Question

Let \[ A=\begin{bmatrix} 1 & 2\\ -2 & 1 \end{bmatrix} \] and \[ B^{-1}=\begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} \] If \[ (AB^{-1})^{-1}=\begin{bmatrix} a & b\\ c & d \end{bmatrix}, \] then \(2b+5c+10d=\)

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(2\)

Hint:

For matrix questions, first perform the multiplication carefully and then use the inverse formula for a \(2\times 2\) matrix.

Answer 1

The Correct Option is B

Step 1: Find \(AB^{-1}\).
\[ A=\begin{bmatrix} 1 & 2\\ -2 & 1 \end{bmatrix}, \quad B^{-1}=\begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} \] \[ AB^{-1} = \begin{bmatrix} 1 & 2\\ -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} \] \[ = \begin{bmatrix} 1(1)+2(0) & 1(1)+2(2)\\ -2(1)+1(0) & -2(1)+1(2) \end{bmatrix} \] \[ = \begin{bmatrix} 1 & 5\\ -2 & 0 \end{bmatrix} \]

Step 2: Find inverse of \(AB^{-1}\).
Let \[ M=AB^{-1}= \begin{bmatrix} 1 & 5\\ -2 & 0 \end{bmatrix} \] For a matrix \[ \begin{bmatrix} p & q\\ r & s \end{bmatrix}, \] its inverse is \[ \frac{1}{ps-qr} \begin{bmatrix} s & -q\\ -r & p \end{bmatrix} \] Here, \[ \det M=1(0)-5(-2)=10 \] So, \[ M^{-1}=\frac{1}{10} \begin{bmatrix} 0 & -5\\ 2 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 0 & -\frac{1}{2}\\ \frac{1}{5} & \frac{1}{10} \end{bmatrix} \]

Step 3: Compare with the given matrix.
\[ (AB^{-1})^{-1}= \begin{bmatrix} a & b\\ c & d \end{bmatrix} = \begin{bmatrix} 0 & -\frac{1}{2}\\ \frac{1}{5} & \frac{1}{10} \end{bmatrix} \] Thus, \[ b=-\frac{1}{2}, \quad c=\frac{1}{5}, \quad d=\frac{1}{10} \]

Step 4: Calculate \(2b+5c+10d\).
\[ 2b+5c+10d = 2\left(-\frac{1}{2}\right) +5\left(\frac{1}{5}\right) +10\left(\frac{1}{10}\right) \] \[ =-1+1+1 \] \[ =1 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{1} \]