Adjoint of a matrix

Every square matrix carries a hidden partner built entirely from its own cofactors: the adjoint. Take every entry of a matrix $A$ , replace it with its cofactor, and transpose the result — that single procedure produces $\operatorname{adj}(A)$ , a matrix that looks unrelated to $A$ at first glance but is secretly locked to it by one clean identity, $A\cdot\operatorname{adj}(A) = |A|\,I$ . This topic sits in the Determinants chapter of CBSE Class 12 Mathematics, right after minors and cofactors and immediately before the formula for the inverse of a matrix. The adjoint is not tested for its own sake — it is the bridge. Without it, there is no algebraic route from "I know the determinant of $A$ " to "I can write down $A^{-1}$ ." Everything below builds toward using that identity confidently, on both $2\times2$ and $3\times3$ matrices, with clean bookkeeping of signs.

The core insight

The adjoint of a square matrix is the transpose of its cofactor matrix, and it satisfies $A\cdot\operatorname{adj}(A) = |A|\,I$ , which is exactly the identity that makes the matrix inverse formula work.

The core idea

Step 1 — form the cofactor matrix. For a square matrix $A = [a_{ij}]$ , compute the cofactor $C_{ij} = (-1)^{i+j}M_{ij}$ of every entry, where $M_{ij}$ is the minor obtained by deleting row $i$ and column $j$ . Arrange these cofactors in the same grid positions as the original entries to get the cofactor matrix $[C_{ij}]$ .

Step 2 — transpose it. The adjoint is defined as the transpose of that cofactor matrix:

\operatorname{adj}(A) = [C_{ij}]^{T}

where: $C_{ij}$ is the cofactor of the $(i,j)$ entry of $A$ , and the superscript $T$ means rows and columns are swapped. The transpose step is not optional decoration — skipping it is the single most common mistake on this topic, and it changes the answer whenever the cofactor matrix is not symmetric.

A fast shortcut for $2\times2$ . If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the general definition collapses to a one-line rule: swap the main-diagonal entries and negate the off-diagonal entries.

\operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

The identity that matters most. For any square matrix $A$ of order $n$ with identity matrix $I$ of the same order,

A \cdot \operatorname{adj}(A) = \operatorname{adj}(A) \cdot A = |A|\, I

where: $|A|$ is the determinant of $A$ (a scalar) and $I$ is the $n\times n$ identity matrix. Read this as a promise: multiplying $A$ by its own adjoint, in either order, always collapses to a plain multiple of the identity matrix — never anything messier. That collapse is precisely why, dividing both sides by $|A|$ when $|A| \neq 0$ , the adjoint lets you isolate a genuine two-sided inverse for $A$ in the very next topic. The figure below tracks the whole pipeline for a $2\times2$ matrix, from raw entries to the final identity check.

Figure

Pipeline from a matrix through its cofactors and transpose to the adjoint identity

Derivation AThe identity A adj(A) equals determinant A times I, for a 2 by 2 matrix

Showing $A\cdot\operatorname{adj}(A) = |A|\,I$ holds for every $2\times2$ matrix — not just a lucky example — by multiplying out $A$ and its adjoint symbolically and confirming both off-diagonal entries vanish while both diagonal entries reduce to the same determinant, so the product is forced into the exact shape $|A|\,I$ with no extra assumptions.

Step 1. Write $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and its adjoint using the shortcut, $\operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .

Step 2. Multiply $A \cdot \operatorname{adj}(A)$ entry by entry. The $(1,1)$ entry is $a\cdot d + b\cdot(-c) = ad - bc$ . The $(1,2)$ entry is $a\cdot(-b) + b\cdot a = -ab + ab = 0$ .

Step 3. The $(2,1)$ entry is $c\cdot d + d\cdot(-c) = cd - cd = 0$ , and the $(2,2)$ entry is $c\cdot(-b) + d\cdot a = ad - bc$ . So

A\cdot\operatorname{adj}(A) = \begin{bmatrix} ad-bc & 0 \\ 0 & ad-bc \end{bmatrix}

Step 4. Recognise $ad - bc$ as exactly $|A|$ , the $2\times2$ determinant. So $A\cdot\operatorname{adj}(A) = (ad-bc)\begin{bmatrix}1&0\\0&1\end{bmatrix} = |A|\,I$ — every off-diagonal term cancelled by construction, and both surviving diagonal terms are forced to equal $|A|$ . The same cancellation pattern, only with more terms, is what makes the identity true for $3\times3$ and larger matrices as well; the syllabus asks you to trust and use that general fact, not re-derive it each time.

