Question

If \(A=\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix}\) and \(A^{-1}=\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & c \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}\), then ...

• A. \(a=1,c=1\)
• B. \(a=1,c=-1\)
• C. \(a=1,c=2\)
• D. \(a=-1,c=2\)

Hint:

Use the condition that A times its inverse is the identity matrix and compare suitable entries.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Since \(A^{-1}\) is the inverse of \(A\), the product \(AA^{-1}=I\).

Step 2: Find c
Multiply row 1 of \(A\), \((0,1,2)\), by column 3 of \(A^{-1}\), \(\left(\dfrac{1}{2},c,\dfrac{1}{2}\right)\).
The entry is off-diagonal, so \(0+c+1=0\), giving \(c=-1\).

Step 3: Find a
Multiply row 3 of \(A\), \((3,a,1)\), by column 1 of \(A^{-1}\), \(\left(\dfrac{1}{2},-4,\dfrac{5}{2}\right)\).
This entry is \(0\), so \(\dfrac{3}{2}-4a+\dfrac{5}{2}=0\), giving \(a=1\).

Step 4: Conclusion
So \(a=1\) and \(c=-1\).

Final Answer:
The values are a equal to 1 and c equal to minus 1. \[ \boxed{\text{(B) }a=1,c=-1} \]