JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 19)
The number of elements in the set $$\left\{ {A = \left( {\matrix{
a & b \cr
0 & d \cr
} } \right):a,b,d \in \{ - 1,0,1\} \,and\,{{(I - A)}^3} = I - {A^3}} \right\}$$, where I is 2 $$\times$$ 2 identity matrix, is :
Answer
8
Explanation
$${(I - A)^3} = {I^3} - {A^3} - 3A(I - A) = I - {A^3}$$
$$ \Rightarrow 3A(I - A) = 0$$ or $${A^2} = A$$
$$ \Rightarrow \left[ {\matrix{ {{a^2}} & {ab + bd} \cr 0 & {{d^2}} \cr } } \right] = \left[ {\matrix{ a & b \cr 0 & d \cr } } \right]$$
$$ \Rightarrow {a^2} = a,b(a + d - 1) = 0,{d^2} = d$$
If b $$\ne$$ 0, a + d = 1 $$\Rightarrow$$ 4 ways
If b = 0, a = 0, 1 & d = 0, 1 $$\Rightarrow$$ 4 ways
$$\Rightarrow$$ Total 8 matrices
$$ \Rightarrow 3A(I - A) = 0$$ or $${A^2} = A$$
$$ \Rightarrow \left[ {\matrix{ {{a^2}} & {ab + bd} \cr 0 & {{d^2}} \cr } } \right] = \left[ {\matrix{ a & b \cr 0 & d \cr } } \right]$$
$$ \Rightarrow {a^2} = a,b(a + d - 1) = 0,{d^2} = d$$
If b $$\ne$$ 0, a + d = 1 $$\Rightarrow$$ 4 ways
If b = 0, a = 0, 1 & d = 0, 1 $$\Rightarrow$$ 4 ways
$$\Rightarrow$$ Total 8 matrices
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