The number of elements in the set {A=begin{bm | vedclass

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(A) Given the condition $(I-A)^3 = I-A^3$. Expanding the left side: $I^3 - 3I^2A + 3IA^2 - A^3 = I - A^3$. Since $I^2 = I$ and $IA = AI = A$,this simp

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$8$

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(A) Given the condition $(I-A)^3 = I-A^3$. Expanding the left side: $I^3 - 3I^2A + 3IA^2 - A^3 = I - A^3$. Since $I^2 = I$ and $IA = AI = A$,this simplifies to $I - 3A + 3A^2 - A^3 = I - A^3$. Subtracting $I$ and adding $A^3$ to both sides,we get $3A^2 - 3A = 0$,which implies $A^2 = A$. Let $A = \begin{bmatrix} a & b \ 0 & d \end{bmatrix}$. Then $A^2 = \begin{bmatrix} a & b \ 0 & d \end{bmatrix} \begin{bmatrix} a & b \ 0 & d \end{bmatrix} = \begin{bmatrix} a^2 & ab+bd \ 0 & d^2 \end{bmatrix}$. Equating $A^2 = A$,we get: $a^2 = a \Rightarrow a \in \{0, 1\}$. $d^2 = d \Rightarrow d \in \{0, 1\}$. $ab + bd = b \Rightarrow b(a + d - 1) = 0$. Case $1$: If $b = 0$,then $a \in \{0, 1\}$ and $d \in \{0, 1\}$. This gives $2 	imes 2 = 4$ matrices. Case $2$: If $b 
eq 0$,then $a + d - 1 = 0$,so $a + d = 1$. Possible pairs $(a, d)$ are $(1, 0)$ and $(0, 1)$. For each pair,$b$ can be $\{-1, 1\}$ (since $b 
eq 0$). This gives $2 	imes 2 = 4$ matrices. Total number of matrices = $4 + 4 = 8$.

The number of elements in the set $\{A=\begin{bmatrix} a & b \\ 0 & d \end{bmatrix} : a, b, d \in \{-1, 0, 1\} \text{ and } (I-A)^3 = I-A^3 \}$,where $I$ is the $2 \times 2$ identity matrix,is:

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