If A and B are square matrices of order 2, then det (A + B) = 0 is possible only when

Question:

If $A$ and $B$ are square matrices of order 2, then $\operatorname{det}(A+B)=0$ is possible only when

(a) $\operatorname{det}(A)=0$ or $\operatorname{det}(B)=0$

(b) $\operatorname{det}(A)+\operatorname{det}(B)=0$

(c) $\operatorname{det}(A)=0$ and $\operatorname{det}(B)=0$

(d) $A+B=0$

Solution:

(d) $A+B=O$

Let $A=\left[a_{i j}\right]$ and $B=\left[b_{i j}\right]$ be a square matrix of order 2 .

As their orders are same, $\mathrm{A}+\mathrm{B}$ is defined as

$\mathrm{A}+\mathrm{B}=\left[\mathrm{a}_{\mathrm{ij}}+\mathrm{b}_{\mathrm{ij}}\right]$

$\Rightarrow|\mathrm{A}+\mathrm{B}|=\left|\mathrm{a}_{\mathrm{i} j}+\mathrm{b}_{\mathrm{i} j}\right|$

Now,

$|A+B|=0$

$\Rightarrow\left|a_{i j}+b_{i j}\right|=0$

$\Rightarrow\left[a_{i j}+b_{i j}\right]=0$      [each corrsponding term is 0]

$\Rightarrow A+B=0$

Leave a comment