The Adjoint

Chapter 3 defined $A^{-1}$ by the property $AB = BA = I$ but gave no way to compute it. The missing tool is the adjoint.

Definition. The adjoint of a square matrix $A = [a_{ij}]$ is the transpose of its cofactor matrix:

$$\operatorname{adj} A = \begin{pmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{pmatrix}$$

— compute every cofactor, then flip across the diagonal. Forgetting the transpose is the signature error of this section.

The $2 \times 2$ shortcut. For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$: swap the diagonal entries, negate the off-diagonal ones:

$$\operatorname{adj} A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$

The master identity

Theorem 1. For any square matrix $A$ of order $n$:

$$\boxed{\;A(\operatorname{adj} A) = (\operatorname{adj} A)A = \vert A \vert I\;}$$

Why it works: multiply row $i$ of $A$ into column $i$ of $\operatorname{adj} A$ — that pairs each row with its own cofactors, giving $\vert A \vert$ on the diagonal. Every off-diagonal entry pairs a row with someone else's cofactors — the alien-cofactor rule from Section 3 makes them all zero.

Singular and nonsingular

A square matrix is singular if $\vert A \vert = 0$ and nonsingular if $\vert A \vert \neq 0$. For example $\begin{pmatrix} 1 & 2 \\ 4 & 8 \end{pmatrix}$ is singular ($8 - 8 = 0$), while $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ is nonsingular ($\vert A \vert = -2$).

Theorem 2. If $A$ and $B$ are nonsingular matrices of the same order, so are $AB$ and $BA$.

Theorem 3 (product rule). $\vert AB \vert = \vert A \vert \, \vert B \vert$ — the determinant of a product is the product of the determinants.

Taking determinants across $A(\operatorname{adj} A) = \vert A \vert I$ gives a JEE favourite as a free corollary:

$$\vert \operatorname{adj} A \vert = \vert A \vert^{\,n-1} \qquad (A \text{ nonsingular of order } n)$$

— so for order $3$, $\vert \operatorname{adj} A \vert = \vert A \vert^2$.

The inverse formula

If $\vert A \vert \neq 0$, divide the master identity by $\vert A \vert$:

$$\boxed{\;A^{-1} = \frac{1}{\vert A \vert}\operatorname{adj} A\;}$$

A matrix is invertible exactly when it is nonsingular. Two consequences worth carrying to the exam: $(AB)^{-1} = B^{-1}A^{-1}$ (reversal, as with transposes), and taking determinants of $AA^{-1} = I$ gives $\det(A^{-1}) = \dfrac{1}{\det A}$.

Solved Examples

Example 1 — Adjoint of a $2 \times 2$

Find $\operatorname{adj} A$ for $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$.

Step 1 — cofactors: $A_{11} = 4$, $A_{12} = -1$, $A_{21} = -3$, $A_{22} = 2$.

Step 2 — transpose the cofactor matrix: $$\operatorname{adj} A = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$$

Answer: as above — exactly the shortcut: diagonal $2, 4$ swapped, off-diagonal $3, 1$ negated in place.

Example 2 — Verify the master identity, then invert

For $A = \begin{pmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{pmatrix}$, verify that $A(\operatorname{adj} A) = \vert A \vert I$ and find $A^{-1}$.

Step 1 — determinant: expanding along $R_1$: $1(16-9) - 3(4-3) + 3(3-4) = 7 - 3 - 3 = 1$.

Step 2 — cofactor matrix, then transpose: computing all nine cofactors and transposing: $$\operatorname{adj} A = \begin{pmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix}$$

Step 3 — multiply: $A(\operatorname{adj} A) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = 1 \cdot I = \vert A \vert I$. ✓

Step 4 — inverse: since $\vert A \vert = 1$, $A^{-1} = \frac{1}{1}\operatorname{adj} A = \operatorname{adj} A$.

Answer: $A^{-1} = \begin{pmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix}$.

Example 3 — Inverse of a $2 \times 2$ by the formula

Find $A^{-1}$ for $A = \begin{pmatrix} 2 & -2 \\ 4 & 3 \end{pmatrix}$, if it exists.

Step 1 — determinant: $\vert A \vert = 6 + 8 = 14 \neq 0$, so the inverse exists.

Step 2 — adjoint by the shortcut: $\operatorname{adj} A = \begin{pmatrix} 3 & 2 \\ -4 & 2 \end{pmatrix}$.

Step 3 — divide: $$A^{-1} = \frac{1}{14}\begin{pmatrix} 3 & 2 \\ -4 & 2 \end{pmatrix}$$

Answer: as above. Quick check: $A \cdot A^{-1}$ has $(1,1)$ entry $\frac{1}{14}(6 + 8) = 1$. ✓

Example 4 — Inverse from a matrix equation

Show that $A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$ satisfies $A^2 - 4A + I = O$, and use this to find $A^{-1}$.

Step 1 — verify: $A^2 = \begin{pmatrix} 7 & 12 \\ 4 & 7 \end{pmatrix}$, so $A^2 - 4A + I = \begin{pmatrix} 7-8+1 & 12-12 \\ 4-4 & 7-8+1 \end{pmatrix} = O$. ✓

Step 2 — rearrange for the inverse: $A^2 - 4A + I = O$ gives $I = 4A - A^2 = A(4I - A)$, so $A^{-1} = 4I - A$.

Step 3 — compute: $$A^{-1} = \begin{pmatrix} 4-2 & -3 \\ -1 & 4-2 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$$

Answer: $A^{-1} = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$ — no adjoint needed when a polynomial identity is on offer; the same trick at order $3$ saves enormous time.