Matrix Inverse and Determinant — AP® Precalculus Notes

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

For numbers, every nonzero $a$ has a reciprocal $\frac{1}{a}$ with $a \cdot \frac{1}{a} = 1$ . For matrices the role of $1$ is played by the identity $I$ , and the role of "nonzero" is played by one number computed from the entries: the determinant.

The determinant

For a $2 \times 2$ matrix,

\det \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] = ad - bc,

the product down the main diagonal minus the product up the other diagonal. For $A = \left[ \begin{matrix} 4 & 3 \\ 2 & 2 \end{matrix} \right]$ , $\det A$ $= 4(2) - 3(2)$ $= 8 - 6$ $= 2$ .

What the determinant measures

Draw the two columns of the matrix as vectors from the origin. They make a parallelogram, and its area is the absolute value of the determinant.

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. — **Figure 1.** Columns $\left\langle 3, 1 \right\rangle$ and $\left\langle 1, 2 \right\rangle$ : area $\left| 3(2) - 1(1) \right| = 5$ . LevlPrep original diagram.

The sign records order: swapping the two columns gives the same parallelogram but changes the sign of the determinant. Doubling one column doubles the area and the determinant; multiplying the whole $2 \times 2$ matrix by $3$ triples both columns, so the determinant is multiplied by $3 \cdot 3 = 9$ . When the determinant is $0$ , the columns lie along one line and the "parallelogram" is flat.

The inverse

The inverse of a square matrix $A$ is the matrix $A^{-1}$ with $AA^{-1} = A^{-1}A = I$ . For a $2 \times 2$ matrix with $\det A \ne 0$ :

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

In words: swap the diagonal entries, change the signs of the other two, divide every entry by the determinant.

Worked example 1. Find the inverse of $A = \left[ \begin{matrix} 4 & 3 \\ 2 & 2 \end{matrix} \right]$ and check it.

$\det A = 2$ . Swap and negate: $\left[ \begin{matrix} 2 & -3 \\ -2 & 4 \end{matrix} \right]$ . Divide by $2$ : $A^{-1} = \left[ \begin{matrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{matrix} \right]$ .

Check one entry of $AA^{-1}$ : row 1 of $A$ times column 2 of $A^{-1}$ is $4\left(-\frac{3}{2}\right) + 3(2)$ $= -6 + 6$ $= 0$ , as it should be. The full product is $I$ .

When there is no inverse

If $\det A = 0$ , the formula would divide by zero, and in fact no inverse exists. Take $B = \left[ \begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix} \right]$ , with $\det B = 12 - 12 = 0$ . Then $B\left[ \begin{matrix} 1 \\ -2 \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right]$ , and also $B\left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right]$ . Two different columns give the same result, so no matrix could send $\left[ \begin{matrix} 0 \\ 0 \end{matrix} \right]$ back to the right one.

Solving a system with the inverse

The system $4x + 3y = 7$ , $2x + 2y = 4$ is the matrix equation $AX = B$ with $X = \left[ \begin{matrix} x \\ y \end{matrix} \right]$ and $B = \left[ \begin{matrix} 7 \\ 4 \end{matrix} \right]$ .

Worked example 2. Solve it with $A^{-1}$ .

Multiply both sides on the left by $A^{-1}$ : $X$ $= A^{-1}B$ $= \left[ \begin{matrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{matrix} \right]\left[ \begin{matrix} 7 \\ 4 \end{matrix} \right]$ .

Row 1: $7 - 6 = 1$ . Row 2: $-7 + 8 = 1$ . So $x = 1$ , $y = 1$ . Check: $4 + 3 = 7$ and $2 + 2 = 4$ .

The order matters: $A^{-1}B$ , not $BA^{-1}$ , because $A$ multiplies $X$ from the left.

$\det A$	Columns	Inverse	$AX = B$
not $0$	a true parallelogram	exists	exactly one solution, $A^{-1}B$
$0$	on one line	does not exist	no solution or many