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(B) Given $A = \begin{bmatrix} 1 & 0 & 0 \ a & -1 & 0 \ b & c & 1 \end{bmatrix}$.We are given that $A^2 = I$.$A^2 = \begin{bmatrix} 1 & 0 & 0 \ a &

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$b = -\frac{ac}{2}$

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(B) Given $A = \begin{bmatrix} 1 & 0 & 0 \ a & -1 & 0 \ b & c & 1 \end{bmatrix}$.We are given that $A^2 = I$.$A^2 = \begin{bmatrix} 1 & 0 & 0 \ a & -1 & 0 \ b & c & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ a & -1 & 0 \ b & c & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Calculating the product:Row $1$: $(1)(1) + (0)(a) + (0)(b) = 1$,$(1)(0) + (0)(-1) + (0)(c) = 0$,$(1)(0) + (0)(0) + (0)(1) = 0$.Row $2$: $(a)(1) + (-1)(a) + (0)(b) = 0$,$(a)(0) + (-1)(-1) + (0)(c) = 1$,$(a)(0) + (-1)(0) + (0)(1) = 0$.Row $3$: $(b)(1) + (c)(a) + (1)(b) = 2b + ac$,$(b)(0) + (c)(-1) + (1)(c) = 0$,$(b)(0) + (c)(0) + (1)(1) = 1$.Thus,$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 2b + ac & 0 & 1 \end{bmatrix}$.Equating this to the identity matrix $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$,we get $2b + ac = 0$.Therefore,$b = -\frac{ac}{2}$.

If $A = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix}$ is such that $A^2 = I$,then

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