Question

If \(AB=\begin{bmatrix} 4&3\\5&4 \end{bmatrix}\) and \(A^{-1}=\begin{bmatrix} 3&-2\\-1&1 \end{bmatrix}\), then B=

• A. \(\begin{bmatrix} 2&1\\1&2 \end{bmatrix}\)
• B. \(\begin{bmatrix} 1&2\\2&1 \end{bmatrix}\)
• C. \(\begin{bmatrix} 1&2\\1&1 \end{bmatrix}\)
• D. \(\begin{bmatrix} 1&1\\1&2 \end{bmatrix}\)
• E. \(\begin{bmatrix} 2&1\\1&1 \end{bmatrix}\)

Answer 1

Answer: (E)

We are given the matrix equation:

\(AB = \begin{bmatrix} 4 & 3 \\ 5 & 4 \end{bmatrix}\) 

and

\(A^{-1} = \begin{bmatrix} 3 & -2 \\ -1 & 1 \end{bmatrix}\)

To find \( B \), we can multiply both sides of the equation \( AB = \begin{bmatrix} 4 & 3 \\ 5 & 4 \end{bmatrix} \) by \( A^{-1} \) on the right. This gives:

\(AB \cdot A^{-1} = \begin{bmatrix} 4 & 3 \\ 5 & 4 \end{bmatrix} \cdot A^{-1}\)

Since \( A \cdot A^{-1} = I \) (the identity matrix), this simplifies to:

\(B = \begin{bmatrix} 4 & 3 \\ 5 & 4 \end{bmatrix} \cdot A^{-1}\)

Now perform the matrix multiplication:

\(B = \begin{bmatrix} 4 & 3 \\ 5 & 4 \end{bmatrix} \cdot \begin{bmatrix} 3 & -2 \\ -1 & 1 \end{bmatrix}\)

The result of the multiplication is:

\(B = \begin{bmatrix} (4 \cdot 3 + 3 \cdot -1) & (4 \cdot -2 + 3 \cdot 1) \\ (5 \cdot 3 + 4 \cdot -1) & (5 \cdot -2 + 4 \cdot 1) \end{bmatrix}\)

\(B = \begin{bmatrix} (12 - 3) & (-8 + 3) \\ (15 - 4) & (-10 + 4) \end{bmatrix}\)

\(B = \begin{bmatrix} 9 & -5 \\ 11 & -6 \end{bmatrix}\)

The correct value of \( B \) is:

\(\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\)