A and B are two non- singular square matrices of each 3 × 3 such that AB = A and left| {A +

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3 and 4 .Determinants and Matrices

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$A$ and $B$ are two non- singular square matrices of each $3 \times 3$ such that $AB = A$ and $\left| {A + B} \right| \ne 0$, then

A

$\left| {A + B} \right| = 2$

B

$\left| {A + B} \right| = 8$

C

$\left| {A -B} \right| = 1$

D

$\left| {A -B} \right| = 2$

Solution

We have $A^{2}=(A B) A=A(B A)=A B=A$

Similarly $\mathrm{B}^{2}=\mathrm{B}$

$(A+B)^{2}=A^{2}+B^{2}+A B+B A$

$= A+B+A+B=2(A+B) $

$ \Rightarrow|A+B|^{2}=8|A+B| \Rightarrow|A+B|=8 $ and

$(A-B)^{2}=A^{2}+B^{2}-A B-B A=A+B-A-B=0 $

$\Rightarrow|A-B|=0 $

Standard 12

Mathematics

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