Q.Let $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that

Concept: Adjoint Matrix Property — For any invertible matrix $A$, the adjoint satisfies $(\text{adj } A)^{-1} = \text{adj}(A^{-1})$.

Step 1: Compute $\det A$

Using expansion along first row:

$\det A = 1(3\cdot5 - 1\cdot1) - 2(2\cdot5 - 1\cdot1) + 1(2\cdot1 - 3\cdot1)$

$= 1(15 - 1) - 2(10 - 1) + 1(2 - 3) = 14 - 18 - 1 = -5 \neq 0$, so $A$ is invertible.

Step 2: Verify $(A^{-1})^{-1} = A$

By definition, $A^{-1}A = I$. Taking inverse of both sides gives $(A^{-1})^{-1} = A$. This holds for any invertible matrix — no computation needed.

Step 3: Verify $(\text{adj } A)^{-1} = \text{adj}(A^{-1})$

We use the identity $A^{-1} = \frac{\text{adj } A}{\det A}$. Then $\text{adj } A = (\det A) A^{-1}$. …

Setup. Expanding along the first row,

The cofactor matrix is symmetric here, so

Part (i): $(\text{adj }A)^{-1} = \text{adj}(A^{-1})$.

Use the standard identity $\text{adj}(A) = \det(A)\,A^{-1}$, valid for any invertible $A$.

• Left side: $(\text{adj }A)^{-1} = \big(\det(A)\,A^{-1}\big)^{-1} = \dfrac{1}{\det(A)}\,A.$
• Right side: $\text{adj}(A^{-1}) = \det(A^{-1})\,(A^{-1})^{-1} = \dfrac{1}{\det(A)}\,A.$ …

Method: Verifying a Matrix-Inverse Identity Using $\operatorname{adj}(A) = \det(A)\,A^{-1}$

When asked to "verify" a general identity like $(\operatorname{adj}A)^{-1} = \operatorname{adj}(A^{-1})$ for a specific matrix, the fast, reliable route is to substitute the standard adjoint-inverse relationship algebraically, rather than computing $A^{-1}$, $\operatorname{adj}(A^{-1})$, and $(\operatorname{adj}A)^{-1}$ all separately from scratch.

Steps

Step 1: Compute $\det(A)$ once

This single number is all you need to connect every quantity in the identity — compute it carefully via cofactor expansion, since every later step depends on it.

Step 2: Recall the identity $\operatorname{adj}(A) = \det(A)\,A^{-1}$

This follows directly from $A^{-1} = \dfrac{1}{\det A}\operatorname{adj}(A)$ rearranged — it holds for any invertible matrix, not just this specific $A$.

Step 3: Rewrite BOTH sides of the identity to be verified in terms of $A$, $A^{-1}$, and $\det(A)$ …

Common Mistakes

Mistake 1: Computing $A^{-1}$, $\operatorname{adj}(A^{-1})$, and $(\operatorname{adj}A)^{-1}$ separately as three full numeric computations

Why it's wrong: this brute-force route needs several full $3\times3$ adjoint/inverse computations, each carrying its own risk of a sign or transpose slip, when the identity $\operatorname{adj}(A)=\det(A)A^{-1}$ proves the result algebraically in a few lines. Correct approach: substitute the standard identity and simplify symbolically; only fall back to full numeric computation if asked to "verify by direct calculation" explicitly.

Mistake 2: Treating $\det(A^{-1})$ as $\det(A)$ instead of $1/\det(A)$

Why it's wrong: this is a distinct, well-known fact ($\det(A^{-1}) = 1/\det(A)$) that's easy to misremember as just $\det(A)$, and using the wrong value derails the algebraic verification of $\operatorname{adj}(A^{-1})$. Correct approach: explicitly write $\det(A^{-1}) = 1/\det(A)$ before substituting it into any adjoint formula. …