Chapter 4 · Determinants

Example 15: $A^2-4A+I=O$ ⁠; use it to find $A^{-1}$

ExamplesInverse of a matrix · Section 4.5

The question

Show that the matrix

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

satisfies the equation $A^2 - 4A + I = O$ ⁠, where $I$ is the $2\times2$ identity matrix and $O$ is the $2\times2$ zero matrix. Using this equation, find $A^{-1}$ ⁠.

Exam answer

\begin{aligned} A^2 &= \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \\ &= \begin{bmatrix} 4+3 & 6+6 \\ 2+2 & 3+4 \end{bmatrix} \\ &= \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \end{aligned}

\begin{aligned} &A^2 - 4A + I \\ &= \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \end{aligned}

Hence $A^2 - 4A + I = O$ ⁠.

Now $|A| = 4 - 3 = 1 \ne 0$ ⁠, so $A^{-1}$ exists. Post-multiplying by $A^{-1}$ and using $AA^{-1} = I$ ⁠,

\begin{aligned} &(A^2 - 4A + I)A^{-1} = OA^{-1} \\ &\Rightarrow\ A - 4I + A^{-1} = O \\ &\Rightarrow\ A^{-1} = 4I - A \\ &\qquad = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \end{aligned}

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

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

The one idea this problem is built on

A matrix equation can be multiplied through by $A^{-1}$ ⁠, just as a number equation can be divided by a non-zero number. Multiplying $A^2 - 4A + I = O$ by $A^{-1}$ lowers every power of $A$ by one: $A^2$ becomes $A$ ⁠, $A$ becomes $I$ ⁠, and $I$ becomes $A^{-1}$ ⁠. What is left is a simple formula for $A^{-1}$ with no cofactors at all.

Given, to show and find, and the plan

Given: The $2\times2$ matrix $A$ ⁠; $I$ and $O$ are the $2\times2$ identity and zero matrices.
To show: $A^2 - 4A + I = O$ ⁠.
To find: $A^{-1}$ ⁠, using that equation.
Plan: Compute $A^2$ entry by entry, then $A^2 - 4A + I$ entry by entry. Check $|A| \ne 0$ so that $A^{-1}$ exists. Post-multiply the equation by $A^{-1}$ and simplify to $A^{-1} = 4I - A$ ⁠. Check the answer.

Basics you need

1Squaring a matrix

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

2Scaling, adding, identity and zero

$4A$ multiplies every entry by 4. $I$ has 1s on the diagonal and 0s elsewhere; $O$ has all entries 0. Sums are taken entry by entry. Basics: scaling and adding matrices

3Rules for products

$(PQ)R = P(QR)$ (associative), $(P + Q)R = PR + QR$ (distributive), $PI = IP = P$ ⁠, and $AA^{-1} = A^{-1}A = I$ ⁠. The order of factors must be kept. Basics: transpose and identity

4When $A^{-1}$ exists

A square matrix has an inverse exactly when its determinant is not 0 (Theorem 4, Section 4.5). Basics: why the inverse divides by |A|

Solution

Compute $A^2 = AA$

$A$ is $2\times2$ ⁠, so $AA$ is defined and is $2\times2$ ⁠. Row $i$ of $A$ times column $j$ of $A$ ⁠:
$\begin{aligned} (A^2)_{11} &= 2\cdot2 + 3\cdot1 = 7,\\ (A^2)_{12} &= 2\cdot3 + 3\cdot2 = 12,\\ (A^2)_{21} &= 1\cdot2 + 2\cdot1 = 4,\\ (A^2)_{22} &= 1\cdot3 + 2\cdot2 = 7. \end{aligned}$ $A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}.$
The equation involves $A^2$ ⁠, so it must be computed first, and correctly as a product. Squaring each entry would give $4, 9, 1, 4$ ⁠, which is wrong.

Multiplication of matrices, Chapter 3, Section 3.4.5. Basics: matrix facts the chapter uses
Compute $A^2 - 4A + I$

Here
$4A = \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix}, \qquad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$
Entry by entry, $(A^2) - (4A) + (I)$ ⁠:
$\begin{aligned} (1,1)&: \ 7 - 8 + 1 = 0,\\ (1,2)&: \ 12 - 12 + 0 = 0,\\ (2,1)&: \ 4 - 4 + 0 = 0,\\ (2,2)&: \ 7 - 8 + 1 = 0. \end{aligned}$ $A^2 - 4A + I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O.$
So $A$ satisfies $A^2 - 4A + I = O$ .

All three matrices are $2\times2$ ⁠, so the sum is defined. The number 1 in " $+\,I$ " only reaches the diagonal entries, because $I$ has 0 off the diagonal. Every entry came out 0, which is what " $= O$ " means.

Scalar multiples and sums are entry by entry (Chapter 3, Sections 3.4.1 and 3.4.2). Basics: scaling and adding matrices
Make sure $A^{-1}$ exists

$|A| = 2\cdot2 - 3\cdot1 = 4 - 3 = 1 \ne 0$ ⁠. So $A$ is non-singular and $A^{-1}$ exists (Theorem 4).

The next step multiplies by $A^{-1}$ ⁠. Doing that with a matrix that has no inverse would be like dividing by zero. The textbook writes this reason in brackets: "Post multiplying by $A^{-1}$ because $|A| \ne 0$ ".

Theorem 4, Section 4.5. Lesson: inverse of a matrix
Post-multiply the equation by $A^{-1}$

Move $I$ to the other side: $A^2 - 4A = -I$ ⁠, that is $AA - 4A = -I$ ⁠. Multiply both sides on the right by $A^{-1}$ ⁠:
$AA(A^{-1}) - 4AA^{-1} = -IA^{-1}.$
Now simplify each term:
$\begin{aligned} AA(A^{-1}) &= A(AA^{-1}) = AI = A,\\ 4AA^{-1} &= 4I,\\ -IA^{-1} &= -A^{-1}. \end{aligned}$
So $A - 4I = -A^{-1}$ ⁠, and multiplying by $-1$ ⁠:
$A^{-1} = 4I - A.$
Multiplying $(AA - 4A)$ by $A^{-1}$ on the right uses the distributive law, giving $AAA^{-1} - 4AA^{-1}$ ⁠. Associativity lets us group $A(AA^{-1})$ ⁠, and $AA^{-1} = I$ ⁠. We must multiply both sides on the same side (here the right), because matrix products do not commute in general.

Associative and distributive laws, $AI = A$ ⁠, Chapter 3, Section 3.4.6. Basics: transpose and identity
Compute $4I - A$
$4I - A = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}.$
Entry by entry: $4 - 2 = 2$ ⁠, $0 - 3 = -3$ ⁠, $0 - 1 = -1$ ⁠, $4 - 2 = 2$ ⁠. Hence
$A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}.$
$4I$ has 4 on the diagonal and 0 elsewhere, so subtracting $A$ changes the sign of the off-diagonal entries and replaces each diagonal entry $a$ by $4 - a$ ⁠.
Check the inverse

Row times column, $AA^{-1}$ ⁠:
$\begin{aligned} (1,1)&: \ 2\cdot2 + 3(-1) = 1,\\ (1,2)&: \ 2(-3) + 3\cdot2 = 0,\\ (2,1)&: \ 1\cdot2 + 2(-1) = 0,\\ (2,2)&: \ 1(-3) + 2\cdot2 = 1. \end{aligned}$
And $A^{-1}A$ ⁠:
$\begin{aligned} (1,1)&: \ 2\cdot2 + (-3)\cdot1 = 1,\\ (1,2)&: \ 2\cdot3 + (-3)\cdot2 = 0,\\ (2,1)&: \ (-1)\cdot2 + 2\cdot1 = 0,\\ (2,2)&: \ (-1)\cdot3 + 2\cdot2 = 1. \end{aligned}$
Both products equal $I$ ⁠. The adjoint formula agrees: $\operatorname{adj}A$ (swap the two diagonal 2s, change the signs of 3 and 1) is the same matrix, and $|A| = 1$ ⁠, so $\frac{1}{|A|}\operatorname{adj}A = A^{-1}$ as found.

An inverse must give $I$ on both sides. Comparing with the method of Section 4.5 is a second, independent check.
A shortcut, and where the equation comes from

The equation itself already shows that $4I - A$ is the inverse. Rearranged, $A^2 - 4A + I = O$ says $4A - A^2 = I$ ⁠, and
$\begin{aligned} A(4I - A) &= 4A - A^2 = I,\\ (4I - A)A &= 4A - A^2 = I. \end{aligned}$
So $4I - A$ multiplies $A$ to $I$ on both sides; that is the definition of $A^{-1}$ ⁠.

