use det(10A) = 10³ det(A).

The coefficient matrix is
$A = \begin{bmatrix} 0.2 & 0.7 & 0.1 \\ 0.5 & 0.3 & 0.2 \\ 0.1 & 0.4 & 0.5 \end{bmatrix}$
Expand along row 1 with signs $+, -, +$ :
$\det A = 0.2(0.3 \cdot 0.5 - 0.2 \cdot 0.4) - 0.7(0.5 \cdot 0.5 - 0.2 \cdot 0.1) + 0.1(0.5 \cdot 0.4 - 0.3 \cdot 0.1)$
$= 0.2(0.15 - 0.08) - 0.7(0.25 - 0.02) + 0.1(0.20 - 0.03) = 0.2(0.07) - 0.7(0.23) + 0.1(0.17)$
$= 0.014 - 0.161 + 0.017 = -0.13$ . Answer: −0.13
Check with whole numbers:
$10A = \begin{bmatrix} 2 & 7 & 1 \\ 5 & 3 & 2 \\ 1 & 4 & 5 \end{bmatrix}$
has $\det = 2(15 - 8) - 7(25 - 2) + 1(20 - 3) = 14 - 161 + 17 = -130$ , and $\det(10A) = 10^3 \det A$ , so $\det A = -130/1000 = -0.13$ ✓.
Since $\det A \neq 0$ , the marks $m, p, c$ are uniquely determined, which is why Q11 has a single answer.

GA 13A matrix whose square is 9I

Numerical Answer Type (NAT).

Let A be a $(4 \times 4)$ matrix such that

(A + 3I)(A - 3I) = 0,

and suppose that $\det(A) > 0$ . Find $\det(A)$ .

Solution

Idea

Expand the product to get A² = 9I, take determinants, and let the sign condition pick between +81 and −81.

Take determinants on both sides, then use the sign condition

Multiply out. $A$ commutes with $I$ , so $(A + 3I)(A - 3I) = A^2 - 3A + 3A - 9I = A^2 - 9I$ . The condition says $A^2 - 9I = 0$ , i.e. $A^2 = 9I$ .
Take determinants: $\det(A^2) = \det(A)^2$ and $\det(9I) = 9^4 = 6561$ (the identity has 4 rows, each scaled by 9). So $\det(A)^2 = 6561$ , giving $\det A = \pm 81$ .
The question says $\det A > 0$ , so $\det A = 81$ . Answer: 81
Sanity check: $A = 3I$ satisfies $A^2 = 9I$ and has $\det = 3^4 = 81$ ✓. But $\text{diag}(3, 3, 3, -3)$ also satisfies $A^2 = 9I$ and has $\det = -81$ , which is why the condition $\det A > 0$ is needed.

Next · All partsRevision Sheet and Exam Checklist

All parts

Revision Sheet and Exam Checklist

The whole week on one sheet, followed by the exam checklist.

Vectors and matrices

Topic	Key point
Vector	$n$ entries, a vector in $\mathbb{R}^n$
Vector add / scale	$(a_i) + (b_i)$ , $c(a_1, \dots, a_n) = (ca_1, \dots, ca_n)$ coordinate-wise
Matrix order	$m \times n$ : $m$ rows, $n$ columns
Add matrices	$(A + B)_{ij} = A_{ij} + B_{ij}$ , same order
Matrix product	$A_{m \times n} B_{n \times p}$ is $m \times p$ ; $(AB)_{ij} = \sum_k A_{ik}B_{kj}$
Identity	1s on the diagonal, 0s elsewhere
Trap	$AB \neq BA$ in general

Systems and determinants

Topic	Key point
System	$Ax = b$ ; unique, infinitely many, or none
$1 \times 1$ , $2 \times 2$	$\det[a] = a$ ; $ad - bc$
$3 \times 3$	expand using signed $2 \times 2$ minors
Transpose, product	$\det A^T = \det A$ ; $\det AB = \det A \det B$
Inverse	$\det A^{-1} = 1/\det A$
Row swap / add / scale	sign flips / unchanged / times $t$
Minor, cofactor	$M_{ij}$ ; $C_{ij} = (-1)^{i+j}M_{ij}$

Exam checklist

Represent vectors as rows or columns; add and scale them coordinate-wise; interpret them in $\mathbb{R}^2$ .
Identify the order and the square, diagonal, scalar and identity matrices.
Check dimensions before multiplying; entries by row × column; remember $AB \neq BA$ .
Write a system as $Ax = b$ and tell unique, infinite and no-solution apart.
Compute $1 \times 1$ , $2 \times 2$ , $3 \times 3$ determinants; use the properties and track row operations.
Find minors and cofactors, and pick an efficient row or column.

