Question

Let \(a \in \mathbb{R}\) and \(A\) be a matrix of order \(3 \times 3\) such that \(\det(A) = -4\) and 

\[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}. \]

If \[ \det\!\left((a+1)\,\operatorname{adj}\!\big((a-1)A\big)\right) \] is \(2^m3^n\), then \(m+n\) equals:

• A. \(14\)
• B. \(07\)
• C. \(18\)
• D. \(20\)

Hint:

Always separate scalar multiplication and adjoint properties before substituting values.

Answer 1

The Correct Option is B

Concept: We use properties: \[ \det(kA)=k^n\det(A)\quad (n=3) \] \[ \det(\operatorname{adj}(A))=\det(A)^{n-1} \]

Step 1: Find \(\det((a-1)A)\). \[ \det((a-1)A)=(a-1)^3\det(A) \] \[ =(a-1)^3(-4) \]

Step 2: Use adjoint determinant property. For \(3\times 3\): \[ \det(\operatorname{adj}(B))=\det(B)^2 \] So: \[ \det(\operatorname{adj}((a-1)A))=\left[(a-1)^3(-4)\right]^2 \] \[ =(a-1)^6 \cdot 16 \]

Step 3: Include scalar \((a+1)\). \[ \det((a+1)\operatorname{adj}((a-1)A)) =(a+1)^3 \cdot (a-1)^6 \cdot 16 \] \[ =2^4 (a+1)^3 (a-1)^6 \]

Step 4: Use given matrix condition to find \(a\). From: \[ A+I= \begin{bmatrix} 1 & a & 1\\ 2 & 1 & 0\\ a & 1 & 2 \end{bmatrix} \] Taking determinant gives: \[ a=2 \]

Step 5: Substitute value. \[ (a+1)^3=3^3,\quad (a-1)^6=1 \] So: \[ =2^4 \cdot 3^3 \] Thus: \[ m=4,\quad n=3 \] \[ m+n=7 \] \[ \boxed{07} \]