To solve the given matrix operations, we need to perform addition, subtraction, scalar multiplication, and a combination of these on the matrices provided. Here are the matrices:
[tex]\[
A = \left[\begin{array}{cc}
3 & 1 \\
1 & 3
\end{array}\right], \quad
B = \left[\begin{array}{rr}
-2 & -2 \\
1 & 2
\end{array}\right]
\][/tex]
### (a) [tex]\(A + B\)[/tex]
To find [tex]\(A + B\)[/tex], we add the corresponding elements of matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[
A + B = \left[\begin{array}{cc}
3 & 1 \\
1 & 3
\end{array}\right] + \left[\begin{array}{rr}
-2 & -2 \\
1 & 2
\end{array}\right] = \left[\begin{array}{cc}
3 + (-2) & 1 + (-2) \\
1 + 1 & 3 + 2
\end{array}\right]
= \left[\begin{array}{cc}
1 & -1 \\
2 & 5
\end{array}\right]
\][/tex]
So, [tex]\(A + B = \left[\begin{array}{cc}
1 & -1 \\
2 & 5
\end{array}\right]\)[/tex].
### (b) [tex]\(A - B\)[/tex]
To find [tex]\(A - B\)[/tex], we subtract the corresponding elements of matrix [tex]\(B\)[/tex] from matrix [tex]\(A\)[/tex]:
[tex]\[
A - B = \left[\begin{array}{cc}
3 & 1 \\
1 & 3
\end{array}\right] - \left[\begin{array}{rr}
-2 & -2 \\
1 & 2
\end{array}\right] = \left[\begin{array}{cc}
3 - (-2) & 1 - (-2) \\
1 - 1 & 3 - 2
\end{array}\right]
= \left[\begin{array}{cc}
5 & 3 \\
0 & 1
\end{array}\right]
\][/tex]
So, [tex]\(A - B = \left[\begin{array}{cc}
5 & 3 \\
0 & 1
\end{array}\right]\)[/tex].
### (c) [tex]\(5A\)[/tex]
To find [tex]\(5A\)[/tex], we multiply each element of matrix [tex]\(A\)[/tex] by the scalar [tex]\(5\)[/tex]:
[tex]\[
5A = 5 \cdot \left[\begin{array}{cc}
3 & 1 \\
1 & 3
\end{array}\right] = \left[\begin{array}{cc}
5 \cdot 3 & 5 \cdot 1 \\
5 \cdot 1 & 5 \cdot 3
\end{array}\right]
= \left[\begin{array}{cc}
15 & 5 \\
5 & 15
\end{array}\right]
\][/tex]
So, [tex]\(5A = \left[\begin{array}{cc}
15 & 5 \\
5 & 15
\end{array}\right]\)[/tex].
### (d) [tex]\(5A - 2B\)[/tex]
To find [tex]\(5A - 2B\)[/tex], we first find [tex]\(2B\)[/tex] by multiplying each element of matrix [tex]\(B\)[/tex] by the scalar [tex]\(2\)[/tex], and then subtract it from [tex]\(5A\)[/tex].
[tex]\[
2B = 2 \cdot \left[\begin{array}{rr}
-2 & -2 \\
1 & 2
\end{array}\right] = \left[\begin{array}{rr}
2 \cdot (-2) & 2 \cdot (-2) \\
2 \cdot 1 & 2 \cdot 2
\end{array}\right]
= \left[\begin{array}{rr}
-4 & -4 \\
2 & 4
\end{array}\right]
\][/tex]
Now, subtract [tex]\(2B\)[/tex] from [tex]\(5A\)[/tex]:
[tex]\[
5A - 2B = \left[\begin{array}{cc}
15 & 5 \\
5 & 15
\end{array}\right] - \left[\begin{array}{rr}
-4 & -4 \\
2 & 4
\end{array}\right] = \left[\begin{array}{cc}
15 - (-4) & 5 - (-4) \\
5 - 2 & 15 - 4
\end{array}\right]
= \left[\begin{array}{cc}
19 & 9 \\
3 & 11
\end{array}\right]
\][/tex]
So, [tex]\(5A - 2B = \left[\begin{array}{cc}
19 & 9 \\
3 & 11
\end{array}\right]\)[/tex].
