Question

If \(\begin{bmatrix} 2 & 1 \\ 3 & 2\end{bmatrix}\) A \(\begin{bmatrix} -3 & 2 \\ 5 & -3\end{bmatrix}\) =\(\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\), then A =?

• A. \(\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\)
• B. \(\begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix}\)
• C. \(\begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix}\)
• D. \(\begin{bmatrix} 1 & 1 \\ 1 & 0\end{bmatrix}\)

Answer 1

The Correct Option is B

The second matrix is \(\begin{bmatrix} -3 & 2 \\ 5 & -3\end{bmatrix}\), which results in the equation:
 \(\begin{bmatrix} 2 & 1 \\ 3 & 2\end{bmatrix}\) A \(\begin{bmatrix} -3 & 2 \\ 5 & -3\end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\)
Now, let's perform the matrix multiplication:
[21A - 32(-32) + 32(5) 21A + 32(-3) + 32(5)] = \(\begin{bmatrix} 10 & 0 \\ \end{bmatrix}\)
Simplifying:
\(\begin{bmatrix} 21A + 1024 - 96& 21A - 96 + 160 \\ \end{bmatrix}\) = \(\begin{bmatrix} 10 & 0 \\ \end{bmatrix}\)
[21A + 928 21A + 64] = \(\begin{bmatrix} 10 & 0 \\ \end{bmatrix}\) 
Equating :
21A + 928 = 10 ………… (1)
21A + 64 = 0 …………… (2)
Solving Equation 2 for A: 
21A = -64 
A = \(-\frac {64}{21}\)
Hence. Matrix A is \(\begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix}\)