Question

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

• A. \(b=\dfrac{ac}{2}\)
• B. \(b=-\dfrac{ac}{2}\)
• C. \(b=\dfrac{a+c}{2}\)
• D. \(b=\sqrt{ac}\)

Hint:

When a matrix satisfies \(A^2=I\), first compute \(A^2\) carefully and compare corresponding entries with the identity matrix.

Answer 1

The Correct Option is B

Step 1: State the goal.
We are given $A=\begin{bmatrix}1 & 0 & 0\\ a & -1 & 0\\ b & c & 1\end{bmatrix}$ and the condition $A^2=I$. We want a relation among $a,b,c$.
Step 2: Multiply $A$ by itself.
Compute $A^2=A\cdot A$ entry by entry; the diagonal already gives $1$, so we focus on the lower-left entries.
Step 3: Find the $(2,1)$ entry.
Row 2 of $A$ is $(a,-1,0)$ times column 1 $(1,a,b)^T$: $a+(-1)a+0=0$, which is automatically fine.
Step 4: Find the $(3,1)$ entry.
Row 3 $(b,c,1)$ times column 1 $(1,a,b)^T$ gives $b+ca+b=2b+ac$.
Step 5: Set it to the identity value.
In $I$ the $(3,1)$ entry is $0$, so $2b+ac=0$, giving $b=-\dfrac{ac}{2}$.
Step 6: Conclude.
This matches option (2).
\[ \boxed{b=-\dfrac{ac}{2}} \]