Exam tips: last-minute checklist

Write orders under matrices: $(m \times n)(n \times p)$
Equations are rows of $A$ ; unknowns are columns
Test a square system with $\det A$ first
Use $\det(cA) = c^n \det A$ ( $n$ = number of rows)
Track row swaps (sign) and row scaling (factor)
Expand along the row or column with most zeros
Substitute your answer back into every equation

Do not

Assume $AB = BA$
Multiply matrices entry by entry
Use $\det(A + B) = \det A + \det B$
Write $\det(2A) = 2\det A$ for a $3 \times 3$
Forget the checkerboard signs in cofactors
Say det = 0 means no solution
Read a row of a table as a column

Which formula for which question?

Question	Use
Order of $AB$ or "does $AB$ exist?"	inner numbers match, outer numbers give the order
Missing scalar in $cA + dB = \ldots$	compare coordinates, start with a known one
Number of solutions	$\det A \neq 0$ : unique; otherwise check the equations
$\det$ of a product or power	$\det(AB) = \det A \det B$ , $\det A^n = (\det A)^n$
$\det(cA)$	$c^n \det A$
Fast $\det$ of a big matrix	row-reduce to triangular, or expand along a row of zeros
$A$ satisfies $A^2 = A$ , $A^2 = 9I$ , $A^T = -A$ , $A^3 = 8I$	take $\det$ of both sides (5.1 D); even power gives $\pm$ , odd power does not
For which $k$ is it singular / not unique	$\det A = 0$ , solve for $k$

Mathematics forData Science 2

Chapters

Vectors

1.1Vectors

1.2Vector addition and scalar multiplication

1.3Geometric representation in R2\mathbb{R}^2R2

1.4Vectors in a physical context

1.5Going deeper: vectors as lists, arrows and combinations

1.6Exam tips and common mistakes

Practice questions (graded-assignment style)

Matrices

2.1Order of a matrix

2.2Special matrices

2.3Matrix addition, scalar multiplication and multiplication

Important properties

2.4Going deeper: when AB exists, and three ways to see it

2.5Transpose, trace and powers: the rules you will be tested on

A. The transpose

B. Symmetric and skew-symmetric matrices

C. The trace

D. Powers, and the identities that fail without commutativity

2.6Exam tips and common mistakes

Practice questions (graded-assignment style)

Systems of Linear Equations

3.1Linear equation and system of linear equations

3.2Matrix representation

3.3Possible types of solutions

3.4Going deeper: how to read a system and its solutions

3.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Determinants — Part 1

4.1First-, second- and third-order determinant

Upper triangular matrices

4.2Going deeper: what a determinant tells you, and how to compute it fast

4.3Exam tips and common mistakes

Practice questions (graded-assignment style)

Determinants — Part 2

5.1Important determinant properties

A. Products, powers, inverses and scaling

B. The determinant is linear in one row at a time

C. When is det⁡=0\det = 0det=0? A checklist

D. Determinant of a special matrix: take det⁡\detdet of both sides

E. Determinant, invertibility and solutions: one idea

F. Cofactors: the "alien cofactor" fact

G. A shortcut when all rows have the same sum

H. Statements that look true but are false

Which property for which question?

5.2Effect of elementary row and column operations

5.3Minors and cofactors

5.4Expansion using minors and cofactors

5.5Going deeper: row operations, zero determinants and smart expansion

5.6Exam tips and common mistakes

Practice questions (graded-assignment style)

5.7Practice: determinant properties

Practice questions: determinant properties (graded-assignment style)

Extra Practice Problems

Revision Sheet and Exam Checklist

Vectors and matrices

Systems and determinants

Exam checklist

Exam tips: last-minute checklist

Which formula for which question?

Mathematics for
Data Science 2

1.3Geometric representation in $\mathbb{R}^2$

C. When is $\det = 0$ ? A checklist

D. Determinant of a special matrix: take $\det$ of both sides