Prove the fundamental adjoint-determinant identity for square matrices.

Prove that A(adj A) = |A|·I — Matrix Identity

Question

Prove that for any square matrix $A$ of order $n$ :

A \cdot \text{adj}(A) = |A| \cdot I

where $\text{adj}(A)$ is the adjoint (classical adjoint) of $A$ and $I$ is the identity matrix of order $n$ .

Solution — Step by Step

Let $A = [a_{ij}]$ be an $n \times n$ matrix. The adjoint $\text{adj}(A)$ is the transpose of the cofactor matrix, so $[\text{adj}(A)]_{ij} = C_{ji}$ , where $C_{ji}$ is the cofactor of the element $a_{ji}$ in $A$ .

The $(i, k)$ entry of the product $A \cdot \text{adj}(A)$ is:

[A \cdot \text{adj}(A)]_{ik} = \sum_{j=1}^{n} a_{ij} \cdot [\text{adj}(A)]_{jk} = \sum_{j=1}^{n} a_{ij} \cdot C_{kj}

When $i = k$ , we get $\sum_{j=1}^{n} a_{ij} C_{ij}$ — this is precisely the cofactor expansion of $|A|$ along row $i$ .

[A \cdot \text{adj}(A)]_{ii} = \sum_{j=1}^{n} a_{ij} C_{ij} = |A|

Every diagonal entry equals $|A|$ . This is just the definition of the determinant via row expansion.

When $i \neq k$ , we’re computing $\sum_{j=1}^{n} a_{ij} C_{kj}$ . This is the expansion of a determinant where row $k$ has been replaced by row $i$ — giving a matrix with two identical rows.

A matrix with two identical rows has determinant zero. So:

[A \cdot \text{adj}(A)]_{ik} = 0 \quad \text{for } i \neq k

Combining steps 2 and 3:

[A \cdot \text{adj}(A)]_{ik} = \begin{cases} |A| & \text{if } i = k \\ 0 & \text{if } i \neq k \end{cases}

This is exactly the matrix $|A| \cdot I$ . Therefore:

\boxed{A \cdot \text{adj}(A) = |A| \cdot I}

Why This Works

The magic is in step 3 — the “alien cofactor” trick. When we sum $a_{ij} C_{kj}$ with $i \neq k$ , we are asking: what would the determinant be if row $k$ were replaced by row $i$ ? Since row $i$ already exists in the matrix, we’d have a repeat row. And any matrix with two identical rows has $|A| = 0$ — you can see this because swapping those two rows changes the sign of the determinant but leaves the matrix unchanged.

This identity is not just a proof exercise. When $|A| \neq 0$ , dividing both sides by $|A|$ gives $A \cdot \frac{\text{adj}(A)}{|A|} = I$ , which means $A^{-1} = \frac{\text{adj}(A)}{|A|}$ . So the entire theory of matrix inverses via cofactors rests on this one identity.

For $|A| = 0$ (singular matrix), the identity still holds — it just tells us $A \cdot \text{adj}(A) = 0$ , which is consistent since $A$ has no inverse.

Alternative Method

For a $2 \times 2$ matrix, we can verify directly — and the pattern makes the general proof more intuitive.

Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ . Then $\text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .

Compute the product:

A \cdot \text{adj}(A) = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix} = (ad-bc)I = |A| \cdot I

This direct computation for $2 \times 2$ is worth memorising — it’s cleaner in MCQ-style questions where the general proof isn’t needed.

Common Mistake

Most students prove $A \cdot \text{adj}(A) = |A| \cdot I$ and forget to also state (or prove) that $\text{adj}(A) \cdot A = |A| \cdot I$ . JEE Advanced and CBSE both sometimes ask for the full identity: $A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| \cdot I$ . The proof for $\text{adj}(A) \cdot A$ is identical in structure — just use cofactor expansion along columns instead of rows. Don’t lose marks by proving only one side.

In JEE Main MCQs, this identity often appears as: “If $A$ is a $3 \times 3$ matrix with $|A| = 5$ , find $|A \cdot \text{adj}(A)|$ .” Direct answer: $|A \cdot \text{adj}(A)| = ||A| \cdot I| = |5I| = 5^3 = 125$ . No need to compute $\text{adj}(A)$ explicitly.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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