Matrix Inverse and Determinant — AP® Precalculus

Part 1 · Hook

Why this matters

A secret message is scrambled by multiplying blocks of numbers by a matrix. To read it, you need the matrix that undoes the scrambling. One number, the determinant, tells you whether such an undo matrix exists, and it also measures how much the matrix stretches areas.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\left[ \begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix} \right]\left[ \begin{matrix} 1 \\ -1 \end{matrix} \right]$ ?

$\left[ \begin{matrix} 1 \\ -2 \end{matrix} \right]$
$\left[ \begin{matrix} 2 \\ -3 \end{matrix} \right]$
$\left[ \begin{matrix} 3 \\ 4 \end{matrix} \right]$
$\left[ \begin{matrix} 1 & -2 \end{matrix} \right]$

Show the answer

Row 1: $2(1) + 1(-1) = 1$ . Row 2: $1(1) + 3(-1) = -2$ .

Correct: $\left[ \begin{matrix} 1 \\ -2 \end{matrix} \right]$ :
$\left[ \begin{matrix} 2 \\ -3 \end{matrix} \right]$ :
$\left[ \begin{matrix} 3 \\ 4 \end{matrix} \right]$ :
$\left[ \begin{matrix} 1 & -2 \end{matrix} \right]$ :

2. What is the area of a parallelogram with base $5$ and height $3$ ?

$15$
$7.5$
$8$
$16$

Show the answer

Base times height: $5 \cdot 3 = 15$ .

Correct: $15$ :
$7.5$ :
$8$ :
$16$ :

Part 4 · See it

See it first

The column vectors <3, 1> and <1, 2> drawn from the origin, and the shaded parallelogram they make, with corners (0, 0), (3, 1), (4, 3) and (1, 2). Its area is 5, the determinant 3 times 2 minus 1 times 1. — The columns of $\left[ \begin{matrix} 3 & 1 \\ 1 & 2 \end{matrix} \right]$ , drawn as $\left\langle 3, 1 \right\rangle$ and $\left\langle 1, 2 \right\rangle$ , make a parallelogram of area $3(2) - 1(1) = 5$ , the determinant. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

For $A = \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]$ , compute $ad - bc$ the determinant $\det A$
Draw the two columns of $A$ as vectors from the originthey make a parallelogram whose area is $\left| \det A \right|$
$\det A \ne 0$ $A$ has an inverse: swap $a$ and $d$ , negate $b$ and $c$ , divide by $\det A$
$\det A = 0$ the columns lie on one line, the parallelogram is flat, and $A$ has no inverse

Part 6 · Key ideas

Key ideas

The determinant of $\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]$ is $ad - bc$ . Its absolute value is the area of the parallelogram made by the two columns.
The inverse matrix $A^{-1}$ satisfies $AA^{-1} = A^{-1}A = I$ . For $\det A \ne 0$ : $A^{-1}$ $= \frac{1}{ad - bc}\left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right]$ .
$\det A = 0$ means no inverse: two different inputs give the same product, so nothing can undo $A$ .
If $AX = B$ and $A$ has an inverse, then $X = A^{-1}B$ .

Part 7 · Misconception

A common mistake

The wrong idea: The inverse of $\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]$ is $\left[ \begin{matrix} \frac{1}{a} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{d} \end{matrix} \right]$ .

What actually happens: Reciprocals of the entries do not undo matrix multiplication. Swap $a$ and $d$ , change the signs of $b$ and $c$ , and divide by $ad - bc$ ; then check that $AA^{-1} = I$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Two matrices

$A = \left[ \begin{matrix} 4 & 3 \\ 2 & 2 \end{matrix} \right], \qquad B = \left[ \begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix} \right]$

1. What is $\det A$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\det A$ $= 4(2) - 3(2)$ $= 8 - 6$ $= 2$ .

Answer: 2

2. Which matrix has no inverse?

$B$ , since $\det B = 0$
$A$ , since $\det A = 2$
Both, since neither is the identity
Neither, since no entry is zero

Show the answer

$\det B$ $= 6(2) - 3(4)$ $= 12 - 12$ $= 0$ , so $B$ has no inverse. $\det A = 2 \ne 0$ , so $A$ does.

Correct: $B$ , since $\det B = 0$ : Determinant $0$ : no inverse.
$A$ , since $\det A = 2$ : A nonzero determinant means $A$ is invertible.
Both, since neither is the identity: A matrix need not be $I$ to have an inverse; it needs a nonzero determinant.
Neither, since no entry is zero: Nonzero entries do not guarantee an inverse; $\det B = 0$ .

3. What is $A^{-1}$ ?

$\left[ \begin{matrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{matrix} \right]$
$\left[ \begin{matrix} 2 & -3 \\ -2 & 4 \end{matrix} \right]$
$\left[ \begin{matrix} \frac{1}{4} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{2} \end{matrix} \right]$
$\left[ \begin{matrix} 1 & \frac{3}{2} \\ 1 & 2 \end{matrix} \right]$

Show the answer

Swap the diagonal, negate the others, divide by $\det A = 2$ : $\frac{1}{2}\left[ \begin{matrix} 2 & -3 \\ -2 & 4 \end{matrix} \right]$ $= \left[ \begin{matrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{matrix} \right]$ .

Correct: $\left[ \begin{matrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{matrix} \right]$ : $\frac{1}{2}$ times the swapped-and-negated matrix.
$\left[ \begin{matrix} 2 & -3 \\ -2 & 4 \end{matrix} \right]$ : Divide by the determinant, $2$ .
$\left[ \begin{matrix} \frac{1}{4} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{2} \end{matrix} \right]$ : Reciprocals of the entries do not undo matrix multiplication.
$\left[ \begin{matrix} 1 & \frac{3}{2} \\ 1 & 2 \end{matrix} \right]$ : Change the signs of the off-diagonal entries.

4. Solve $A\left[ \begin{matrix} x \\ y \end{matrix} \right] = \left[ \begin{matrix} 7 \\ 4 \end{matrix} \right]$ .

$\left[ \begin{matrix} 1 \\ 1 \end{matrix} \right]$
$\left[ \begin{matrix} 7 \\ 4 \end{matrix} \right]$
$\left[ \begin{matrix} 2 \\ -\frac{1}{3} \end{matrix} \right]$
$\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right]$

Show the answer

$A^{-1}\left[ \begin{matrix} 7 \\ 4 \end{matrix} \right]$ $= \left[ \begin{matrix} 1(7) - \frac{3}{2}(4) \\ -1(7) + 2(4) \end{matrix} \right]$ $= \left[ \begin{matrix} 1 \\ 1 \end{matrix} \right]$ . Check: $4 + 3 = 7$ , $2 + 2 = 4$ .

Correct: $\left[ \begin{matrix} 1 \\ 1 \end{matrix} \right]$ : $x = 1$ , $y = 1$ .
$\left[ \begin{matrix} 7 \\ 4 \end{matrix} \right]$ : This is the right side; multiply it by $A^{-1}$ .
$\left[ \begin{matrix} 2 \\ -\frac{1}{3} \end{matrix} \right]$ : Use $A^{-1}$ ; dividing $7$ by entries of $A$ does not solve the system.
$\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right]$ : $4(-1) + 3(1) = -1$ , not $7$ .

5. Write $\det \left[ \begin{matrix} k & 3 \\ 2 & k \end{matrix} \right]$ as an expression in $k$ .

Type an expression.

Show the answer

$ad - bc$ $= k \cdot k - 3 \cdot 2$ $= k^2 - 6$ .

Answer: $k^{2} - 6$

6. What is the inverse of $\left[ \begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix} \right]$ ?

$\left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right]$
$\left[ \begin{matrix} 2 & -1 \\ -5 & 3 \end{matrix} \right]$
$\left[ \begin{matrix} -3 & 1 \\ 5 & -2 \end{matrix} \right]$
$\left[ \begin{matrix} 3 & 1 \\ 5 & 2 \end{matrix} \right]$

Show the answer

$\det = 6 - 5 = 1$ . Swap $2$ and $3$ , negate $1$ and $5$ , divide by $1$ .

Correct: $\left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right]$ : Swap the diagonal, negate the rest, divide by $1$ .
$\left[ \begin{matrix} 2 & -1 \\ -5 & 3 \end{matrix} \right]$ : Swap the diagonal entries $2$ and $3$ .
$\left[ \begin{matrix} -3 & 1 \\ 5 & -2 \end{matrix} \right]$ : Every sign is flipped: this is $-1$ times the inverse. Swap $2$ and $3$ but keep their signs.
$\left[ \begin{matrix} 3 & 1 \\ 5 & 2 \end{matrix} \right]$ : Change the signs of $1$ and $5$ .

Part 9 · Summary

Summary

For a

2 \times 2

matrix the determinant

ad - bc

is the signed area of the parallelogram made by its columns. If it is not

0

, the inverse is

\frac{1}{ad - bc}\left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right]

, which undoes the matrix and solves

AX = B

as

X = A^{-1}B

. If it is

0

, there is no inverse.

Part 10 · Up next

What comes next

Part 11 · Connections