Week 2 Notes

Mathematics for
Data Science 2

Solving systems of linear equations: determinants and Cramer's Rule, the inverse-matrix method, echelon forms, row reduction and Gauss elimination.

7 chapters40 worked examples~47 min read

Cramer's RuleInverse matrixEchelon formsGauss elimination

Chapters

01Linear Equations and a Determinants RecapPart 1
02Cramer's RulePart 2
03Solving with an Invertible Coefficient MatrixPart 3
04The Echelon FormPart 4
05Row ReductionPart 5
06The Gauss Elimination MethodPart 6
07Week 2 Summary SheetAll parts

Linear systems

Linear Equations and a Determinants Recap

What a linear system is, how it is written as Ax = b, and what the determinant of the coefficient matrix says about its solutions.

1.1What is a linear equation?

Definition: linear equation

A linear equation in the variables $x_1, x_2, \dots, x_n$ is an equation of the form

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

where the coefficients $a_1, \dots, a_n$ and the constant $b$ are real numbers. Every variable appears to the power one: no squares, no product of two variables, nothing under a root.

Equation	Linear?	Why
$x + 2y = 1$	Yes	both variables to the power 1
$x + y + z = 1$	Yes	three variables, each to the power 1
$2x^2 + y = 1$	No	$x$ is squared
$3x^2 + 3y^2 + z^2 = 1$	No	all three variables are squared
$xy = 4$	No	a product of two variables (extra example)

1.2A system of equations as Ax = b

A system of linear equations is a collection of such equations that share the same variables. When there are $m$ equations in $n$ unknowns, the whole system packs into one matrix equation $A\mathbf{x} = \mathbf{b}$ :

\underbrace{\begin{bmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\a_{21}&a_{22}&\cdots&a_{2n}\\\vdots&\vdots&\ddots&\vdots\\a_{m1}&a_{m2}&\cdots&a_{mn}\end{bmatrix}}_{A\ (m\times n)}\underbrace{\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}}_{\mathbf{x}\ (n\times 1)} = \underbrace{\begin{bmatrix}b_1\\b_2\\\vdots\\b_m\end{bmatrix}}_{\mathbf{b}\ (m\times 1)}

$A$ is the coefficient matrix, $m \times n$ : one row per equation, one column per unknown.
$\mathbf{x}$ is the column of unknowns and $\mathbf{b}$ is the column of constants on the right-hand side.

Figure 1. The system 2x₁ + 3x₂ = 6, −2x₁ + 2x₂ = 3, 4x₁ + 6x₂ = 12 written as Ax = b

Example 1Writing a system as Ax = b

Write $2x_1 + 3x_2 = 6,\ -2x_1 + 2x_2 = 3,\ 4x_1 + 6x_2 = 12$ as $A\mathbf{x} = \mathbf{b}$ and state the order of $A$ .

Solution

Read the coefficients row by row: $A = \begin{bmatrix}2&3\\-2&2\\4&6\end{bmatrix}$ .
The unknowns are $\mathbf{x} = \begin{bmatrix}x_1\\x_2\end{bmatrix}$ and the constants are $\mathbf{b} = \begin{bmatrix}6\\3\\12\end{bmatrix}$ .
So $A\mathbf{x} = \mathbf{b}$ holds, and $A$ has 3 rows and 2 columns: its order is $3 \times 2$ .

1.3Determinant recap

Cramer's Rule and the inverse-matrix method both work only when $\det(A) \neq 0$ , so the determinant has to be back in hand.

2 × 2 matrix

\det\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} = a_{11}a_{22} - a_{12}a_{21}

3 × 3 matrix: expand along row 1

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

Invertible matrix

A square matrix $A$ is invertible (non-singular) exactly when $\det(A) \neq 0$ . That one number is the gatekeeper: every method that follows starts by checking it.

Example 2Two determinants

Compute $\det\begin{bmatrix}2&1\\3&4\end{bmatrix}$ and $\det\begin{bmatrix}1&-1&1\\2&1&-2\\-1&-2&4\end{bmatrix}$ .

Solution

$2 \times 2$ : $2 \cdot 4 - 1 \cdot 3 = 8 - 3 = 5$ .
$3 \times 3$ , along row 1: $1\,(1\cdot 4 - (-2)(-2)) - (-1)\,(2\cdot 4 - (-2)(-1)) + 1\,(2(-2) - 1\cdot(-1))$ .
$= 1(4 - 4) + 1(8 - 2) + 1(-4 + 1) = 0 + 6 - 3 = 3$ .

1.4Seeing it geometrically

For two equations in two unknowns, each equation draws a line in the $xy$ -plane, and solving the system means finding where the lines meet. The determinant of the coefficient matrix decides the story.

Figure 2. Crossing lines have one solution; the same line has infinitely many; parallel lines have none

$\det(A) \neq 0$ : the lines cross at exactly one point, a unique solution.
$\det(A) = 0$ , same line: $x + y = 3$ and $2x + 2y = 6$ are multiples of each other, so every point of the line is a solution: infinitely many.
$\det(A) = 0$ , parallel lines: they never meet, so there is no solution.

Important

$\det(A) = 0$ alone cannot tell you whether you are in the "infinitely many" case or the "no solution" case. Telling those two apart is exactly what row reduction (Parts 4 to 6) is for.

1.5Going deeper

Square needed for a determinant. A determinant exists only for a square matrix, so $\det(A) \neq 0$ is a test for systems with as many equations as unknowns ( $m = n$ ). Rectangular systems need Gauss elimination (Part 6).
Three variables. Each equation is a plane in 3D. The same three outcomes appear: planes meeting at a point, along a line or sheet, or not meeting at all (extra).
Consistent / inconsistent (extra). A system with at least one solution is called consistent; with none, inconsistent.

1.6Exam tips and common mistakes

Exam tips

In "is it linear?" questions look only at powers and products of variables, not at the numbers.
The order of $A$ is (number of equations) $\times$ (number of unknowns).

Common mistakes

Calling $2x^2 + y = 1$ linear because the coefficients are numbers.
Writing the order of $A$ as $n \times m$ instead of $m \times n$ .
Concluding "no solution" just because $\det(A) = 0$ .

Practice questions (graded-assignment style)

Q1Which are linear?

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

Which of the following is/are a linear equation?

2x^2 + y = 1

x + 2y = 1

x + y + z = 1

3x^2 + 3y^2 + z^2 = 1

Solution

Idea

Variables only to the first power.

Options 2 and 3 have every variable to the power 1.
Options 1 and 4 contain squares of variables, so they are not linear.
Answer: options 2 and 3.

Q2Order of A

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

For $2x_1 + 3x_2 = 6,\ -2x_1 + 2x_2 = 3,\ 4x_1 + 6x_2 = 12$ with $A\mathbf{x} = \mathbf{b}$ , what is the order of $A$ ?

2 \times 3

3 \times 2

3 \times 3

2 \times 2

Solution

Idea

Rows = equations, columns = unknowns.

There are 3 equations, so 3 rows. There are 2 unknowns, so 2 columns.
The order is $3 \times 2$ .
Answer: option 2.

Q3Which system has which kind of solution?

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

Consider $2x_1 + 3x_2 = 6,\ -2x_1 + kx_2 = d,\ 4x_1 + 6x_2 = 12$ . Choose the correct statements.

No solution if

k = -3,\ d = 0

A unique solution if

k = 3,\ d = 0

Infinitely many solutions if

k = -3,\ d = 6

Infinitely many solutions if

k = -3,\ d = -6

Solution

Idea

The third equation is 2 times the first, so it adds nothing new. Look at the first two.

For $k = -3$ , the second equation is $-1$ times the first ( $-2x_1 - 3x_2 = d$ vs $2x_1 + 3x_2 = 6$ ). It agrees only when $d = -6$ .
$d = 0$ gives $-2x_1 - 3x_2 = 0$ against $2x_1 + 3x_2 = 6$ : contradiction, no solution (option 1 true).
$d = 6$ also contradicts (option 3 false); $d = -6$ makes the two equations identical, so infinitely many (option 4 true).
For $k = 3$ the first two equations are independent ( $\det\begin{bmatrix}2&3\\-2&3\end{bmatrix} = 12 \neq 0$ ), so one solution (option 2 true).
Answer: options 1, 2 and 4.

Q4A determinant

Numerical Answer Type (NAT).

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

Solution

Idea

Expand along column 1: only one non-zero entry.

Column 1 is $(1, 0, 0)$ , so $\det = 1 \cdot \det\begin{bmatrix}1&3\\0&2\end{bmatrix} = 1 \cdot (2 - 0) = 2$ .
(For a triangular matrix the determinant is the product of the diagonal: $1 \cdot 1 \cdot 2 = 2$ .)
Answer: 2

Q5Same line

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

The system $x + y = 1,\ 2x + 2y = 2$ has:

No solution

Exactly one solution

Infinitely many solutions

Exactly two solutions

Solution

Idea

Is the second equation a multiple of the first?

