Chapter 4 · Determinants

Exercise 4.4 · Q13: Show $A^2-5A+7I=O$ ⁠; hence find $A^{-1}$

Exercise 4.4Adjoint and inverse of a matrix · Section 4.5

The question

A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix},

show that $A^2 - 5A + 7I = O$ ⁠. Hence find $A^{-1}$ ⁠.

Exam answer

\begin{aligned} A^2 &= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\ &= \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} \\ &= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \end{aligned}

\begin{aligned} &A^2 - 5A + 7I \\ &= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} \\ &\quad + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \end{aligned}

Hence $A^2 - 5A + 7I = O$ ⁠.

Now $|A| = 6 + 1 = 7 \ne 0$ ⁠, so $A^{-1}$ exists. Post-multiplying by $A^{-1}$ ⁠,

\begin{aligned} &(A^2 - 5A + 7I)A^{-1} = OA^{-1} \\ &\Rightarrow\ A - 5I + 7A^{-1} = O \\ &\Rightarrow\ A^{-1} = \frac{1}{7}(5I - A) \\ &\qquad = \frac{1}{7}\bigg(\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\bigg) \end{aligned}

$\therefore\ A^{-1} = \dfrac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A matrix equation such as $A^2 - 5A + 7I = O$ can be rearranged like ordinary algebra, as long as the order of multiplication is respected. Multiplying it by $A^{-1}$ lowers every power of $A$ by one and turns $I$ into $A^{-1}$ ⁠, so $A^{-1}$ can be read off as $\frac{1}{7}(5I - A)$ without computing any cofactors.

Given, to find, and the plan

Given: The $2\times2$ matrix $A$ above; $I$ is the $2\times2$ identity and $O$ the $2\times2$ zero matrix.
To show / find: $A^2 - 5A + 7I = O$ ⁠; then $A^{-1}$ ⁠.
Plan: (1) Compute $A^2$ ⁠. (2) Compute $5A$ and $7I$ and combine, entry by entry. (3) Check $|A| \neq 0$ so $A^{-1}$ exists. (4) Multiply the equation by $A^{-1}$ and solve for $A^{-1}$ ⁠. (5) Check by $AA^{-1} = I$ and against the adjoint.

Basics you need

1Squaring a matrix

$A^2$ means $A\cdot A$ ⁠, a matrix product (row times column), not the squares of the entries. Basics: matrix facts the chapter uses

2Scalar multiples and sums

$5A$ multiplies every entry by 5; matrices of the same order are added entry by entry. Basics: scaling and adding matrices

3Identity and inverse

$AI = IA = A$ ⁠, and $AA^{-1} = A^{-1}A = I$ ⁠. $A^{-1}$ exists exactly when $|A| \neq 0$ (Theorem 4). Basics: why the inverse divides by |A|

4The method of Example 15

The book shows, for another matrix, how an equation of this kind gives $A^{-1}$ ⁠. Example 15

Solution

Compute $A^2 = A\cdot A$

$A$ is $2\times2$ ⁠, so $A\cdot A$ is defined and is $2\times2$ ⁠.
$\begin{aligned} c_{11} &= 3\cdot3 + 1\cdot(-1) = 9 - 1 = 8\\ c_{12} &= 3\cdot1 + 1\cdot2 = 3 + 2 = 5\\ c_{21} &= (-1)\cdot3 + 2\cdot(-1) = -3 - 2 = -5\\ c_{22} &= (-1)\cdot1 + 2\cdot2 = -1 + 4 = 3 \end{aligned}$ $A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$
Each entry is a row of the first $A$ times a column of the second $A$ ⁠. For $c_{21}$ ⁠: row 2 is $-1, 2$ and column 1 is $3, -1$ ⁠, giving $(-1)3 + 2(-1)$ ⁠. Squaring the entries instead ( $9, 1, 1, 4$ ⁠) would be wrong.

