Let's solve for the variables [tex]\(a\)[/tex] and [tex]\(b\)[/tex] in the given matrix equation:
[tex]\[
\begin{pmatrix}
a & -2 \\
0 & -4
\end{pmatrix}
\begin{pmatrix}
3 \\
b
\end{pmatrix}
=
\begin{pmatrix}
16 \\
-8
\end{pmatrix}
\][/tex]
We perform the matrix multiplication on the left-hand side:
[tex]\[
\begin{pmatrix}
a & -2 \\
0 & -4
\end{pmatrix}
\begin{pmatrix}
3 \\
b
\end{pmatrix}
=
\begin{pmatrix}
a \cdot 3 + (-2) \cdot b \\
0 \cdot 3 + (-4) \cdot b
\end{pmatrix}
=
\begin{pmatrix}
3a - 2b \\
-4b
\end{pmatrix}
\][/tex]
Now, set this equal to the right-hand side of the given matrix equation:
[tex]\[
\begin{pmatrix}
3a - 2b \\
-4b
\end{pmatrix}
=
\begin{pmatrix}
16 \\
-8
\end{pmatrix}
\][/tex]
This gives two equations:
1. [tex]\(3a - 2b = 16\)[/tex]
2. [tex]\(-4b = -8\)[/tex]
We can solve the second equation first for [tex]\(b\)[/tex]:
[tex]\[
-4b = -8 \\
b = \frac{-8}{-4} \\
b = 2
\][/tex]
Now substitute [tex]\(b = 2\)[/tex] into the first equation:
[tex]\[
3a - 2b = 16 \\
3a - 2(2) = 16 \\
3a - 4 = 16 \\
3a = 16 + 4 \\
3a = 20 \\
a = \frac{20}{3} \\
a = \frac{20}{3}
\][/tex]
So, the solution is:
[tex]\[ a = \frac{20}{3} \quad \text{and} \quad b = 2 \][/tex]
These values satisfy both equations obtained from the matrix multiplication. Therefore, the solution is:
[tex]\[
\left(a = \frac{20}{3}, \, b = 2\right)
\][/tex]
