Inverse Matrices

The inverse of a matrix $A$A, denoted $A^{-1}$A−1, is the matrix such that $AA^{-1} = A^{-1}A = I$AA−1=A−1A=I. Not every matrix has an inverse — only non-singular matrices (those with non-zero determinant) are invertible.

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The 2×2 Inverse

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$A=(ac​bd​) with $\det(A) = ad - bc \neq 0$det(A)=ad−bc=0:

$$A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$A−1=ad−bc1​(d−c​−ba​)

The recipe: swap the diagonal, negate the off-diagonal, divide by the determinant.

Worked Example 1: Find the inverse of $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$A=(35​12​).

$\det(A) = 6 - 5 = 1$det(A)=6−5=1.

$$A^{-1} = \frac{1}{1}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$$A−1=11​(2−5​−13​)=(2−5​−13​)

Check: $AA^{-1} = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$AA−1=(35​12​)(2−5​−13​)=(10​01​) ✓

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The 3×3 Inverse

For a 3×3 matrix, the inverse is found using:

$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$A−1=det(A)1​adj(A)

where $\text{adj}(A)$adj(A) is the adjugate (or adjoint) matrix — the transpose of the matrix of cofactors.

Step-by-Step Method:

1. Find the determinant $\det(A)$det(A). If it is 0, stop — no inverse exists.
2. Find the matrix of minors: each entry is the determinant of the 2×2 submatrix obtained by deleting the corresponding row and column.
3. Apply the sign pattern (checkerboard of $+$+ and $-$−) to get the matrix of cofactors.
4. Transpose the cofactor matrix to get $\text{adj}(A)$adj(A).
5. Divide by $\det(A)$det(A).

Worked Example 2: Find the inverse of $A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{pmatrix}$A=​102​210​011​​.

Step 1: $\det(A) = 1(1-0) - 2(0-2) + 0 = 1 + 4 = 5$det(A)=1(1−0)−2(0−2)+0=1+4=5.

Step 2: Matrix of minors:

$$\begin{pmatrix} 1 & -2 & -2 \\ 2 & 1 & -4 \\ 2 & 1 & 1 \end{pmatrix}$$​122​−211​−2−41​​

Step 3: Apply signs:

$$\begin{pmatrix} 1 & 2 & -2 \\ -2 & 1 & 4 \\ 2 & -1 & 1 \end{pmatrix}$$​1−22​21−1​−241​​

Step 4: Transpose:

$$\text{adj}(A) = \begin{pmatrix} 1 & -2 & 2 \\ 2 & 1 & -1 \\ -2 & 4 & 1 \end{pmatrix}$$adj(A)=​12−2​−214​2−11​​