Pair up, multiply, add. Basics: matrix facts the chapter uses
Combine $A^2 - 5A + 7I$
$5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}, \quad 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$
All three matrices are $2\times2$ ⁠, so the combination is defined. Entry by entry:
$\begin{aligned} (1,1)&:\ 8 - 15 + 7 = 0\\ (1,2)&:\ 5 - 5 + 0 = 0\\ (2,1)&:\ -5 - (-5) + 0 = 0\\ (2,2)&:\ 3 - 10 + 7 = 0 \end{aligned}$
So $A^2 - 5A + 7I = O$ ⁠, as required.

The identity matrix $7I$ has $7$ only on the diagonal, so it changes only the diagonal entries. The off-diagonal entries of $A^2$ and $5A$ already agree, which is why they cancel to $0$ ⁠.

Add and subtract matrices of the same order entry by entry. Basics: scaling and adding matrices
Make sure $A^{-1}$ exists
$|A| = \begin{vmatrix} 3 & 1 \\ -1 & 2 \end{vmatrix} = 3\cdot2 - 1\cdot(-1) = 7$
$|A| = 7 \neq 0$ ⁠, so $A$ is non-singular and $A^{-1}$ exists (Theorem 4).

The next step multiplies by $A^{-1}$ ⁠. That is only allowed if $A^{-1}$ exists, so we confirm it first, exactly as the book does in Example 15 ("because $|A| \neq 0$ ").

Lesson: inverse of a matrix
Multiply the equation by $A^{-1}$

Start from $A^2 - 5A + 7I = O$ ⁠, that is $AA - 5A + 7I = O$ ⁠. Multiply every term on the right by $A^{-1}$ ⁠:
$\begin{aligned} AA A^{-1} - 5AA^{-1} + 7IA^{-1} &= OA^{-1}\\ A(AA^{-1}) - 5I + 7A^{-1} &= O\\ AI - 5I + 7A^{-1} &= O\\ A - 5I + 7A^{-1} &= O \end{aligned}$
Solve for $A^{-1}$ ⁠:
$\begin{aligned} 7A^{-1} &= 5I - A\\ A^{-1} &= \frac{1}{7}(5I - A) \end{aligned}$
Matrix multiplication distributes over addition, $(X + Y)Z = XZ + YZ$ ⁠, so post-multiplying the whole left side means post-multiplying each term. Associativity lets us regroup $(AA)A^{-1}$ as $A(AA^{-1})$ ⁠. Then $AA^{-1} = I$ ⁠, $IA^{-1} = A^{-1}$ ⁠, $AI = A$ and $OA^{-1} = O$ ⁠. Every term was multiplied on the same side (the right), which matters because matrix products are not commutative in general.

$XI = X$ ⁠, $XX^{-1} = I$ ⁠. Basics: transpose and identity
Compute $\frac{1}{7}(5I - A)$
$\begin{aligned} 5I - A &= \begin{bmatrix} 5 - 3 & 0 - 1 \\ 0 - (-1) & 5 - 2 \end{bmatrix}\\[1mm] &= \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \end{aligned}$ $A^{-1} = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$
$5I$ has $5$ on the diagonal and $0$ elsewhere, so only the diagonal entries of $-A$ gain a $5$ ⁠. As a cross-check, the adjoint method gives the same thing: $\operatorname{adj}A$ swaps $3$ and $2$ and negates $1$ and $-1$ ⁠, giving the matrix with rows $2, -1$ and $1, 3$ ⁠, and $\frac{1}{|A|} = \frac17$ ⁠.

Lesson: the adjoint
Check: $AA^{-1} = I$

Multiply $A$ by the integer matrix, then divide by 7:
$\begin{aligned} (1,1)&:\ 3\cdot2 + 1\cdot1 = 7\\ (1,2)&:\ 3(-1) + 1\cdot3 = 0\\ (2,1)&:\ (-1)2 + 2\cdot1 = 0\\ (2,2)&:\ (-1)(-1) + 2\cdot3 = 7 \end{aligned}$ $AA^{-1} = \frac{1}{7}\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$
The product is $7I$ before dividing, which is $|A|\,I$ ⁠, exactly what $A(\operatorname{adj}A) = |A|\,I$ (Theorem 1) predicts.

