Adjoint and inverse

In the previous subtopic we built the cofactor matrix of a square $A$. Its transpose is called the adjoint of $A$ (also adjugate; in older Indian textbooks called the classical adjoint). The adjoint satisfies $A \cdot \text{adj}\,A = (\det A)\,I$, which gives an explicit formula for the inverse:

$A^{-1} = \frac{1}{\det A}\,\text{adj}\,A, \quad \text{provided } \det A \neq 0.$

This is the adjoint method for inverting a matrix.

Definition

The adjoint of a square matrix $A$ is the transpose of its cofactor matrix:

$\text{adj}(A) = \text{cof}(A)^T.$

So $(\text{adj}\,A)_{ij} = C_{ji}$, where $C_{ji}$ is the $(j, i)$ cofactor of $A$.

The main identity

Theorem. For any square $A$,

$A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = (\det A) I.$

Proof sketch. The $(i, k)$ entry of $A \cdot \text{adj}(A)$ is $\sum_j a_{ij} (\text{adj} A)_{jk} = \sum_j a_{ij} C_{kj}$. By cofactor orthogonality this equals $\det A$ if $i = k$, and $0$ otherwise. So the product is $(\det A) I$.

The inverse formula

Dividing both sides by $\det A$ (when nonzero):

$A^{-1} = \frac{1}{\det A}\,\text{adj}(A).$

This is the adjoint formula for the inverse. It is explicit, clean, and central to proofs.

Worked computation

Find the inverse of $A = \begin{pmatrix} 2 & 3 & 1 \\ 1 & 2 & 2 \\ 3 & 1 & -1 \end{pmatrix}$.

Step 1. Compute $\det A$ by expansion along row 1.

$\det A = 2(2 \cdot -1 - 2 \cdot 1) - 3(1 \cdot -1 - 2 \cdot 3) + 1(1 \cdot 1 - 2 \cdot 3)$ $= 2(-4) - 3(-7) + 1(-5) = -8 + 21 - 5 = 8$.

So $\det A = 8 \neq 0$. The matrix is invertible.

Step 2. Compute the cofactors.

$C_{11} = +\det\begin{pmatrix} 2 & 2 \\ 1 & -1 \end{pmatrix} = -4$.

$C_{12} = -\det\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix} = -(-1 - 6) = 7$.

$C_{13} = +\det\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} = -5$.

$C_{21} = -\det\begin{pmatrix} 3 & 1 \\ 1 & -1 \end{pmatrix} = -(-3 - 1) = 4$.

$C_{22} = +\det\begin{pmatrix} 2 & 1 \\ 3 & -1 \end{pmatrix} = -5$.

$C_{23} = -\det\begin{pmatrix} 2 & 3 \\ 3 & 1 \end{pmatrix} = -(2 - 9) = 7$.

$C_{31} = +\det\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix} = 4$.

$C_{32} = -\det\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} = -(4 - 1) = -3$.

$C_{33} = +\det\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} = 1$.

Step 3. Assemble the cofactor matrix and transpose:

$\text{cof}(A) = \begin{pmatrix} -4 & 7 & -5 \\ 4 & -5 & 7 \\ 4 & -3 & 1 \end{pmatrix}$.

$\text{adj}(A) = \begin{pmatrix} -4 & 4 & 4 \\ 7 & -5 & -3 \\ -5 & 7 & 1 \end{pmatrix}$.

Step 4. $A^{-1} = \tfrac{1}{8}\,\text{adj}(A) = \tfrac{1}{8}\begin{pmatrix} -4 & 4 & 4 \\ 7 & -5 & -3 \\ -5 & 7 & 1 \end{pmatrix}$.

Useful identities about the adjoint

For an $n \times n$ matrix $A$:

1. $A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = (\det A) I$.
2. $\text{adj}(A^T) = (\text{adj}\,A)^T$.
3. $\det(\text{adj}\,A) = (\det A)^{n - 1}$.
4. $\text{adj}(AB) = \text{adj}(B) \cdot \text{adj}(A)$.
5. $\text{adj}(\lambda A) = \lambda^{n - 1}\,\text{adj}(A)$.

The reversal in (4) parallels $(AB)^{-1} = B^{-1}A^{-1}$ and is often tested in JEE.

Worked examples

Example 1. Find the adjoint and inverse of $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.