$2x + 2y = 2$ is $2 \times (x + y = 1)$ : the same line.
Every point of the line $x + y = 1$ is a solution, so there are infinitely many; $\det\begin{bmatrix}1&1\\2&2\end{bmatrix} = 0$ .
Answer: option 3.

Revision overview

Linear equationa₁x₁ + … + aₙxₙ = b
Variables to power 1 only

Systemm equations, n unknowns
Ax = b with A of order m × n

Determinant2 × 2: ad − bc
3 × 3: expand along a row

Solving linear equationsWeek 2 · Part 1

det(A) ≠ 0A invertible
Unique solution

det(A) = 0Infinitely many or none
Cannot tell which

Geometry (2 unknowns)Lines cross, coincide
or are parallel

Revision: what a linear system is and what the determinant tells you

Next · Square systemsCramer's Rule

Square systems

Cramer's Rule

A direct formula for the unknowns as ratios of determinants, for square systems with a non-zero determinant.

2.1When can Cramer's Rule be used?

Cramer's Rule gives a direct formula for solving $A\mathbf{x} = \mathbf{b}$ , with no elimination. It applies only when both conditions hold:

The coefficient matrix $A$ must be square.
$\det(A)$ must be non-zero, i.e. $A$ is invertible.

If either condition fails, Cramer's Rule does not apply and you use Gauss elimination (Part 6) instead.

2.2Order 2

For $a_{11}x_1 + a_{12}x_2 = b_1,\ a_{21}x_1 + a_{22}x_2 = b_2$ with $A = \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}$ , replace one column of $A$ at a time by $\mathbf{b}$ :

Cramer's Rule, order 2

A_{x_1} = \begin{bmatrix}b_1&a_{12}\\b_2&a_{22}\end{bmatrix},\quad A_{x_2} = \begin{bmatrix}a_{11}&b_1\\a_{21}&b_2\end{bmatrix}

x_1 = \frac{\det(A_{x_1})}{\det(A)},\qquad x_2 = \frac{\det(A_{x_2})}{\det(A)}

The idea

$A_{x_i}$ is $A$ with its $i$ -th column replaced by $\mathbf{b}$ . The unknown $x_i$ is the ratio of the new determinant to the old one.

Example 3Cramer's Rule, order 2

Solve $2x_1 + x_2 = 1,\ 3x_1 + 4x_2 = -1$ .

Solution

$A = \begin{bmatrix}2&1\\3&4\end{bmatrix}$ , $\det(A) = 2(4) - 1(3) = 5 \neq 0$ , so the rule applies.
$A_{x_1} = \begin{bmatrix}1&1\\-1&4\end{bmatrix}$ , $\det(A_{x_1}) = 4 - (-1) = 5$ .
$A_{x_2} = \begin{bmatrix}2&1\\3&-1\end{bmatrix}$ , $\det(A_{x_2}) = -2 - 3 = -5$ .
$x_1 = \dfrac{5}{5} = 1$ , $x_2 = \dfrac{-5}{5} = -1$ .
Check: $2(1) + (-1) = 1$ and $3(1) + 4(-1) = -1$ .

Figure 3. Example 3: the replaced column is highlighted

2.3Order 3

The pattern extends directly. For $a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = b_1$ and so on, build $A_{x_1}, A_{x_2}, A_{x_3}$ by putting $\mathbf{b}$ in place of the 1st, 2nd and 3rd columns of $A$ . Then

x_1 = \frac{\det(A_{x_1})}{\det(A)},\quad x_2 = \frac{\det(A_{x_2})}{\det(A)},\quad x_3 = \frac{\det(A_{x_3})}{\det(A)}

Example 4Cramer's Rule, order 3

Solve $x_1 - x_2 + x_3 = 1,\ 2x_1 + x_2 - 2x_3 = -1,\ -x_1 - 2x_2 + 4x_3 = 1$ .

Solution

$A = \begin{bmatrix}1&-1&1\\2&1&-2\\-1&-2&4\end{bmatrix}$ , $\mathbf{b} = \begin{bmatrix}1\\-1\\1\end{bmatrix}$ , and $\det(A) = 3 \neq 0$ (Example 2).
Replace columns by $\mathbf{b}$ : $A_{x_1} = \begin{bmatrix}1&-1&1\\-1&1&-2\\1&-2&4\end{bmatrix}$ , $A_{x_2} = \begin{bmatrix}1&1&1\\2&-1&-2\\-1&1&4\end{bmatrix}$ , $A_{x_3} = \begin{bmatrix}1&-1&1\\2&1&-1\\-1&-2&1\end{bmatrix}$ .
Their determinants are $\det(A_{x_1}) = -1$ , $\det(A_{x_2}) = -7$ , $\det(A_{x_3}) = -3$ .
$x_1 = -\dfrac{1}{3}$ , $x_2 = -\dfrac{7}{3}$ , $x_3 = \dfrac{-3}{3} = -1$ .
Check in equation 1: $-\tfrac{1}{3} + \tfrac{7}{3} - 1 = 2 - 1 = 1$ .

2.4Order n: the general pattern

Cramer's Rule, order n

For an $n \times n$ system $A\mathbf{x} = \mathbf{b}$ with $\det(A) \neq 0$ , let $A_{x_i}$ be the matrix obtained from $A$ by replacing its $i$ -th column with $\mathbf{b}$ . Then

x_i = \frac{\det(A_{x_i})}{\det(A)},\qquad i = 1, 2, \dots, n

It needs $n + 1$ determinants of order $n$ : one for $A$ and one for each $A_{x_i}$ .
A bonus fact (extra). You can find just one unknown without solving for the others: compute only $\det(A)$ and that one $\det(A_{x_i})$ .
Why it is slow for big $n$ (extra). Computing $n + 1$ determinants is expensive, which is why elimination is preferred for large systems.

2.5Exam tips and common mistakes

Exam tips

Check $\det(A) \neq 0$ first. If it is 0, stop: Cramer's Rule does not apply.
Replace the column, not the row, and keep the other columns in their original order.
When a question gives some unknowns, plug them into the rows first; it often fixes the parameter without any determinant.

Common mistakes

Putting $\mathbf{b}$ in the wrong column.
Dividing by $\det(A_{x_i})$ instead of $\det(A)$ .
Using Cramer's Rule on a non-square system.

Practice questions (graded-assignment style)

Q1Which matrices are A_{x}?

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

For $x_1 + x_3 = 1,\ -x_1 + x_2 - x_3 = 1,\ -x_2 + x_3 = 1$ with $A = \begin{bmatrix}1&0&1\\-1&1&-1\\0&-1&1\end{bmatrix}$ , $\mathbf{b} = \begin{bmatrix}1\\1\\1\end{bmatrix}$ , and $A_{x_i}$ the matrix got by replacing the $i$ -th column of $A$ with $\mathbf{b}$ , choose the correct options.

A_{x_1} = \begin{bmatrix}1&1&0\\-1&1&-1\\0&-1&1\end{bmatrix}

A_{x_1} = \begin{bmatrix}1&0&1\\1&1&-1\\1&-1&1\end{bmatrix}

A_{x_2} = \begin{bmatrix}1&0&1\\-1&1&-1\\0&-1&1\end{bmatrix}

A_{x_3} = \begin{bmatrix}1&0&1\\-1&1&1\\0&-1&1\end{bmatrix}

Solution

Idea

Replace the whole column, keep the rest unchanged.

$A_{x_1}$ : column 1 becomes $(1, 1, 1)^T$ , so the rows start with 1, 1, 1 and the other columns stay: option 2 is true; option 1 is not $A$ with a column replaced by $\mathbf{b}$ , so it is false.
$A_{x_2}$ : column 2 must become $(1, 1, 1)^T$ ; option 3 is just $A$ again, so it is false.
$A_{x_3}$ : column 3 becomes $(1, 1, 1)^T$ : option 4 is true.
Answer: options 2 and 4.

Q2Solving with the values

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

For the same system, $\det(A) = 1$ and $\mathbf{x} = A^{-1}\mathbf{b} = (-2, 2, 3)^T$ . Which are correct?

x_1 = -2

x_2 = -2

x_3 = 3

None of the above

Solution

Idea

Read the entries of the solution vector.

$x_1 = -2$ and $x_3 = 3$ are entries 1 and 3; $x_2 = 2$ , not $-2$ .
Check in equation 1: $x_1 + x_3 = -2 + 3 = 1$ . Equation 2: $2 + 2 - 3 = 1$ . Equation 3: $-2 + 3 = 1$ .
Answer: options 1 and 3.

Q3Find a²

Numerical Answer Type (NAT).

