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Mathematics Question on Matrices

If [21 32]\begin{bmatrix} 2 & 1 \\\ 3 & 2\end{bmatrix} A [−32 5−3]\begin{bmatrix} -3 & 2 \\\ 5 & -3\end{bmatrix} =[10 01]\begin{bmatrix} 1 & 0 \\\ 0 & 1\end{bmatrix}, then A =?

A

[11 01]\begin{bmatrix} 1 & 1 \\\ 0 & 1\end{bmatrix}

B

[10 11]\begin{bmatrix} 1 & 0 \\\ 1 & 1\end{bmatrix}

C

[11 11]\begin{bmatrix} 1 & 1 \\\ 1 & 1\end{bmatrix}

D

[11 10]\begin{bmatrix} 1 & 1 \\\ 1 & 0\end{bmatrix}

Answer

[10 11]\begin{bmatrix} 1 & 0 \\\ 1 & 1\end{bmatrix}

Explanation

Solution

The second matrix is [−32 5−3]\begin{bmatrix} -3 & 2 \\\ 5 & -3\end{bmatrix}, which results in the equation:
[21 32]\begin{bmatrix} 2 & 1 \\\ 3 & 2\end{bmatrix} A [−32 5−3]\begin{bmatrix} -3 & 2 \\\ 5 & -3\end{bmatrix} = [10 01]\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)] = [100 ]\begin{bmatrix} 10 & 0 \\\ \end{bmatrix}
Simplifying:
[21A+1024−9621A−96+160 ]\begin{bmatrix} 21A + 1024 - 96& 21A - 96 + 160 \\\ \end{bmatrix} = [100 ]\begin{bmatrix} 10 & 0 \\\ \end{bmatrix}
[21A + 928 21A + 64] = [100 ]\begin{bmatrix} 10 & 0 \\\ \end{bmatrix}
Equating :
21A + 928 = 10 ………… (1)
21A + 64 = 0 …………… (2)
Solving Equation 2 for A:
21A = -64
A = −6421-\frac {64}{21}
Hence. Matrix A is [10 11]\begin{bmatrix} 1 & 0 \\\ 1 & 1\end{bmatrix}