Find the matrix $A$ such that $\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix} A = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$

$A = \begin{bmatrix} 1 & -2 & -5 \\ 3 & 4 & 0 \end{bmatrix}$

The correct answer is Option (2) → $A = \begin{bmatrix} 1 & -2 & -5 \\ 3 & 4 & 0 \end{bmatrix}$ ##

We have,

$\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix}_{3 \times 2} A = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}_{3 \times 3}$

From the given equation, it is clear that order of $A$ should be $2 \times 3$.

Let $A = \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}_{2 \times 3}$

$∴\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix} = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2a - d & 2b - e & 2c - f \\ a + 0 \cdot d & b + 0 \cdot e & c + 0 \cdot f \\ -3a + 4d & -3b + 4e & -3c + 4f \end{bmatrix} = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2a - d & 2b - e & 2c - f \\ a & b & c \\ -3a + 4d & -3b + 4e & -3c + 4f \end{bmatrix} = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$

Since, both matrices are equal.

$∴$ On comparing corresponding elements of both sides, we get

$a = 1, b = -2, c = -5$

And $2a - d = -1 \Rightarrow d = 2a + 1 = 3; \quad [∵a = 1]$

$2b - e = -8 \Rightarrow e = 2(-2) + 8 = 4 \quad [∵b = -2]$

$2c - f = -10 \Rightarrow f = 2c + 10 = 0 \quad [∵c = -5]$

$∴A = \begin{bmatrix} 1 & -2 & -5 \\ 3 & 4 & 0 \end{bmatrix}$