$A\mathbf{x} = \mathbf{b}$ with $A = \begin{bmatrix}1&a&0\\a&1&2\\1&0&a\end{bmatrix}$ , $\mathbf{x} = \begin{bmatrix}x_1\\0\\x_3\end{bmatrix}$ , $\mathbf{b} = \begin{bmatrix}1\\0\\0\end{bmatrix}$ . Find $a^2$ , given $a \neq 0,\ \sqrt{3},\ -\sqrt{3}$ .

Solution

Idea

Use the known second entry $x_2 = 0$ in each row.

Row 1: $x_1 + a \cdot 0 + 0 \cdot x_3 = 1$ , so $x_1 = 1$ .
Row 2: $a x_1 + 0 + 2x_3 = 0$ , so $a + 2x_3 = 0$ and $x_3 = -a/2$ .
Row 3: $x_1 + a x_3 = 0$ , so $1 + a(-a/2) = 0$ , i.e. $1 - a^2/2 = 0$ and $a^2 = 2$ .
Answer: 2

Q4Find 3a

Numerical Answer Type (NAT).

$A\mathbf{x} = \mathbf{b}$ with $A = \begin{bmatrix}1&2a+3\\3a+2&1\end{bmatrix}$ , $\mathbf{x} = \begin{bmatrix}x_1\\0\end{bmatrix}$ , $\mathbf{b} = \begin{bmatrix}1\\1\end{bmatrix}$ . Find $3a$ , given $\det(A) \neq 0$ .

Solution

Idea

Put $x_2 = 0$ into both rows.

Row 1: $x_1 = 1$ .
Row 2: $(3a + 2)(1) + 1 \cdot 0 = 1$ , so $3a + 2 = 1$ and $3a = -1$ .
Answer: -1

Q5Cramer by hand (extra)

Numerical Answer Type (NAT).

Solve $2x + y = 7,\ x - y = 2$ with Cramer's Rule and give $x$ .

Solution

Idea

$\det(A)$ , then $\det(A_{x_1})$ .

$\det(A) = 2(-1) - 1(1) = -3$ .
$A_{x_1} = \begin{bmatrix}7&1\\2&-1\end{bmatrix}$ , $\det = -7 - 2 = -9$ .
$x = \dfrac{-9}{-3} = 3$ . (And $y = \dfrac{\det\begin{bmatrix}2&7\\1&2\end{bmatrix}}{-3} = \dfrac{-3}{-3} = 1$ .)
Answer: 3

Revision overview

ConditionsA is square
det(A) ≠ 0

Build A_{xᵢ}Replace the i-th column
of A by b

OrdersOrder 2: 3 determinants
Order 3: 4 determinants
Order n: n + 1

Cramer's RuleWeek 2 · Part 2

Formulaxᵢ = det(A_{xᵢ}) / det(A)
i = 1, 2, …, n

If it failsdet(A) = 0 or not square
Use Gauss elimination

CheckPut x back into every
original equation

Revision: solving a square system with determinants

Next · Square systemsSolving with an Invertible Coefficient Matrix

Square systems

Solving with an Invertible Coefficient Matrix

The same systems solved by finding the inverse of A, and a picture of what a unique solution looks like in 3D.

3.1The idea

Cramer's Rule needs $n + 1$ determinants. This method reaches straight for the inverse of $A$ . Start from $A\mathbf{x} = \mathbf{b}$ and multiply both sides on the left by $A^{-1}$ :

A^{-1}A\mathbf{x} = A^{-1}\mathbf{b} \;\Longrightarrow\; \mathbf{x} = A^{-1}\mathbf{b}

This works because $A^{-1}A = I$ . The applicability conditions are exactly the same as for Cramer's Rule: $A$ must be square and $\det(A) \neq 0$ .

Figure 4. The inverse-matrix method in four steps

Find the inverse $A^{-1}$ (adjugate and determinant: $A^{-1} = \dfrac{\operatorname{adj}(A)}{\det(A)}$ ).
Compute the product $A^{-1}\mathbf{b}$ ; it is a column matrix.
Read off $x_1, x_2, \dots, x_n$ by comparing with $\mathbf{x}$ .

Order matters. It is

A^{-1}\mathbf{b}

, with

A^{-1}

on the left. Matrix multiplication is not commutative, so

\mathbf{b}A^{-1}

is a different (and usually undefined) product.

3.2Seeing it in 3D

With three variables each equation is a plane, and an invertible coefficient matrix means all three planes meet at exactly one point. For example, $x + y + z = 1$ , $x = 1$ , $y = -2$ meet only at $(1, -2, 2)$ .

Figure 5. Three planes with an invertible coefficient matrix meet in a single point

3.3Cramer's Rule or the inverse?

	Cramer's Rule	Inverse-matrix method
Conditions	$A$ square, $\det(A) \neq 0$	$A$ square, $\det(A) \neq 0$
What you compute	$n + 1$ determinants	$A^{-1}$ , then $A^{-1}\mathbf{b}$
Strength	one unknown at a time (extra)	all unknowns at once; reuse $A^{-1}$ for many $\mathbf{b}$ (extra)

3.4Going deeper

Why the answer is unique. If $A$ is invertible, $\mathbf{x} = A^{-1}\mathbf{b}$ is forced: any solution has to equal it. So an invertible $A$ gives exactly one solution for every $\mathbf{b}$ .
The 2 × 2 shortcut (extra). For $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ : swap the diagonal entries, negate the off-diagonal ones, divide by $ad - bc$ . That is how $\operatorname{adj}$ of Example 5 came out.
Adjugate = inverse when $\det = 1$ . In Q1 below $\det(A) = 1$ , so $A^{-1}$ and $\operatorname{adj}(A)$ are the same matrix.

3.5Exam tips and common mistakes

Exam tips

Always check the answer: $A\mathbf{x}$ must give $\mathbf{b}$ back.
If $\det(A) = 1$ , the adjugate is the inverse, so you can skip the division.

Common mistakes

Forgetting to divide the adjugate by $\det(A)$ .
Multiplying $\mathbf{b}A^{-1}$ instead of $A^{-1}\mathbf{b}$ .
Sign slips while building the adjugate (signs alternate in a checkerboard).

Practice questions (graded-assignment style)

Q1Inverse of A (Part 1)

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

Consider $2x_1 - x_2 = 3,\ x_1 - x_3 = 3,\ x_2 - x_3 = 2$ , with $A = \begin{bmatrix}2&-1&0\\1&0&-1\\0&1&-1\end{bmatrix}$ , $\mathbf{x} = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ , $\mathbf{b} = \begin{bmatrix}3\\3\\2\end{bmatrix}$ . Choose the correct options.

A^{-1} = \begin{bmatrix}2&-1&0\\1&-1&0\\1&0&-1\end{bmatrix}

A^{-1} = \begin{bmatrix}1&-1&1\\1&-2&2\\1&-2&1\end{bmatrix}

The adjoint of

A

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

\det(A) = 1

Solution

Idea

Compute $\det(A)$ first, then check each option with $AA^{-1} = I$ .

$\det(A) = 2(0 + 1) + 1(-1 - 0) + 0 = 2 - 1 = 1$ , so option 4 is true.
With $\det(A) = 1$ , $A^{-1} = \operatorname{adj}(A) = \begin{bmatrix}1&-1&1\\1&-2&2\\1&-2&1\end{bmatrix}$ : options 2 and 3 are true.
Option 1 fails: row 1 of $A$ times column 1 of option 1 is $2 \cdot 2 + (-1) \cdot 1 = 3 \neq 1$ .
Answer: options 2, 3 and 4.

Q2Solution (Part 2)

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

For the same system, which are correct?

x_1 = -2

x_2 = 1

x_3 = -1

None of the above

Solution

Idea

Use $\mathbf{x} = A^{-1}\mathbf{b}$ .

$A^{-1}\mathbf{b} = \begin{bmatrix}1&-1&1\\1&-2&2\\1&-2&1\end{bmatrix}\begin{bmatrix}3\\3\\2\end{bmatrix} = \begin{bmatrix}2\\1\\-1\end{bmatrix}$ .
So $x_1 = 2$ (not $-2$ ), $x_2 = 1$ , $x_3 = -1$ .
Check: $2(2) - 1 = 3$ , $2 - (-1) = 3$ , $1 - (-1) = 2$ .
Answer: options 2 and 3.

Q3A 2 × 2 solve

Numerical Answer Type (NAT).

Use the inverse method on $2x_1 + x_2 = 1,\ 5x_1 + 3x_2 = 2$ and find $x_1 + 2x_2$ .

Solution

Idea

$\det(A) = 1$ , so the adjugate is the inverse.