Worked example

Find $\operatorname{adj}(A)$ for $A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$ , and verify $A \cdot \operatorname{adj}(A) = |A|\,I$ .

Step 1. Apply the $2\times2$ shortcut: swap the diagonal entries ( $3$ and $2$ ), negate the off-diagonal entries ( $1$ and $5$ ).

\operatorname{adj}(A) = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}

Step 2. Compute the determinant: $|A| = (3)(2) - (1)(5) = 6 - 5 = 1$ .

Step 3. Multiply $A \cdot \operatorname{adj}(A)$ directly.

\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} 3(2)+1(-5) & 3(-1)+1(3) \\ 5(2)+2(-5) & 5(-1)+2(3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Step 4. Compare with $|A|\,I = 1\cdot\begin{bmatrix}1&0\\0&1\end{bmatrix}$ . The two matrices match, so $\operatorname{adj}(A) = \begin{bmatrix}2&-1\\-5&3\end{bmatrix}$ and the identity $A\cdot\operatorname{adj}(A) = |A|\,I$ is verified.

Competitive stretch

For $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix}$ , the nine cofactors work out to $C_{11}=1,\ C_{12}=0,\ C_{13}=-3,\ C_{21}=0,\ C_{22}=-5,\ C_{23}=0,\ C_{31}=-2,\ C_{32}=0,\ C_{33}=1$ . Arranging these gives the cofactor matrix $\begin{bmatrix}1&0&-3\\0&-5&0\\-2&0&1\end{bmatrix}$ , and transposing swaps the $(1,3)$ and $(3,1)$ positions to give $\operatorname{adj}(A) = \begin{bmatrix}1&0&-2\\0&-5&0\\-3&0&1\end{bmatrix}$ — a useful reminder that the transpose step does real work even when most entries look symmetric already.

Common traps

Forgetting the transpose step entirely — the adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself; skipping it is wrong whenever the cofactor matrix is not symmetric.
Sign errors in the cofactors — always apply the checkerboard pattern $(-1)^{i+j}$ carefully rather than guessing signs from memory.
For the $2\times2$ shortcut, swapping the wrong pair — the main diagonal entries ( $a, d$ ) get swapped, while the off-diagonal entries ( $b, c$ ) get negated, not swapped.
Applying the adjoint formula to a non-square matrix — like the determinant, the adjoint exists only for square matrices.

Ask the tutor: Try a $3\times3$ matrix of your own and ask the AI tutor to check your cofactor signs and the final transpose step entry by entry.

Why do we transpose the cofactor matrix instead of just using it directly?

Because only the transposed arrangement makes the identity $A\cdot\operatorname{adj}(A) = |A|\,I$ come out correctly. The multiplication that produces the diagonal entries $|A|$ requires each row of $A$ to line up with the matching column of cofactors, and that alignment only happens after transposing. Using the raw cofactor matrix without transposing generally gives a matrix that is not a plain multiple of the identity, so the transpose is a required step, not a stylistic choice.

Do I always need to compute all nine cofactors for a three by three matrix, or is there a shortcut like the two by two case?

There is no shortcut for $3\times3$ and larger — you must compute every cofactor $C_{ij}$ using the minor and the $(-1)^{i+j}$ sign, then transpose. The $2\times2$ shortcut (swap diagonal, negate off-diagonal) is a special case that only works because a $2\times2$ cofactor matrix always has this exact swap-and-negate structure; it does not generalise to bigger matrices, so practise the full cofactor-then-transpose method for $3\times3$ examples.

How is the adjoint actually used in the next topic, finding the inverse?

Rearranging $A\cdot\operatorname{adj}(A) = |A|\,I$ and dividing both sides by the scalar $|A|$ (valid only when $|A| \neq 0$ ) gives $A\cdot\left(\dfrac{1}{|A|}\operatorname{adj}(A)\right) = I$ , which is exactly the defining property of an inverse. So $A^{-1} = \dfrac{1}{|A|}\operatorname{adj}(A)$ . Everything you practise here — cofactors, the transpose, and the identity — is assembled directly into that formula in the very next topic, so accuracy on the adjoint carries straight through.

Adjoint of a matrix

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads