Q.$(AB)^{-1} = A^{-1} \cdot B^{-1}$, where $A$ and $B$ are invertible matrices satisfying commutative property with respect to multiplication.

Concept & Intuition

The adjugate (or classical adjoint) of a matrix is intimately tied to its inverse via the formula $A^{-1} = \frac{\operatorname{Adj}(A)}{\det A}$. This relationship is symmetric: if we replace $A$ by $A^{-1}$, we get $\operatorname{Adj}(A^{-1}) = (\det A^{-1}) A$. But since $\det A^{-1} = 1/\det A$, this simplifies to $\frac{1}{\det A} A$. That looks like option (B) at first glance — but wait: we must check whether this equals the inverse of the adjugate. Using the same fundamental formula on $\operatorname{Adj}(A)$ itself reveals the elegant identity $\operatorname{Adj}(A^{-1}) = (\operatorname{Adj} A)^{-1}$. Let’s verify step by step.

Step-by-step reasoning

1. Recall the fundamental adjugate-inverse relation For any invertible matrix $A$,

This is the definition of the adjugate: it’s the transpose of the cofactor matrix, and it satisfies $A \cdot \operatorname{Adj}(A) = \det A \cdot I$.

2. Apply the same relation to $A^{-1}$ Since $A^{-1}$ is also invertible, we have

But $(A^{-1})^{-1} = A$, so

Rearranging gives

3. Simplify $\det(A^{-1})$ For any invertible matrix, $\det(A^{-1}) = \frac{1}{\det A}$. Thus

This matches option (B) — but we must check if it’s also equal to the inverse of the adjugate.

4. Find $(\operatorname{Adj} A)^{-1}$ From the fundamental relation again, $\operatorname{Adj}(A) = (\det A) A^{-1}$. Taking the inverse of both sides: $(\operatorname{Adj} A)^{-1} = \left( (\det A) A^{-1} \right)^{-1} = \frac{1}{\det A} (A^{-1})^{-1} = \frac{1}{\det A} A.$ …

The key here is that the equation $(A-2I)(A-3I)=0$ is a polynomial relation satisfied by the matrix $A$. Since $A$ is non-singular (invertible), we can manipulate this relation algebraically, just like we would with numbers, but carefully respecting matrix multiplication (which is not commutative, though here all terms involve $A$ and $I$, which commute).

1. Expand the given equation. Multiply out:

So we have:

2. Rearrange to isolate $A^{-1}$. Since $A$ is invertible, multiply the whole equation by $A^{-1}$ on the left (or right — it commutes with $I$):

This gives:

Hence:

3. Substitute into the target expression. We need $\frac{1}{5}A + \frac{6}{5}A^{-1}$. Replace $A^{-1}$:

Simplify the second term: …

Equate real and imaginary parts of $(2x-y+1) + i(x-2y-1) = 2 - 3i$:

Solve: from the first, $y = 2x - 1$. Substitute into the second:

$x - 2(2x-1) = -2 \implies -3x + 2 = -2 \implies x = \tfrac{4}{3},\quad y = \tfrac{5}{3}$ …

We start with the matrix

We are given

and need $|A|$.

Concept & Intuition

The adjugate (classical adjoint) satisfies two key identities for an $n \times n$ matrix $M$:

• $\operatorname{Adj}(kM) = k^{n-1} \operatorname{Adj}(M)$ for a scalar $k$.
• $\operatorname{Adj}(\operatorname{Adj}(M)) = |M|^{n-2} M$ (for $n \ge 2$).

Here $n=3$, so $n-1=2$ and $n-2=1$. That means the second property simplifies to $\operatorname{Adj}(\operatorname{Adj}(M)) = |M| M$.

We also know $\operatorname{Adj}(M) = |M| M^{-1}$ when $M$ is invertible.

The plan: work from the inside out, applying these properties step by step, and take determinants at the end.

Step-by-step solution

1. Find $|P|$ Compute determinant of $P$:

So $|P| = -4$.

2. Find $\operatorname{Adj}(P^{-1})$ Since $P$ is invertible ($|P| \neq 0$), $P^{-1} = \frac{1}{|P|} \operatorname{Adj}(P)$. But we need $\operatorname{Adj}(P^{-1})$. Using $\operatorname{Adj}(M) = |M| M^{-1}$:

Because $|P^{-1}| = 1/|P|$ and $(P^{-1})^{-1} = P$.

So

3. Multiply by $-2$ inside the outer adjugate

   We have the matrix $M = -2\left(\operatorname{Adj}(P^{-1})\right) = -2 \cdot \left(-\frac{1}{4} P\right) = \frac{1}{2} P$.

   So $M = \frac{1}{2} P$.

4. Take the adjugate of $M$

   $A = \operatorname{Adj}(M) = \operatorname{Adj}\left(\frac{1}{2} P\right)$. …