$A = \begin{bmatrix}2&1\\5&3\end{bmatrix}$ , $\det(A) = 6 - 5 = 1$ , $A^{-1} = \begin{bmatrix}3&-1\\-5&2\end{bmatrix}$ .
$\mathbf{x} = \begin{bmatrix}3&-1\\-5&2\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}$ .
So $x_1 + 2x_2 = 1 + 2(-1) = -1$ .
Answer: -1

Q4The right formula

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

If $A$ is square and invertible, the solution of $A\mathbf{x} = \mathbf{b}$ is:

\mathbf{x} = \mathbf{b}A^{-1}

\mathbf{x} = A^{-1}\mathbf{b}

\mathbf{x} = A\mathbf{b}^{-1}

\mathbf{x} = \mathbf{b} - A

Solution

Idea

Multiply both sides of $A\mathbf{x} = \mathbf{b}$ by $A^{-1}$ on the left.

$A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}$ and $A^{-1}A = I$ , so $\mathbf{x} = A^{-1}\mathbf{b}$ .
$\mathbf{b}^{-1}$ does not exist (a column is not a square matrix), and $\mathbf{b}A^{-1}$ has the wrong shape.
Answer: option 2.

Revision overview

StartAx = b
A square, det(A) ≠ 0

MultiplyA⁻¹ on the left
A⁻¹A = I

InverseA⁻¹ = adj(A) / det(A)

Inverse methodWeek 2 · Part 3

Resultx = A⁻¹b
A column matrix

GeometryThree planes meet
at one point

CheckAx must give b back

Revision: solving a square system with the inverse matrix

Next · Any systemThe Echelon Form

Any system

The Echelon Form

The shapes that a matrix is reduced to, so that systems of any size, square or not, can be solved.

4.1Why we need it

Cramer's Rule and the inverse method share a blind spot: they work only when $A$ is square and invertible. Real systems are often rectangular, or square but singular. The echelon form is the tool that handles those too. It is really two related shapes.

4.2Row echelon form (REF)

Row echelon form

A matrix is in row echelon form if:

The first non-zero entry (the leading entry) of each row is 1.
The leading 1 of each row sits strictly to the right of the leading 1 in the row above.
Any all-zero rows sit at the bottom, below every non-zero row.

4.3Reduced row echelon form (RREF)

Reduced row echelon form

A matrix is in reduced row echelon form if it satisfies everything above, plus one more condition:

The leading 1 in each row is the only non-zero entry in its entire column.

RREF is REF with extra clean-up

Every matrix in RREF is automatically in REF. RREF just adds the column clean-up condition on top.

Figure 6. A REF (left) and the RREF it reduces to (right); the dashed staircase marks the zeros

4.4Going deeper

Leading 1 = pivot (extra). The leading 1 of a row is also called a pivot, and its column a pivot column. This word is used a lot in Part 6.
RREF is unique (extra). A matrix can have many REFs, but its RREF is one single matrix, whichever row operations you use to reach it.
Echelon form and determinants. A square matrix with a row of zeros has $\det = 0$ . A square matrix whose RREF is the identity has $\det \neq 0$ (it is invertible), as in Q2 below.

4.5Exam tips and common mistakes

Exam tips

Test the three REF conditions in order: leading 1s, staircase, zero rows. Only then test the RREF column condition.
Look only at the entry above and below each leading 1 for RREF; other entries are free.

Common mistakes

Requiring the leading entry to be 1 but forgetting the staircase direction (it goes right as you go down).
Calling a matrix RREF when a non-leading entry above a leading 1 is non-zero.
Putting a zero row above a non-zero row.

Practice questions (graded-assignment style)

Q1Reading a matrix

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

Let $A = \begin{bmatrix}0&1&3\\1&0&2\\0&0&0\end{bmatrix}$ (row 1 is the topmost row). Identify the correct statements.

The first non-zero element in the first row is 3.

The first non-zero element in the second row is 1.

There is a non-zero element in the third row.

Since there is a row with all elements zero,

\det(A) = 0

Solution

Idea

Read each row from the left.

Row 1 is $(0, 1, 3)$ : its first non-zero entry is 1, not 3 (option 1 false).
Row 2 is $(1, 0, 2)$ : it begins with 1 (option 2 true). Row 3 is entirely zero (option 3 false).
A square matrix with a zero row has $\det = 0$ (option 4 true).
Answer: options 2 and 4.

Q2Row operations and RREF

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

Let $A = \begin{bmatrix}1&-1&1\\0&1&1\\-1&0&1\end{bmatrix}$ , a $3 \times 3$ matrix. Which statements are true?

A

can be transformed by row operations into

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

, which is in REF.

The RREF of

A

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

The RREF of

A

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

A

can be transformed into

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

, which is in REF.

Solution

Idea

Compute $\det(A)$ : non-zero means the RREF is the identity.

$R_3 + R_1$ gives $\begin{bmatrix}1&-1&1\\0&1&1\\0&-1&2\end{bmatrix}$ , then $R_3 + R_2$ gives $\begin{bmatrix}1&-1&1\\0&1&1\\0&0&3\end{bmatrix}$ , then $\tfrac{1}{3}R_3$ gives the matrix of option 1: true.
$\det(A) = 3 \neq 0$ , so the RREF is the identity matrix $I$ : option 3 is true and option 2 is false.
The identity is also in REF, and $A$ can reach it, so option 4 is true.
Answer: options 1, 3 and 4.

Q3A missing column

Numerical Answer Type (NAT).

The RREF of a matrix $A$ is $\begin{bmatrix}1&0&-1\\0&1&1\end{bmatrix}$ . The first and third columns of $A$ itself are $\begin{bmatrix}-1\\2\end{bmatrix}$ and $\begin{bmatrix}-1\\1\end{bmatrix}$ . If the second column of $A$ is $\begin{bmatrix}m_1\\m_2\end{bmatrix}$ , find $m_1 + m_2$ .

Solution

Idea

Row operations act on every column in the same way: $E A = \text{RREF}$ .

From columns 1 and 3: $E\begin{bmatrix}-1&-1\\2&1\end{bmatrix} = \begin{bmatrix}1&-1\\0&1\end{bmatrix}$ , giving $E = \begin{bmatrix}3&2\\-2&-1\end{bmatrix}$ .
The second column of the RREF is $\begin{bmatrix}0\\1\end{bmatrix}$ , so $E\begin{bmatrix}m_1\\m_2\end{bmatrix} = \begin{bmatrix}0\\1\end{bmatrix}$ .
Check the logic: $E$ turns $(-1, 2)^T$ into $(1, 0)^T$ and $(-1, 1)^T$ into $(-1, 1)^T$ .
$E^{-1} = \begin{bmatrix}-1&-2\\2&3\end{bmatrix}$ , so $\begin{bmatrix}m_1\\m_2\end{bmatrix} = \begin{bmatrix}-2\\3\end{bmatrix}$ and $m_1 + m_2 = 1$ .
Answer: 1

Q4Which is RREF?

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

Which of the following matrices is in reduced row echelon form?

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

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

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

\begin{bmatrix}1&0&3\\0&0&0\\0&1&0\end{bmatrix}

Solution

Idea

Check staircase, zero rows and clean columns.

Option 1 is the identity: RREF.
Option 2: the 2 above the second leading 1 breaks the column condition (REF only).
Option 3: the leading 1 of row 2 is to the left of the leading 1 of row 1. Not REF.
Option 4: the zero row is above a non-zero row. Not REF.
Answer: option 1.

Revision overview

REFLeading entry is 1
Staircase to the right
Zero rows at the bottom

RREFREF plus
leading 1 is alone
in its column

RelationRREF is always REF
REF need not be RREF

Echelon formsWeek 2 · Part 4

WhyWorks for non-square
and singular matrices

PivotA leading 1
Its column is a pivot column

With detZero row: det = 0
RREF = I: det ≠ 0

Revision: recognising row echelon and reduced row echelon form

Next · Any systemRow Reduction

Any system

Row Reduction

The three elementary row operations, what they do to the determinant, and the algorithm that reaches REF and RREF.

5.1The three elementary row operations

Getting a matrix into echelon form uses three elementary row operations. Each has a predictable effect on the determinant, which is worth remembering.

Figure 7. The three row operations and what each does to the determinant

Type	Operation	Effect on $\det$
1 · interchange	swap two rows	$\det(B) = -\det(A)$
2 · scale	multiply a row by a constant $c$	$\det(B) = c \cdot \det(A)$
3 · add a multiple	add a multiple of one row to another	$\det(B) = \det(A)$ , unchanged

5.2The row-reduction algorithm

To carry any $m \times n$ matrix into REF and then RREF:

Step 1. Find the left-most non-zero column.
Step 2. Use a Type 1 or Type 2 operation to make its top entry 1.
Step 3. Use Type 3 operations to zero out every entry below that 1.
Step 4. Repeat on the submatrix below and to the right of the current row, until no non-zero rows are left. The matrix is now in REF.
Step 5. For each column with a leading 1, use Type 3 operations to clear every other entry in that column. The matrix is now in RREF.

Forward, then backward. Steps 1 to 4 sweep downward and rightward and produce zeros below each leading 1 (REF). Step 5 sweeps back up and produces zeros above them (RREF).

