Question

If $A$ be a non-singular matrix of order $2,$ such that $|A+|A|\text{adj}\left(A\right)|=0,$ then which of the following option(s) is/are always correct ?
(where $\text{adj}\left(A\right)$ is the adjoint of matrix $A$ )

• A. $|A|=1$
• B. the trace of matrix $A$ is $0.$
  (the trace of a square matrix is the sum of elements on the main diagonal)
• C. $|A-|A|\text{adj}\left(A\right)|=2$
• D. $|A-|A|\text{adj}\left(A\right)|=4$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $|A|=1$
B the trace of matrix $A$ is $0.$
(the trace of a square matrix is the sum of elements on the main diagonal)
D $|A-|A|\text{adj}\left(A\right)|=4$

Let $A=\left[\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right]$
$\Rightarrow |A|=a_1{a}_4-a_2{a}_3=k\left(\text{say}\right)$

$\text{adj}\left(A\right)=\left[\begin{array}{cc}{a}_4& -a_2\\ -a_3& a_1\end{array}\right]$

$|A+|A|\text{adj}\left(A\right)|=0$
$\Rightarrow \left[\begin{array}{cc}{a}_1+k{a}_4& a_2-k{a}_2\\ a_3-k{a}_3& a_4+k{a}_1\end{array}\right]=0$
$\Rightarrow k[1+a_4^2+a_1^2+k^2+2a_2{a}_3]=0$
$\Rightarrow (a_4+a_1{)}^2+1+k^2-2(a_1{a}_4-a_2{a}_3)=0\Rightarrow (a_4+a_1{)}^2+1+k^2-2k=0$
$\Rightarrow k=|A|=1\&a_4+a_1=0$

$|A-|A|\text{adj}\left(A\right)|=|A-\text{adj}\left(A\right)|(\because |A|=1)$
$=\left[\begin{array}{cc}{a}_1-a_4& 2a_2\\ 2a_3& a_4-a_1\end{array}\right]$
$=2a_1{a}_4-a_1^2-a_4^2-4a_2{a}_3$
$=4a_1{a}_4-(2a_1{a}_4+a_1^2+a_4^2)-4a_2{a}_3$
$=4(a_1{a}_4-a_2{a}_3)-(a_1+a_4{)}^2$
$=4-0$
$=4$