If $A$, $B$ and $C$ are matrices of order $2 imes2$ each and $2A+B+C=\begin{bmatrix}1&2\3&0\end{bmatrix}$, $A+B+C=\begin{bmatrix}0&1\2&1\end{bmatrix}$, $A+B-C=\begin{bmatrix}1&2\1&0\end{bmatrix}$, find $A$, $B$ and $C$.

Question

If $A$, $B$ and $C$ are matrices of order $2	imes2$ each and $2A+B+C=\begin{bmatrix}1&2\3&0\end{bmatrix}$, $A+B+C=\begin{bmatrix}0&1\2&1\end{bmatrix}$, $A+B-C=\begin{bmatrix}1&2\1&0\end{bmatrix}$, find $A$, $B$ and $C$.

Accepted Answer

>

> [!TLDR]
> Subtracting the given matrix equations pairwise isolates $A$, then $C$, then $B$ by back-substitution.
>
> Given: (i) $2A+B+C=\begin{bmatrix}1&2\3&0\end{bmatrix}$, (ii) $A+B+C=\begin{bmatrix}0&1\2&1\end{bmatrix}$, (iii) $A+B-C=\begin{bmatrix}1&2\1&0\end{bmatrix}$.
>
> **Find $A$:** (i) $-$ (ii): $(2A+B+C)-(A+B+C)=A$:
> $$A=\begin{bmatrix}1&2\3&0\end{bmatrix}-\begin{bmatrix}0&1\2&1\end{bmatrix}=\begin{bmatrix}1&1\1&-1\end{bmatrix}.$$
>
> **Find $C$:** (ii) $-$ (iii): $(A+B+C)-(A+B-C)=2C$:
> $$2C=\begin{bmatrix}0&1\2&1\end{bmatrix}-\begin{bmatrix}1&2\1&0\end{bmatrix}=\begin{bmatrix}-1&-1\1&1\end{bmatrix}\ \Rightarrow\ C=\begin{bmatrix}-\frac12&-\frac12\\frac12&\frac12\end{bmatrix}.$$
>
> **Find $B$:** from (ii), $B=\begin{bmatrix}0&1\2&1\end{bmatrix}-A-C$:
> $$A+C=\begin{bmatrix}1&1\1&-1\end{bmatrix}+\begin{bmatrix}-\frac12&-\frac12\\frac12&\frac12\end{bmatrix}=\begin{bmatrix}\frac12&\frac12\\frac32&-\frac12\end{bmatrix}.$$ …

Q.If $A$ , $B$ and $C$ are matrices of order $2\times2$ each and $2A+B+C=\begin{bmatrix}1&2\\3&0\end{bmatrix}$ , $A+B+C=\begin{bmatrix}0&1\\2&1\end{bmatrix}$ , $A+B-C=\begin{bmatrix}1&2\\1&0\end{bmatrix}$ , find $A$ , $B$ and $C$ .

Concept understanding — Matrix Equation Solving

Solving a System of Equations by the Matrix Method

Writing the system as $AX=B$

Solving when $A$ is invertible

When $\det(A)=0$

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