5.3Going deeper

Row operations keep the solutions. Applying them to $[A \mid \mathbf{b}]$ never changes the solution set of $A\mathbf{x} = \mathbf{b}$ . Each step is reversible, so equations are replaced by equivalent ones (extra).
Why Type 2 uses $c \neq 0$ (extra). Multiplying a row by 0 would destroy information and cannot be undone.
Each operation is a matrix (extra). Doing a row operation is the same as multiplying on the left by an elementary matrix. That is why $E A = \text{RREF}$ in Q3 of Part 4 acts on every column in the same way.

5.4Exam tips and common mistakes

Exam tips

Write the operation next to every step (R₂ − 3R₁ and so on) so mistakes are easy to trace.
To avoid fractions, you may swap rows to get a 1 early, or scale only when needed.
Never change two rows at once with a swap and an add in the same step.

Common mistakes

Believing that scaling a row leaves $\det$ unchanged (Type 2 changes it by $c$ ).
Using the already-changed row while updating a second row in the same step.
Forgetting Step 5 and calling a REF a RREF.

Practice questions (graded-assignment style)

Q1Non-zero rows

Numerical Answer Type (NAT).

Let $A = \begin{bmatrix}0&1&3\\1&0&2\\1&-1&-1\end{bmatrix}$ and let $R$ be its reduced row echelon form. How many non-zero rows does $R$ have?

Solution

Idea

Row-reduce $A$ .

Swap $R_1 \leftrightarrow R_2$ : $\begin{bmatrix}1&0&2\\0&1&3\\1&-1&-1\end{bmatrix}$ .
$R_3 - R_1$ : $\begin{bmatrix}1&0&2\\0&1&3\\0&-1&-3\end{bmatrix}$ ; then $R_3 + R_2$ : $\begin{bmatrix}1&0&2\\0&1&3\\0&0&0\end{bmatrix}$ .
This is already the RREF, with 2 non-zero rows.
Answer: 2

Q2Using the RREF

Numerical Answer Type (NAT).

For the same $A$ , consider $A\mathbf{x} = \mathbf{b}$ with $\mathbf{x} = (x_1, x_2, x_3)^T$ and $\mathbf{b} = (4, 3, -1)^T$ . If $x_1 = 0$ is given, find $x_2$ .

Solution

Idea

Use the first two rows of $A\mathbf{x} = \mathbf{b}$ .

Row 2 of $A\mathbf{x} = \mathbf{b}$ : $x_1 + 2x_3 = 3$ . With $x_1 = 0$ , $x_3 = \tfrac32$ .
Row 1: $x_2 + 3x_3 = 4$ , so $x_2 = 4 - \tfrac92 = -\tfrac12$ .
Check row 3: $x_1 - x_2 - x_3 = 0 + \tfrac12 - \tfrac32 = -1$ . $\checkmark$
Answer: -0.5

Q3Row operations and the determinant

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

Let $A$ and $B$ be $3 \times 3$ matrices. Eqn 1: $\det(A) = -\det(B)$ . Eqn 2: $\det(A) = -c\det(B)$ for some $c \in \mathbb{R}$ . Eqn 3: $\det(A) = \det(B)$ . Choose the correct options.

B

comes from

A

via a Type 1 operation, Eqn 1 holds.

B

comes from

A

via a Type 1 operation followed by a Type 2 operation, Eqn 2 holds for some

c

B

comes from

A

via a Type 2 operation, Eqn 3 holds.

B

comes from

A

via a Type 3 operation, Eqn 3 holds.

Solution

Idea

Recall the effect of each type on $\det$ .

Type 1 flips the sign: option 1 is true.
Type 1 then Type 2 gives $\det(B) = -c\det(A)$ (with $c$ the scaling factor), which is Eqn 2 for some $c$ : option 2 is true.
Type 2 generally changes the determinant (it multiplies it by $c$ ), so option 3 is false.
Type 3 leaves $\det$ unchanged: option 4 is true.
Answer: options 1, 2 and 4.

Q4Swap and scale (extra)

Numerical Answer Type (NAT).

$\det(A) = 6$ . Swap two rows of $A$ , then multiply one row of the result by 3. What is the determinant of the final matrix?

Solution

Idea

Apply the two effects one after another.

Swap: $6 \to -6$ .
Scale a row by 3: $-6 \to -18$ .
Answer: -18

Revision overview

Type 1Swap two rows
det changes sign

Type 2Scale a row by c ≠ 0
det × c

Type 3Add a multiple of
one row to another
det unchanged

Row reductionWeek 2 · Part 5

Steps 1 to 4Left-most column, make 1,
zero below, repeat
Result: REF

Step 5Clear above each leading 1
Result: RREF

CheckTrack det through each step
as a sanity check

Revision: carrying a matrix to REF and RREF

Next · Any systemThe Gauss Elimination Method

Any system

The Gauss Elimination Method

Solving any linear system by row-reducing its augmented matrix, and reading off whether the answer is unique, infinite, or absent.

6.1Why Gauss elimination?

Cramer's Rule and the inverse method both need a square, invertible $A$ . Gauss elimination has no such requirement. It works whether $A$ is square or rectangular, invertible or singular, and it tells you outright whether the system has a unique solution, infinitely many, or none at all.

6.2Homogeneous and non-homogeneous systems

Homogeneous: Ax = 0

Every right-hand side is zero. It always has the trivial solution $\mathbf{x} = \mathbf{0}$ . Then exactly one of two things happens: $\mathbf{0}$ is the only solution, or there are infinitely many solutions besides $\mathbf{0}$ .

Non-homogeneous: Ax = b

Some $b_i \neq 0$ . It can land in any of three buckets: a unique solution, infinitely many, or none. Gauss elimination decides which.

What is guaranteed for Ax = 0

More unknowns than equations ( $n > m$ ): there are always infinitely many solutions besides $\mathbf{0}$ , because a free variable must exist.
$A$ square and invertible: $\mathbf{0}$ is the only solution.

Correction to the source (extra). The source says

\mathbf{0}

is the only solution "whenever the number of equations exceeds the number of variables". That is not always true:

x + y = 0,\ 2x + 2y = 0,\ 3x + 3y = 0

has 3 equations and 2 unknowns but infinitely many solutions. The exact test is:

\mathbf{0}

is the only solution exactly when every variable column has a leading 1 in the RREF.

6.3The augmented matrix

For $A\mathbf{x} = \mathbf{b}$ , the augmented matrix $[A \mid \mathbf{b}]$ glues the column $\mathbf{b}$ onto $A$ as an extra last column:

[A \mid \mathbf{b}] = \left[\begin{array}{cccc|c}a_{11}&a_{12}&\cdots&a_{1n}&b_1\\a_{21}&a_{22}&\cdots&a_{2n}&b_2\\\vdots&\vdots&\ddots&\vdots&\vdots\\a_{m1}&a_{m2}&\cdots&a_{mn}&b_m\end{array}\right]

6.4The algorithm

Figure 8. Gauss elimination in four moves

Step 1. Form the augmented matrix $[A \mid \mathbf{b}]$ .
Step 2. Row-reduce it all the way to RREF, with the operations of Part 5.
Step 3. Split the result as $[R \mid \mathbf{c}]$ , where $R$ is the RREF of $A$ and $\mathbf{c}$ is the transformed last column.
Step 4. Rewrite it as the equation $R\mathbf{x} = \mathbf{c}$ .
Step 5. Solve $R\mathbf{x} = \mathbf{c}$ directly. A variable with no leading 1 in its column is a free (independent) variable; the others are dependent, written in terms of the free ones.

6.5Reading the outcome

Figure 9. The three possible RREFs of an augmented matrix

RREF of [A \| b]	Outcome	Why
every variable column has a leading 1, no bad row	unique solution	each variable is fixed
some variable column has no leading 1, no bad row	infinitely many	free variables can take any value
a row $(0\ \ 0\ \cdots\ 0 \mid 1)$	no solution	it says $0 = 1$

6.6Choosing a method

Figure 10. Which method fits which system

6.7Going deeper

Free variables. The number of free variables is (number of unknowns) $-$ (number of leading 1s in the coefficient part). One free variable gives a line of solutions, two give a plane, and so on (extra).
Parameter problems. When a system has a letter like $c$ or $d$ , reduce as far as the letters allow, then split into cases (Q3 below).
The same pivot idea (extra). "Leading 1 in every variable column" means unique; "a free column" means infinitely many; "0 = 1 row" means none. These three checks answer almost every question on this topic.

6.8Rank, the inverse by Gauss–Jordan, and the rules of the inverse

Four ideas that the graded assignments lean on and that grow straight out of row reduction: rank, the rank test for solutions, finding $A^{-1}$ by Gauss–Jordan, and the algebra of inverses.

A. Rank

Rank

The rank of a matrix $A$ , written $\operatorname{rank}(A)$ , is the number of non-zero rows in its row echelon form, which is the same as the number of leading 1s (pivots) in its RREF.

