Q.Determine the matrices $A$ and $B$ if they satisfy

[!FORMULA] $2A - B + \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1\end{pmatrix} = O \qquad \text{and} \qquad A - 2B = \begin{pmatrix} 3 & 2 & 8 \\ -2 & 1 & -7\end{pmatrix}$

Step 1. Rewrite the two given relations in a standard form.

$2A-B+\begin{pmatrix}6&-6&0\\-4&2&1\end{pmatrix}=O \Rightarrow 2A-B=\begin{pmatrix}-6&6&0\\4&-2&-1\end{pmatrix}=P$ … (I)

$A-2B=\begin{pmatrix}3&2&8\\-2&1&-7\end{pmatrix}=Q$ … (II)

Step 2. Eliminate $B$. From (I), $B=2A-P$. Substitute into (II):

$A-2(2A-P)=Q \Rightarrow A-4A+2P=Q \Rightarrow -3A=Q-2P \Rightarrow A=\dfrac{2P-Q}{3}$.

Step 3. Compute $2P-Q$.

$2P=\begin{pmatrix}-12&12&0\\8&-4&-2\end{pmatrix}$

$2P-Q=\begin{pmatrix}-12-3 & 12-2 & 0-8\\ 8-(-2) & -4-1 & -2-(-7)\end{pmatrix}=\begin{pmatrix}-15&10&-8\\10&-5&5\end{pmatrix}$

Step 4. Divide by $3$ to get $A$.

$A=\dfrac13\begin{pmatrix}-15&10&-8\\10&-5&5\end{pmatrix}=\begin{pmatrix}-5 & \frac{10}3 & -\frac83 \\ \frac{10}3 & -\frac53 & \frac53\end{pmatrix}$

Step 5. Recover $B$ from $B=2A-P$.

$2A=\begin{pmatrix}-10 & \frac{20}3 & -\frac{16}3\\ \frac{20}3 & -\frac{10}3 & \frac{10}3\end{pmatrix}$, so …

• Forgetting to move the constant matrix to the other side before eliminating $B$ …