Q.Find matrices $A$ and $B$, if $2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix}$ and $A - 2B = \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix}$.

Given $2A - B = \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix}$ ...(1) and $A - 2B = \begin{bmatrix} 3 & 2 & 8 \\ -2 & 1 & -7 \end{bmatrix}$ ...(2).

Multiply (1) by 2: $4A - 2B = \begin{bmatrix} 12 & -12 & 0 \\ -8 & 4 & 2 \end{bmatrix}$ ...(3)

Subtract (2) from (3): $3A = \begin{bmatrix} 12-3 & -12-2 & 0-8 \\ -8-(-2) & 4-1 & 2-(-7) \end{bmatrix} = \begin{bmatrix} 9 & -14 & -8 \\ -6 & 3 & 9 \end{bmatrix}$

$A = \begin{bmatrix} 3 & -\frac{14}{3} & -\frac{8}{3} \\ -2 & 1 & 3 \end{bmatrix}$

From (1): $B = 2A - \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 6 & -\frac{28}{3} & -\frac{16}{3} \\ -4 & 2 & 6 \end{bmatrix} - \begin{bmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{10}{3} & -\frac{16}{3} \\ 0 & 0 & 5 \end{bmatrix}$ …

Eliminate B by doubling the first equation and subtracting the second to solve for A, then back-su …

• Forgetting to double EVERY entry of equation (1) before subtracting equation (2). …