Example. $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 3 \\ 3 & 6 & 4 \end{bmatrix}$ . Then $R_2 - 2R_1 = (0, 0, 1)$ and $R_3 - 3R_1 = (0, 0, 1)$ ; subtracting once more, $R_3 - R_2 = (0, 0, 0)$ . The RREF is

\begin{bmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \quad\Rightarrow\quad \operatorname{rank}(A) = 2.

Elementary row operations do not change the rank, so any echelon form gives the right count.
For an $m \times n$ matrix, $\operatorname{rank}(A) \le \min(m, n)$ , and $\operatorname{rank}(A) = \operatorname{rank}(A^T)$ .
Rank counts the rows that are genuinely new: here row 2 is not a multiple of row 1, but row 3 is made of rows 1 and 2, so only two rows survive.
For an $n \times n$ matrix: $\operatorname{rank}(A) = n \iff \det A \neq 0$ (RREF is $I$ ). In the example, $\det A = 0$ and $\operatorname{rank} = 2 < 3$ .

B. The rank test for $A\mathbf{x} = \mathbf{b}$

Let $A$ be $m \times n$ ( $n$ unknowns) and compare the rank of $A$ with the rank of the augmented matrix $[A \mid \mathbf{b}]$ . This is just 6.5 written with ranks: a row $(0\ \cdots\ 0 \mid 1)$ raises the rank of $[A \mid \mathbf{b}]$ by one.

Ranks	Outcome
$\operatorname{rank}[A \mid \mathbf{b}] > \operatorname{rank}(A)$	no solution (inconsistent)
$\operatorname{rank}[A \mid \mathbf{b}] = \operatorname{rank}(A) = n$	exactly one solution
$\operatorname{rank}[A \mid \mathbf{b}] = \operatorname{rank}(A) = r < n$	infinitely many, with $n - r$ free variables

The two ranks can differ by at most 1, because

[A \mid \mathbf{b}]

has only one extra column. For a homogeneous system

A\mathbf{x} = \mathbf{0}

the two ranks are always equal, so it is always consistent, and it has a non-trivial solution exactly when

\operatorname{rank}(A) < n

A parameter example. Take $x + y + z = 1,\; x + 2y + 3z = 4,\; x + 3y + kz = m$ . Row-reduce the augmented matrix: $R_2 - R_1$ gives $(0, 1, 2 \mid 3)$ , $R_3 - R_1$ gives $(0, 2, k - 1 \mid m - 1)$ , then $R_3 - 2R_2$ gives

\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & k - 5 & m - 7 \end{array}\right]

$k \neq 5$ : both ranks are 3, so a unique solution, for every $m$ .
$k = 5$ and $m = 7$ : last row is all zeros, both ranks are 2, so infinitely many solutions ( $3 - 2 = 1$ free variable).
$k = 5$ and $m \neq 7$ : last row is $(0\ 0\ 0 \mid m - 7)$ with $m - 7 \neq 0$ ; ranks are 2 and 3, so no solution.

The coefficient determinant of this system is $k - 5$ , which agrees: the system fails to have a unique solution only at $k = 5$ .

C. Finding $A^{-1}$ by Gauss–Jordan

Write $A$ and the identity side by side as $[A \mid I]$ and row-reduce until the left half becomes $I$ . Whatever the same row operations have turned the right half into is $A^{-1}$ :

[A \mid I] \;\xrightarrow{\text{row operations}}\; [I \mid A^{-1}]

Why it works. Each row operation is multiplication on the left by an invertible matrix. If the product of all of them is $E$ , then $EA = I$ and $EI = E$ . So $E = A^{-1}$ , and the right half records $E$ .

Example. $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}$ .

\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 5 & 6 & 0 & 0 & 0 & 1 \end{array}\right] \xrightarrow{R_3 - 5R_1} \left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & -4 & -15 & -5 & 0 & 1 \end{array}\right]

\xrightarrow{R_3 + 4R_2} \left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 4 & 0 & 1 & 0 \\ 0 & 0 & 1 & -5 & 4 & 1 \end{array}\right] \xrightarrow{R_2 - 4R_3} \left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 0 & 20 & -15 & -4 \\ 0 & 0 & 1 & -5 & 4 & 1 \end{array}\right]

\xrightarrow{R_1 - 3R_3} \left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 16 & -12 & -3 \\ 0 & 1 & 0 & 20 & -15 & -4 \\ 0 & 0 & 1 & -5 & 4 & 1 \end{array}\right] \xrightarrow{R_1 - 2R_2} \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -24 & 18 & 5 \\ 0 & 1 & 0 & 20 & -15 & -4 \\ 0 & 0 & 1 & -5 & 4 & 1 \end{array}\right]

So $A^{-1} = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix}$ . Check: $\det A = 1$ (so $\operatorname{adj}(A)$ is the same matrix), and the first row of $A$ times the first column of $A^{-1}$ is $1(-24) + 2(20) + 3(-5) = 1$ , while row 1 times column 2 is $1(18) + 2(-15) + 3(4) = 0$ ✓.

When it fails. If a zero row appears in the left half,

A

is singular and has no inverse. For

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

R_2 - 4R_1 = (0, -3, -6)

R_3 - 7R_1 = (0, -6, -12)

, then

R_3 - 2R_2 = (0, 0, 0)

. Rank 2, so no inverse. Gauss–Jordan therefore also tests invertibility without computing a determinant.

2 × 2 shortcut. $\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \dfrac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ . For $\begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$ , $ad - bc = 1$ , so the inverse is $\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$ .

D. The rules of the inverse

Let $A$ and $B$ be invertible matrices of the same order.

Rule	Why
$(A^{-1})^{-1} = A$	$A$ does what the inverse of $A^{-1}$ must do: $A^{-1}A = I$ .
$(AB)^{-1} = B^{-1}A^{-1}$ (order reverses)	$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I$ . Undo the last step first, like taking off socks and shoes.
$(A^T)^{-1} = (A^{-1})^T$	$A^T(A^{-1})^T = (A^{-1}A)^T = I^T = I$ .
$(cA)^{-1} = \dfrac{1}{c}A^{-1}$ , $c \neq 0$	$(cA)\left(\tfrac1c A^{-1}\right) = AA^{-1} = I$ .
$(A^k)^{-1} = (A^{-1})^k$	Apply the product rule $k$ times.
$\det(A^{-1}) = \dfrac{1}{\det A}$	$\det A \cdot \det A^{-1} = \det I = 1$ .
$AB$ is invertible exactly when both $A$ and $B$ are	$\det(AB) = \det A \det B \neq 0$ needs both factors non-zero.

Not true

$(AB)^{-1} \neq A^{-1}B^{-1}$ in general. It is $B^{-1}A^{-1}$ .
$(A + B)^{-1} \neq A^{-1} + B^{-1}$ . There is no simple formula for the inverse of a sum.
No cancelling without an inverse. $AB = AC$ gives $B = C$ only when $A$ is invertible (multiply by $A^{-1}$ on the left). With $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ , $B = I$ , $C = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ , we get $AB = AC = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ but $B \neq C$ .

Solving matrix equations. For invertible $A$ : $AX = B \Rightarrow X = A^{-1}B$ (multiply on the left), but $XA = B \Rightarrow X = BA^{-1}$ (multiply on the right). The side matters because matrix products do not commute.

E. One square matrix, many equivalent statements

For an $n \times n$ matrix $A$ , each of these says the same thing, and each has a negation that all say the opposite:

$A$ is invertible	$A$ is singular
$\det A \neq 0$	$\det A = 0$
$\operatorname{rank}(A) = n$	$\operatorname{rank}(A) < n$
RREF of $A$ is $I$	RREF has a zero row
$A\mathbf{x} = \mathbf{0}$ has only $\mathbf{x} = \mathbf{0}$	$A\mathbf{x} = \mathbf{0}$ has a non-trivial solution
$A\mathbf{x} = \mathbf{b}$ has exactly one solution for every $\mathbf{b}$	$A\mathbf{x} = \mathbf{b}$ has none or infinitely many, depending on $\mathbf{b}$
Cramer's Rule and $\mathbf{x} = A^{-1}\mathbf{b}$ apply	Neither applies; use Gauss elimination

6.9Exam tips and common mistakes

Exam tips

Always carry the last column along with every row operation.
After solving, plug your answer back into the original equations.
For a solution set with a free variable, write it with a parameter: S = { ... | c ∈ ℝ }.

Common mistakes

Row-reducing only A and forgetting the b column.
Treating a free variable as dependent, or giving only one solution for a system with a free variable.
Missing the row $(0\ \ 0\ \cdots\ 0 \mid 1)$ and claiming a solution exists.

Practice questions (graded-assignment style)

Q1Going back two steps

Numerical Answer Type (NAT).

Let $P = \begin{bmatrix}1&0&1\\0&1&1\\0&0&1\end{bmatrix}$ and $PX_{n-1} = X_n$ . If $X_2 = \begin{bmatrix}7\\8\\3\end{bmatrix}$ , what is the sum of all elements of $X_0$ ?

Solution

Idea

$X_0 = P^{-2}X_2$ .

$P^{-1} = \begin{bmatrix}1&0&-1\\0&1&-1\\0&0&1\end{bmatrix}$ ; check: $P P^{-1} = I$ .
$X_1 = P^{-1}X_2 = (7 - 3,\ 8 - 3,\ 3)^T = (4, 5, 3)^T$ .
$X_0 = P^{-1}X_1 = (4 - 3,\ 5 - 3,\ 3)^T = (1, 2, 3)^T$ . The sum is $6$ .
Answer: 6

Q2How many solutions?

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

The system $x - y + z = 2,\ x + y - z = 3,\ -x + y + z = 4$ has:

No solution.

Infinitely many solutions.

A unique solution.

Finitely many solutions.

Solution

Idea

Find $\det$ of the coefficient matrix.

$\det\begin{bmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{bmatrix} = 4 \neq 0$ , so the coefficient matrix is invertible and the system has a unique solution.
In fact $x = \tfrac52,\ y = \tfrac72,\ z = 3$ (check: $\tfrac52 - \tfrac72 + 3 = 2$ , $\tfrac52 + \tfrac72 - 3 = 3$ , $-\tfrac52 + \tfrac72 + 3 = 4$ ).
Option 4 ("finitely many") is not a distinct case: a linear system has 0, 1 or infinitely many solutions. Option 3 is the one that fits.
Answer: option 3.

Q3A system with parameters

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

Consider $-x_1 + x_2 + 2x_3 = 1,\ 2x_1 + x_2 - 2x_3 = -1,\ 3x_2 + cx_3 = d$ . Choose the correct options.

c = 2

and

d = 1

, the system has infinitely many solutions.

c = 1

and

d = 1

, the system has infinitely many solutions.

c = 1

and

d = 2

, the system has no solution.

c = 3

and

d = 2

, the system has a unique solution.

Solution

Idea

Solve the first two equations for $x_1, x_2$ in terms of $x_3$ , then substitute into the third.

The first two give $x_1 = \tfrac43 x_3 - \tfrac23$ and $x_2 = \tfrac13 - \tfrac23 x_3$ .
Substituting into the third: $3x_2 + cx_3 = 1 - 2x_3 + cx_3 = 1 + (c - 2)x_3 = d$ .
$c = 2$ : the equation is $1 = d$ . For $d = 1$ it always holds, so infinitely many (option 1 true); for $d \neq 1$ it has no solution.
$c \neq 2$ : $x_3 = \dfrac{d - 1}{c - 2}$ , a unique solution. So $c = 1, d = 1$ gives $x_3 = 0$ (unique, option 2 false); $c = 1, d = 2$ gives $x_3 = -1$ (unique, option 3 false); $c = 3, d = 2$ gives $x_3 = 1$ (unique, option 4 true).
Answer: options 1 and 4.

Q4Counting free variables (extra)

Numerical Answer Type (NAT).

The RREF of the augmented matrix of a system in $x_1, x_2, x_3, x_4$ is $\left[\begin{array}{cccc|c}1&0&2&0&5\\0&1&-1&0&2\\0&0&0&1&7\end{array}\right]$ . How many free variables are there?

Solution

Idea

Which variable column has no leading 1?

Leading 1s are in columns 1, 2 and 4.
Column 3 has no leading 1, so $x_3$ is the only free variable: $x_1 = 5 - 2x_3$ , $x_2 = 2 + x_3$ , $x_4 = 7$ .
Answer: 1

Q5A homogeneous system (extra)

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

A homogeneous system has 2 equations in 3 unknowns. Which statement is true?

\mathbf{0}

is the only solution.

It has infinitely many solutions.

It has no solution.

It has exactly two solutions.

Solution

Idea

A homogeneous system always has $\mathbf{0}$ , and unknowns exceed equations.

It is never inconsistent ( $\mathbf{x} = \mathbf{0}$ works), so option 3 is false.
With 3 unknowns and at most 2 leading 1s, some column is free, giving infinitely many solutions.
A linear system never has exactly two solutions. Option 2 is true.
Answer: option 2.

Q6Rank of a matrix

Numerical Answer Type (NAT).

Find the rank of $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 3 \\ 3 & 6 & 4 \end{bmatrix}$ .

Solution

Idea

Row-reduce and count the non-zero rows.

$R_2 - 2R_1 = (0, 0, 1)$ and $R_3 - 3R_1 = (0, 0, 1)$ . Then $R_3 - R_2 = (0, 0, 0)$ .
Echelon form: $\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ , with 2 non-zero rows.
Check: $\det A = 0$ , so the rank must be less than 3 ✓. Rows 1 and 2 are not proportional, so the rank is not 1.
Answer: 2

Q7Number of solutions with two parameters

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

Consider $x + y + z = 1,\; x + 2y + 3z = 4,\; x + 3y + kz = m$ . Select all correct statements.

For

k = 5

and

m = 7

, there are infinitely many solutions.

For

k = 5

and

m = 8

, there is no solution.

For

k = 6

and any

m

, there is exactly one solution.

For every

k \neq 5

, there is no solution.

Solution

Idea

Reduce the augmented matrix and compare the rank of $A$ with the rank of $[A \mid \mathbf{b}]$ .

Row reduction gives the last row $(0\ \ 0\ \ k - 5 \mid m - 7)$ (see the parameter example in 6.9).
Option 1 ✓ $k = 5, m = 7$ : last row is $(0\ 0\ 0 \mid 0)$ . Both ranks are 2, less than 3 unknowns, so infinitely many solutions.
Option 2 ✓ $k = 5, m = 8$ : last row is $(0\ 0\ 0 \mid 1)$ , which says $0 = 1$ . The ranks are 2 and 3, so no solution.
Option 3 ✓ $k = 6$ : the pivot $k - 5 = 1 \neq 0$ , so all three columns have pivots: one solution whatever $m$ is.
Option 4 ✗ For $k \neq 5$ the last pivot exists and the solution is unique, not absent.
Answer: options 1, 2 and 3.

Q8An entry of the inverse by Gauss–Jordan

Numerical Answer Type (NAT).

For $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}$ , find the $(1, 1)$ entry of $A^{-1}$ .

Solution

Idea

Reduce $[A \mid I]$ to $[I \mid A^{-1}]$ and read the entry. (Cross-check with a cofactor.)

The worked reduction in 6.9 gives $A^{-1} = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix}$ .
Cross-check with $A^{-1} = \operatorname{adj}(A)/\det A$ : the $(1,1)$ entry of the adjugate is the cofactor $C_{11} = 1\cdot 0 - 4 \cdot 6 = -24$ , and $\det A = 1\cdot(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1$ . So the entry is $-24/1 = -24$ ✓.
Answer: -24

Q9Determinant of an expression with inverses

Numerical Answer Type (NAT).

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

Solution

Idea

$\det$ of an inverse is the reciprocal, and $\det$ of a product is the product of the $\det$ s. The order inside $(AB)^{-1}$ does not matter for the determinant.

$\det(AB) = 2 \cdot 4 = 8$ , so $\det\big((AB)^{-1}\big) = \dfrac18$ .
$\det(A^2) = 2^2 = 4$ .
$\det\big((AB)^{-1}A^2\big) = \dfrac18 \cdot 4 = \dfrac12 = 0.5$ .
Answer: 0.5

Q10Rules of the inverse

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

$A$ and $B$ are invertible $n \times n$ matrices. Choose the correct statements.

(AB)^{-1} = A^{-1}B^{-1}

always.

(AB)^{-1} = B^{-1}A^{-1}

(A^T)^{-1} = (A^{-1})^T

(A + B)^{-1} = A^{-1} + B^{-1}

(2A)^{-1} = \tfrac12 A^{-1}

CA = CB

for some matrix

C

, then

A = B

Solution

Idea

Test each rule by multiplying the proposed inverse by the matrix and asking whether the result is $I$ .

Option 1 ✗ $(AB)(A^{-1}B^{-1}) = ABA^{-1}B^{-1}$ , which is generally not $I$ because $A$ and $B$ do not commute.
Option 2 ✓ $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = I$ .
Option 3 ✓ $A^T(A^{-1})^T = (A^{-1}A)^T = I$ .
Option 4 ✗ For $A = B = I$ : $(2I)^{-1} = \tfrac12 I$ , but $I^{-1} + I^{-1} = 2I$ .
Option 5 ✓ $(2A)\left(\tfrac12 A^{-1}\right) = I$ .
Option 6 ✗ Cancelling needs $C$ to be invertible. With $C = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ , $A = I$ , $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ : $CA = CB$ but $A \neq B$ .
Answer: options 2, 3 and 5.

Q11Which side does $A^{-1}$ go on?

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

$A$ is an invertible $n \times n$ matrix and $X$ , $B$ are $n \times n$ matrices with $XA = B$ . Then $X$ equals

A^{-1}B

BA^{-1}

AB^{-1}

B^{-1}A

Solution

Idea

$A$ sits on the right of $X$ , so multiply both sides by $A^{-1}$ on the right to cancel it.

$XA = B \Rightarrow XAA^{-1} = BA^{-1} \Rightarrow X = BA^{-1}$ .
Option 1 would be the answer to $AX = B$ , not $XA = B$ . Options 3 and 4 invert $B$ , which need not be invertible.
Numeric check: $A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$ , $X = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$ give $B = XA = \begin{bmatrix} 1 & 2 \\ 5 & 9 \end{bmatrix}$ , and $BA^{-1}$ returns $X$ ✓.
Answer: option 2.

Q12When does a homogeneous system have a non-trivial solution?

Numerical Answer Type (NAT).

For which value of $k$ does $x + y + z = 0,\; x + 2y + 3z = 0,\; x + 3y + kz = 0$ have a non-trivial solution?

Solution

Idea

A square homogeneous system has a non-trivial solution exactly when the coefficient determinant is $0$ .

$\det\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & k \end{bmatrix} = 1(2k - 9) - 1(k - 3) + 1(3 - 2) = 2k - 9 - k + 3 + 1 = k - 5$ .
$k - 5 = 0 \Rightarrow k = 5$ .
Check: at $k = 5$ , $R_3 = 2R_2 - R_1$ , so the rows are dependent and the rank is 2, with $3 - 2 = 1$ free variable.
Answer: 5

Q13A system with more unknowns than equations

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

$A$ is a $3 \times 4$ matrix with $\operatorname{rank}(A) = 3$ . Which is true about $A\mathbf{x} = \mathbf{b}$ ?

For some

\mathbf{b}

there is no solution.

For every

\mathbf{b}

there is exactly one solution.

For every

\mathbf{b}

there are infinitely many solutions.

There are infinitely many solutions only when

\mathbf{b} = \mathbf{0}

Solution

Idea

Compare ranks: $[A \mid \mathbf{b}]$ also has 3 rows, so its rank cannot exceed 3.

$\operatorname{rank}[A \mid \mathbf{b}] \le 3 = \operatorname{rank}(A)$ , and it is at least $\operatorname{rank}(A)$ , so the two ranks are equal: the system is consistent for every $\mathbf{b}$ . That rules out option 1.
The number of free variables is $n - r = 4 - 3 = 1 > 0$ , so each solvable system has infinitely many solutions. That rules out option 2 and option 4.
Answer: option 3.

Q14Counting free variables from the rank

Numerical Answer Type (NAT).

A homogeneous system $A\mathbf{x} = \mathbf{0}$ has a $4 \times 6$ coefficient matrix of rank 3. How many free variables does its general solution have?

Solution

Idea

Free variables = unknowns $-$ rank.

The number of unknowns is the number of columns, $n = 6$ (not the number of rows).
Free variables $= 6 - 3 = 3$ . Pivot columns give the 3 dependent variables, and the other 3 columns are free.
Answer: 3

Revision overview

SystemsAx = 0: homogeneous
Ax = b: non-homogeneous

Augmented[A | b]
b glued as last column

Reduce[A | b] → [R | c]
R is the RREF of A

Gauss eliminationWeek 2 · Part 6

SolveRx = c
Free column: parameter

OutcomesUnique, infinitely many,
or none (0 = 1 row)

Works forAny shape, any rank

Revision: solving any linear system with row reduction

Next · All partsWeek 2 Summary Sheet

All parts

Week 2 Summary Sheet

One page to revise all of Week 2: the methods, their conditions, the row operations and the three possible outcomes.

The big picture

Part 1 · Linear systems

$A\mathbf{x} = \mathbf{b}$ , with $A$ of order $m \times n$ .
$\det(A) \neq 0 \iff A$ invertible.

Part 2 · Cramer's Rule

$x_i = \det(A_{x_i}) / \det(A)$ .
Needs a square, invertible A.

Part 3 · Inverse method

$\mathbf{x} = A^{-1}\mathbf{b}$ , $A^{-1} = \operatorname{adj}(A)/\det(A)$ .
Same conditions as Cramer.

Part 4 · Echelon forms

REF: leading 1s step right, zero rows at the bottom.
RREF: each leading 1 alone in its column.

Part 5 · Row reduction

Three row operations; five-step algorithm.
Swap: $\det$ flips; scale: $\det \times c$ ; add: unchanged.

Part 6 · Gauss elimination

$[A \mid \mathbf{b}] \to [R \mid \mathbf{c}]$ , then solve $R\mathbf{x} = \mathbf{c}$ .
Unique, infinitely many, or none.

Revision sheet

Topic	Key idea
Linear equation	$a_1 x_1 + \cdots + a_n x_n = b$ : every variable to the power 1
Invertible / non-singular	$\det(A) \neq 0$
Cramer's Rule	$x_i = \det(A_{x_i}) / \det(A)$ ; needs a square, invertible $A$
Inverse-matrix method	$\mathbf{x} = A^{-1}\mathbf{b}$ ; needs a square, invertible $A$
Row echelon form	leading 1s step right going down; zero rows at the bottom
Reduced row echelon form	REF, and each leading 1 is the only non-zero entry in its column
Type 1 row operation	swap two rows $\to$ $\det$ flips sign
Type 2 row operation	scale a row by $c$ $\to$ $\det$ scales by $c$
Type 3 row operation	add a multiple of one row to another $\to$ $\det$ unchanged
Augmented matrix	$[A \mid \mathbf{b}]$ : glue $\mathbf{b}$ on as the last column
Gauss elimination	row-reduce $[A \mid \mathbf{b}]$ to $[R \mid \mathbf{c}]$ , then solve $R\mathbf{x} = \mathbf{c}$
Free variable	a variable with no leading 1 in its column; set it to a parameter

Which method when?

Figure 11. Choosing between Cramer / inverse and Gauss elimination

Situation	Use
square $A$ , $\det(A) \neq 0$ , one answer wanted	Cramer's Rule or $A^{-1}\mathbf{b}$
$\det(A) = 0$ , or $A$ not square	Gauss elimination
decide unique / infinitely many / none	Gauss elimination (look at the RREF)
is a matrix in REF / RREF?	check the four conditions in order

Common mistakes

Calling a non-linear equation linear

Squares and products of variables make it non-linear.

Using Cramer on a singular system

If $\det(A) = 0$ or $A$ is not square, use Gauss elimination.

Wrong product order

It is $A^{-1}\mathbf{b}$ , not $\mathbf{b}A^{-1}$ .

REF vs RREF

A REF may still have non-zero entries above a leading 1.

Scaling does not keep det

Only Type 3 (and no other) leaves $\det$ unchanged.

Missing the 0 = 1 row

A row $(0\ \cdots\ 0 \mid 1)$ means no solution.

Mathematics forData Science 2

Chapters

Linear Equations and a Determinants Recap

1.1What is a linear equation?

1.2A system of equations as Ax = b

1.3Determinant recap

1.4Seeing it geometrically

1.5Going deeper

1.6Exam tips and common mistakes

Practice questions (graded-assignment style)

Cramer's Rule

2.1When can Cramer's Rule be used?

2.2Order 2

2.3Order 3

2.4Order n: the general pattern

2.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Solving with an Invertible Coefficient Matrix

3.1The idea

3.2Seeing it in 3D

3.3Cramer's Rule or the inverse?

3.4Going deeper

3.5Exam tips and common mistakes

Practice questions (graded-assignment style)

The Echelon Form

4.1Why we need it

4.2Row echelon form (REF)

4.3Reduced row echelon form (RREF)

4.4Going deeper

4.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Row Reduction

5.1The three elementary row operations

5.2The row-reduction algorithm

5.3Going deeper

5.4Exam tips and common mistakes

Practice questions (graded-assignment style)

The Gauss Elimination Method

6.1Why Gauss elimination?

6.2Homogeneous and non-homogeneous systems

6.3The augmented matrix

6.4The algorithm

6.5Reading the outcome

6.6Choosing a method

6.7Going deeper

6.8Rank, the inverse by Gauss–Jordan, and the rules of the inverse

A. Rank

B. The rank test for Ax=bA\mathbf{x} = \mathbf{b}Ax=b

C. Finding A−1A^{-1}A−1 by Gauss–Jordan

D. The rules of the inverse

E. One square matrix, many equivalent statements

6.9Exam tips and common mistakes

Practice questions (graded-assignment style)

Week 2 Summary Sheet

The big picture

Revision sheet

Which method when?

Common mistakes

Mathematics for
Data Science 2

B. The rank test for $A\mathbf{x} = \mathbf{b}$

C. Finding $A^{-1}$ by Gauss–Jordan