Mathfolis

The Inverse and Determinant of a Matrix

Unit 4 · Functions Involving Parameters, Vectors, and Matrices

What AP Precalc asks here

The determinant of a 2×2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad−bcad - bc. A matrix is invertible if and only if its determinant is nonzero. The inverse formula for an invertible 2×2 is 1det⁡A(d−b−ca)\frac{1}{\det A} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} — swap the diagonal entries and negate the off-diagonal ones, then divide by the determinant.

2×2 determinant and inverse

Determinant
det⁡(abcd)=ad−bc\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc
Inverse (if det ≠ 0)
(abcd)−1=1ad−bc(d−b−ca)\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
Caution: If det = 0, the matrix is singular and has no inverse. Don't try to compute A−1A^{-1} in that case.
Type 1

Compute the determinant

Apply ad − bc.

Example 1
Compute det⁡(2314)\det\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}. (A) 5 (B) 8 (C) 11 (D) 24

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Type 2

Compute the inverse

Swap diagonal entries, negate off-diagonal, divide by determinant.

Example 2
Find the inverse of A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}. (A) (4−2−31)\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} (B) (−211.5−0.5)\begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} (C) (1−2−34)\begin{pmatrix} 1 & -2 \\ -3 & 4 \end{pmatrix} (D) Not invertible

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Type 3

Decide invertibility from the determinant

Compute det; nonzero ⇔ invertible.

Example 3
Is A=(2412)A = \begin{pmatrix} 2 & 4 \\ 1 & 2 \end{pmatrix} invertible? (A) Yes, det = 4 (B) Yes, det = -8 (C) No, det = 0 (D) Cannot determine

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The Inverse and Determinant of a Matrix | AP Precalculus — Mathfolis