# Prove that

Question:

Let $\mathrm{A}=\left[\begin{array}{ccc}0 & 1 & -2 \\ 5 & -1 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{ccc}1 & -3 & -1 \\ 0 & -2 & 5\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{ccc}2 & -5 & 1 \\ -4 & 0 & 6\end{array}\right]$. Compute $5 \mathrm{~A}-3 \mathrm{~B}+4 \mathrm{C}$.

Solution:

$5 A-3 B+4 C=5\left(\left[\begin{array}{ccc}0 & 1 & -2 \\ 5 & -1 & -4\end{array}\right]\right)-3\left(\left[\begin{array}{ccc}1 & -3 & -1 \\ 0 & -2 & 5\end{array}\right]\right)+4\left(\left[\begin{array}{ccc}2 & -5 & 1 \\ -4 & 0 & 6\end{array}\right]\right)$

$=\left(\left[\begin{array}{ccc}0 & 5 & -10 \\ 25 & -5 & -20\end{array}\right]\right)-\left(\left[\begin{array}{ccc}3 & -9 & -3 \\ 0 & -6 & 15\end{array}\right]\right)+\left(\left[\begin{array}{ccc}8 & -20 & 4 \\ -16 & 0 & 24\end{array}\right]\right)$

$=\left[\begin{array}{ccc}-3 & 14 & -7 \\ 25 & 1 & -35\end{array}\right]+\left[\begin{array}{ccc}8 & -20 & 4 \\ -16 & 0 & 24\end{array}\right]$

$=\left[\begin{array}{ccc}5 & -6 & -3 \\ 9 & 1 & -11\end{array}\right]$

Conclusion: $5 \mathrm{~A}-3 \mathrm{~B}+4 \mathrm{C}=\left[\begin{array}{ccc}5 & -6 & -3 \\ 9 & 1 & -11\end{array}\right]$