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: Write the given matrix.
We have \[ A= \begin{bmatrix} 1 & 0 & 0\\ a & -1 & 0\\ b & c & 1 \end{bmatrix} \] Since \[ A^2=I, \] we calculate \(A^2\).

Step 2: Compute \(A^2\).
\[ A^2= \begin{bmatrix} 1 & 0 & 0\\ a & -1 & 0\\ b & c & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ a & -1 & 0\\ b & c & 1 \end{bmatrix} \] Now compute each entry carefully.
First row: \[ (1,0,0) \] Second row: \[ (a,-1,0) \begin{bmatrix} 1\\ a\\ b \end{bmatrix} =a-a=0 \] \[ (a,-1,0) \begin{bmatrix} 0\\ -1\\ c \end{bmatrix} =1 \] \[ (a,-1,0) \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} =0 \] Thus second row becomes \[ (0,1,0) \] Third row:
First entry: \[ (b,c,1) \begin{bmatrix} 1\\ a\\ b \end{bmatrix} =b+ac+b \] \[ =2b+ac \] Second entry: \[ (b,c,1) \begin{bmatrix} 0\\ -1\\ c \end{bmatrix} =-c+c=0 \] Third entry: \[ (b,c,1) \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} =1 \] Hence, \[ A^2= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 2b+ac & 0 & 1 \end{bmatrix} \]

Step 3: Use the condition \(A^2=I\).
Since \[ A^2=I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}, \] comparing corresponding entries gives \[ 2b+ac=0 \] Thus, \[ 2b=-ac \] Therefore, \[ b=-\frac{ac}{2} \]

Step 4: Final conclusion.
Hence, \[ \boxed{b=-\dfrac{ac}{2}} \] which corresponds to option (2).