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

Answer: (B)

2

Answer 2

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