A and B are two non-singular square matrices of each 3×3 such that AB = A and BA=B and |A+B|≠0 then

We have A2= (AB)A = A(BA) = AB = ASimilarlyB2= B: (A+B)2= A2+ B2+ AB + BA = A + B + A + B = 2(A + B)⇒|A + B|2 = 8 |A + B| ⇒ |A + B| = 8 and(A – B)2= A2+ B2– AB – BA = A + B – A – B = 0 ⇒|A – B| = 0

A and B are two non-singular square matrices of each 3×3 such that AB = A and BA=B and |A+B|≠0 then

We have A2= (AB)A = A(BA) = AB = ASimilarlyB2= B: (A+B)2= A2+ B2+ AB + BA = A + B + A + B = 2(A + B)⇒|A + B|2 = 8 |A + B| ⇒ |A + B| = 8 and(A – B)2= A2+ B2– AB – BA = A + B – A – B = 0 ⇒|A – B| = 0

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A and B are two non-singular square matrices of each $3 \times 3$ such that AB = A and $B A = B$ and $| A + B | \neq 0$ then

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$| A - B | = 1$

$| A + B | = 8$

$| A + B | = 2$

$| A + B | = 0$

answer is B.

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Detailed Solution

We have $A^{2} = (A B) A = A (B A) = A B = A$
Similarly
$B^{2} = B : {(A + B)}^{2} = A^{2} + B^{2} + A B + B A = A + B + A + B = 2 (A + B)$ $\Rightarrow | A + B | 2 = 8 | A + B | \Rightarrow | A + B | = 8 a n d$
${(A - B)}^{2} = A^{2} + B^{2} - A B - B A = A + B - A - B = 0$