$\det A = -2$. $\text{cof}(A) = \begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix}$. $\text{adj}(A) = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$. $A^{-1} = -\tfrac{1}{2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}$.

Example 2. Verify $A \cdot \text{adj}(A) = (\det A) I$ for the matrix in Example 1.

$A \cdot \text{adj}(A) = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 4 - 6 & -2 + 2 \\ 12 - 12 & -6 + 4 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} = -2 I$. $\checkmark$

Example 3. If $\det A = 5$ for a $3 \times 3$ matrix, find $\det(\text{adj}\,A)$.

By property (3): $\det(\text{adj}\,A) = 5^{3 - 1} = 25$.

Example 4. If $A$ is a $3 \times 3$ matrix with $\det A = 4$, find $|\text{adj}(2 A)|$.

$\det(2 A) = 2^3 \det A = 32$. $|\text{adj}(2A)| = 32^{3 - 1} = 1024$.

Example 5. Show that for a $2 \times 2$ matrix, $\text{adj}(\text{adj}(A)) = A$.

$\text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. Compute its adjoint: swap diagonal and negate off-diagonal: $\begin{pmatrix} a & b \\ c & d \end{pmatrix} = A$. $\checkmark$ (For higher orders, $\text{adj}(\text{adj}(A)) = (\det A)^{n-2} A$.)

Example 6. Solve $\begin{cases} 2x + 3y + z = 11 \\ x + 2y + 2z = 13 \\ 3x + y - z = 5 \end{cases}$ using the adjoint method.

Write $AX = B$. $A = \begin{pmatrix} 2 & 3 & 1 \\ 1 & 2 & 2 \\ 3 & 1 & -1 \end{pmatrix}$, $B = \begin{pmatrix} 11 \\ 13 \\ 5 \end{pmatrix}$.

From the worked computation above, $A^{-1} = \tfrac{1}{8} \begin{pmatrix} -4 & 4 & 4 \\ 7 & -5 & -3 \\ -5 & 7 & 1 \end{pmatrix}$.

$X = A^{-1} B = \tfrac{1}{8}\begin{pmatrix} -44 + 52 + 20 \\ 77 - 65 - 15 \\ -55 + 91 + 5 \end{pmatrix} = \tfrac{1}{8}\begin{pmatrix} 28 \\ -3 \\ 41 \end{pmatrix}$.

Hmm , let me double-check by substituting into the first equation: $2 \cdot 28/8 + 3 \cdot (-3)/8 + 41/8 = (56 - 9 + 41)/8 = 88/8 = 11$. $\checkmark$. So $X = (7/2, -3/8, 41/8)$. (This system has a non-integer solution; you can re-solve with different RHS for a cleaner answer.)

Try it yourself

1. Find the adjoint of $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.
2. Find the adjoint of $I_3$.
3. Find the inverse of $\begin{pmatrix} 2 & 5 \\ -3 & 1 \end{pmatrix}$ using the adjoint method.
4. Find the inverse of $\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$.
5. Verify $A \cdot \text{adj}(A) = (\det A) I$ for $\begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix}$.
6. For a $3 \times 3$ matrix with $\det = 7$, find $\det(\text{adj}\,A)$.
7. Find $\text{adj}(\text{adj}(I_3))$.
8. Verify $\text{adj}(AB) = \text{adj}(B)\,\text{adj}(A)$ for some $2 \times 2$ matrices.
9. Find $A^{-1}$ for $A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ , note it should be the rotation by $-\theta$.
10. Show that if $A$ is symmetric and invertible, then $\text{adj}(A)$ is symmetric.
11. If $A$ is skew-symmetric of even order, must $\text{adj}(A)$ be invertible?
12. Show that for any invertible $A$, $(A^{-1})^T = (A^T)^{-1}$.
13. Use $\text{adj}$ to invert $\begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$.
14. Show that $\text{adj}(\lambda A) = \lambda^{n - 1} \text{adj}(A)$ for an $n \times n$ matrix.

Pitfalls / Tricks

• Adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself.
• Do not forget the $\tfrac{1}{\det A}$ factor when computing the inverse.
• $\det(\text{adj}\,A) = (\det A)^{n - 1}$ is a JEE staple; memorise.
• The adjoint method scales poorly with size; for $n \ge 4$ row reduction is faster.
• For a $2 \times 2$ matrix, adjoint = swap diagonal entries and negate off-diagonals.

Next we apply these tools to solve linear systems via Cramer's rule.