Question

Let A be a $2 \times 2$ matrix with det $( A )=-1$ and $\operatorname{det}(( A + I )(\operatorname{Adj}( A )+ I ))=4$. Then the sum of the diagonal elements of A can be:

• A. $-1$
• B. 2
• C. 1
• D. $-\sqrt{2}$

Answer 1

The Correct Option is B

The correct option is (B) : 2
Given : Relation det ((A + I)(adj(A) + I)) = 4 , det (A) = -1,
Then, adj A = -A^-1
| (A + I )A^-1 + I | = 4
| -I + A - A^-1 + I | =4
| A - A^-1 | = 4
Let A \(=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \) then A^-1 = \(\begin{bmatrix} -d & b \\ c & -a \end{bmatrix}\)
| A - A^-1 | = \(\begin{bmatrix} a+d & 0 \\ 0 & d+a \end{bmatrix}=4\)
(a + d)² = 4
⇒ a + d = ± 2
⇒ | a + d | = 2