Answer
The matrix is,
$\left[ \begin{array}{*{35}{r}}
5 & 4 \\
1 & 11 \\
\end{array} \right]$
Work Step by Step
To solve the given expression, follow the methods given below:
Matrix addition: consider the two given matrices $X,Y$
$X=\left[ \begin{matrix}
{{x}_{1}} & {{x}_{2}} \\
{{x}_{3}} & {{x}_{4}} \\
\end{matrix} \right]$ and $Y=\left[ \begin{matrix}
{{y}_{1}} & {{y}_{2}} \\
{{y}_{3}} & {{y}_{4}} \\
\end{matrix} \right]$
Now,
$\begin{align}
& X+Y=\left[ \begin{matrix}
{{x}_{1}} & {{x}_{2}} \\
{{x}_{3}} & {{x}_{4}} \\
\end{matrix} \right]+\left[ \begin{matrix}
{{y}_{1}} & {{y}_{2}} \\
{{y}_{3}} & {{y}_{4}} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
{{x}_{1}}+{{y}_{1}} & {{x}_{2}}+{{y}_{2}} \\
{{x}_{3}}+{{y}_{3}} & {{x}_{4}}+{{y}_{4}} \\
\end{matrix} \right]
\end{align}$
Matrix multiplication with a scalar: consider a matrix $Z$ and a scalar value a:
$Z=\left[ \begin{matrix}
{{z}_{1}} & {{z}_{2}} \\
{{z}_{3}} & {{z}_{4}} \\
\end{matrix} \right]$
Now,
$\begin{align}
& aZ=a\left[ \begin{matrix}
{{z}_{1}} & {{z}_{2}} \\
{{z}_{4}} & {{z}_{3}} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
a{{z}_{1}} & a{{z}_{2}} \\
a{{z}_{4}} & a{{z}_{3}} \\
\end{matrix} \right]
\end{align}$
Using the method given above solve the given expression:
$\begin{align}
& 2B+3C=2\left[ \begin{array}{*{35}{r}}
1 & -1 \\
2 & 1 \\
\end{array} \right]+3\left[ \begin{array}{*{35}{r}}
1 & 2 \\
-1 & 3 \\
\end{array} \right] \\
& =\left[ \begin{array}{*{35}{r}}
2 & -2 \\
4 & 2 \\
\end{array} \right]+\left[ \begin{array}{*{35}{r}}
3 & 6 \\
-3 & 9 \\
\end{array} \right] \\
& =\left[ \begin{array}{*{35}{r}}
2+3 & -2+6 \\
4-3 & 2+9 \\
\end{array} \right] \\
& =\left[ \begin{array}{*{35}{r}}
5 & 4 \\
1 & 11 \\
\end{array} \right]
\end{align}$
Hence, the required matrix is, $2B+3C=\left[ \begin{array}{*{35}{r}}
5 & 4 \\
1 & 11 \\
\end{array} \right]$.