Why does such an equation exist? For any $2\times2$ matrix $M$ with top row $a, b$ and bottom row $c, d$ ⁠:
$M^2 = \begin{bmatrix} a^2 + bc & ab + bd \\ ca + dc & cb + d^2 \end{bmatrix},$ $(a + d)M = \begin{bmatrix} a^2 + ad & ab + bd \\ ac + cd & ad + d^2 \end{bmatrix}.$
Subtracting, the off-diagonal entries cancel and each diagonal entry is $bc - ad$ ⁠. So
$M^2 - (a + d)M + (ad - bc)I = O.$
For our $A$ ⁠: $a + d = 2 + 2 = 4$ and $ad - bc = 4 - 3 = 1$ ⁠, which is exactly $A^2 - 4A + I = O$ ⁠.

The shortcut avoids post-multiplying, and even proves that $A$ is invertible without computing $|A|$ ⁠. The general identity explains the numbers 4 and 1 in the question: they are the sum of the diagonal entries and the determinant. (This general fact goes beyond the textbook; it is included so the equation does not look like magic.)

Definition of the inverse, Chapter 3, Section 3.7: $AB = BA = I$ ⁠. Basics: algebra the chapter leans on

Result

$A^2$ has rows $7, 12$ and $4, 7$ ⁠, and $A^2 - 4A + I = O$ entry by entry. Post-multiplying by $A^{-1}$ (allowed since $|A| = 1 \ne 0$ ⁠) gives $A^{-1} = 4I - A$ ⁠, so

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

(This agrees with the textbook's own solution.)

Check it with numbers

Use $A^{-1}$ to solve $2x + 3y = 5$ ⁠, $x + 2y = 3$ (the coefficient matrix is $A$ ⁠). Then $x = 2\cdot5 + (-3)\cdot3 = 1$ and $y = (-1)\cdot5 + 2\cdot3 = 1$ ⁠. Substituting: $2 + 3 = 5$ and $1 + 2 = 3$ ⁠. The inverse undoes $A$ correctly.

Common mistakes

Squaring the entries. $A^2$ is $AA$ ⁠, with $(1,1)$ entry $2\cdot2 + 3\cdot1 = 7$ ⁠, not $2^2 = 4$ ⁠.
Writing $4$ instead of $4I$ ⁠. $A - 4$ has no meaning; a number must be turned into the matrix $4I$ before it is added to or subtracted from a matrix.
Multiplying on different sides. Multiplying one side on the right and the other on the left is not allowed for matrices.
Skipping $|A| \ne 0$ ⁠. Multiplying by $A^{-1}$ is legitimate only after checking that $A^{-1}$ exists (or by using the shortcut of Step 7).

What to take away

From a polynomial equation in $A$ with a non-zero constant term, multiply by $A^{-1}$ to get $A^{-1}$ as a combination of $I$ ⁠, $A$ ⁠, $A^2, \dots$
Here $A^2 - 4A + I = O$ gives $A^{-1} = 4I - A$ ⁠.
Every $2\times2$ matrix $M$ with diagonal entries $a$ and $d$ satisfies $M^2 - (a+d)M + |M|\,I = O$ (a fact beyond the textbook, proved in Step 7).

Every fact used, and where it comes from

Fact	Used in	Source
Matrix product, row times column	Steps 1, 6, 7	Chapter 3, Section 3.4.5 (Basics: matrix facts the chapter uses)
Sums and scalar multiples entry by entry	Steps 2, 5	Chapter 3, Sections 3.4.1, 3.4.2 (Basics: scaling and adding matrices)
$\|A\| \ne 0 \Rightarrow A^{-1}$ exists	Step 3	Theorem 4, Section 4.5 (Lesson: inverse of a matrix)
Associative, distributive laws; $AI = A$	Steps 4, 7	Chapter 3, Section 3.4.6
$A^{-1} = \frac{1}{\|A\|}\operatorname{adj}A$ ⁠; $2\times2$ adjoint	Step 6	Theorem 4, Section 4.5; Remark after Example 12 (Lesson: the adjoint)
Inverse: $AB = BA = I$	Steps 6, 7	Chapter 3, Section 3.7

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Example 15: A2−4A+I=OA^2-4A+I=OA2−4A+I=O⁠; use it to find A−1A^{-1}A−1

Basics you need

Solution

Compute A2=AAA^2 = AAA2=AA

Compute A2−4A+IA^2 - 4A + IA2−4A+I

Make sure A−1A^{-1}A−1 exists

Post-multiply the equation by A−1A^{-1}A−1

Compute 4I−A4I - A4I−A

Check the inverse

A shortcut, and where the equation comes from

Every fact used, and where it comes from

Example 15: $A^2-4A+I=O$ ⁠; use it to find $A^{-1}$

Compute $A^2 = AA$

Compute $A^2 - 4A + I$

Make sure $A^{-1}$ exists

Post-multiply the equation by $A^{-1}$

Compute $4I - A$