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 (integer)
8
Solution
${(I - A)^3} = {I^3} - {A^3} - 3A(I - A) = I - {A^3}$<br><br>$\Rightarrow 3A(I - A) = 0$ or ${A^2} = A$<br><br>$$ \Rightarrow \left[ {\matrix{
{{a^2}} & {ab + bd} \cr
0 & {{d^2}} \cr
} } \right] = \left[ {\matrix{
a & b \cr
0 & d \cr
} } \right]$$<br><br>$\Rightarrow {a^2} = a,b(a + d - 1) = 0,{d^2} = d$<br><br>If b $\ne$ 0, a + d = 1 $\Rightarrow$ 4 ways<br><br>If b = 0, a = 0, 1 & d = 0, 1 $\Rightarrow$ 4 ways<br><br>$\Rightarrow$ Total 8 matrices
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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