Result

$A^2 - 5A + 7I = O$ holds, and

A^{-1} = \frac{1}{7}(5I - A) = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}

(Textbook answer: $\frac{1}{7}$ times the matrix with rows $2, -1$ and $1, 3$ ⁠.)

Check it with numbers

The other order: $A^{-1}A$ has $(1,1)$ entry $\frac17(2\cdot3 + (-1)(-1)) = 1$ and $(1,2)$ entry $\frac17(2\cdot1 + (-1)\cdot2) = 0$ ✓. Also $|A^{-1}|$ should be $\frac{1}{|A|} = \frac17$ ⁠. The entries of $A^{-1}$ are $\frac27, -\frac17$ (row 1) and $\frac17, \frac37$ (row 2), so $|A^{-1}|$ is $\frac27\cdot\frac37 - (-\frac17)\cdot\frac17$ ⁠, that is $\frac{6}{49} + \frac{1}{49} = \frac{7}{49} = \frac17$ ✓.

Common mistakes

Squaring entries. $A^2$ is $A\cdot A$ ⁠; its $(1,1)$ entry is $8$ ⁠, not $9$ ⁠.
Writing $7$ instead of $7I$ ⁠. A number cannot be added to a matrix; the equation has $7I$ ⁠, and multiplying it by $A^{-1}$ gives $7A^{-1}$ ⁠.
Multiplying different terms on different sides. Multiply every term by $A^{-1}$ on the same side.
Getting $A - 5I$ instead of $5I - A$ ⁠. From $A - 5I + 7A^{-1} = O$ ⁠, moving terms gives $7A^{-1} = 5I - A$ ⁠.

What to take away

If $A^2 + pA + qI = O$ with $q \neq 0$ ⁠, then $A^{-1} = -\frac{1}{q}(A + pI)$ ⁠. Here $p = -5$ ⁠, $q = 7$ ⁠.
Always confirm $A^{-1}$ exists before multiplying by it.
Check the answer two ways: $AA^{-1} = I$ ⁠, and against $\frac{1}{|A|}\operatorname{adj}A$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Row-times-column product; $A^2 = AA$	Steps 1, 6	Chapter 3, Section 3.4.5
Scalar multiples, sums, distributive law $(X+Y)Z = XZ + YZ$	Steps 2, 4	Chapter 3, Sections 3.4.2–3.4.6
$A$ invertible $\iff \|A\| \neq 0$ ⁠; $A^{-1} = \frac{1}{\|A\|}\operatorname{adj}A$	Steps 3, 5	Theorem 4, Section 4.5 (Lesson: inverse)
$2\times2$ adjoint rule	Step 5	Remark after Example 12, Section 4.5.1
$A(\operatorname{adj}A) = \|A\|\,I$	Step 6	Theorem 1, Section 4.5
Post-multiplying by $A^{-1}$ to find the inverse	Step 4	Example 15
All entries and the check	Steps 1–6, Check	Recomputed in the site's verification script

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Exercise 4.4 · Q13: Show A2−5A+7I=OA^2-5A+7I=OA2−5A+7I=O⁠; hence find A−1A^{-1}A−1

Basics you need

Solution

Compute A2=A⋅AA^2 = A\cdot AA2=A⋅A

Combine A2−5A+7IA^2 - 5A + 7IA2−5A+7I

Make sure A−1A^{-1}A−1 exists

Multiply the equation by A−1A^{-1}A−1

Compute 17(5I−A)\frac{1}{7}(5I - A)71​(5I−A)

Check: AA−1=IAA^{-1} = IAA−1=I

Every fact used, and where it comes from

Exercise 4.4 · Q13: Show $A^2-5A+7I=O$ ⁠; hence find $A^{-1}$

Compute $A^2 = A\cdot A$

Combine $A^2 - 5A + 7I$

Make sure $A^{-1}$ exists

Multiply the equation by $A^{-1}$

Compute $\frac{1}{7}(5I - A)$

Check: $AA^{-1} = I$