Answer
$\left[ \begin{matrix}
-4 & -\frac{1}{2} \\
3 & 3 \\
\end{matrix} \right]$
Work Step by Step
To solve the provided expression, follow the below methods:
Matrix addition: Consider two 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, $\begin{align}
& aZ=a\left[ \begin{matrix}
{{z}_{1}} & {{z}_{2}} \\
{{z}_{4}} & {{z}_{3}} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
a{{z}_{1}} & a{{z}_{2}} \\
a{{z}_{4}} & a{{z}_{3}} \\
\end{matrix} \right]
\end{align}$
Substitute $ C=\left[ \begin{matrix}
-1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ and $ B=\left[ \begin{matrix}
4 & 1 \\
-6 & -2 \\
\end{matrix} \right]$ in $2C-\frac{1}{2}B $
$\begin{align}
& 2C-\frac{1}{2}B=2\left[ \begin{matrix}
-1 & 0 \\
0 & 1 \\
\end{matrix} \right]-\frac{1}{2}\left[ \begin{matrix}
4 & 1 \\
-6 & -2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2 & 0 \\
0 & 2 \\
\end{matrix} \right]-\left[ \begin{matrix}
2 & \frac{1}{2} \\
-3 & -1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2-2 & 0-\frac{1}{2} \\
0+3 & 2+1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-4 & -\frac{1}{2} \\
3 & 3 \\
\end{matrix} \right]
\end{align}$
Hence, $2C-\frac{1}{2}B=\left[ \begin{matrix}
-4 & -\frac{1}{2} \\
3 & 3 \\
\end{matrix} \right]$
