To solve this problem, we need to perform scalar multiplication of a matrix. Scalar multiplication involves multiplying each element of the matrix by the given scalar.
Given matrix:
[tex]\[
\left[\begin{array}{ccc}
4 & 0 & 1 \\
-1 & 5 & 2
\end{array}\right]
\][/tex]
Given scalar: [tex]\( 3 \)[/tex]
We need to multiply each element of the matrix by [tex]\( 3 \)[/tex].
Step-by-step multiplication:
1. Multiply [tex]\( 4 \)[/tex] by [tex]\( 3 \)[/tex]:
[tex]\[
3 \times 4 = 12
\][/tex]
2. Multiply [tex]\( 0 \)[/tex] by [tex]\( 3 \)[/tex]:
[tex]\[
3 \times 0 = 0
\][/tex]
3. Multiply [tex]\( 1 \)[/tex] by [tex]\( 3 \)[/tex]:
[tex]\[
3 \times 1 = 3
\][/tex]
4. Multiply [tex]\( -1 \)[/tex] by [tex]\( 3 \)[/tex]:
[tex]\[
3 \times -1 = -3
\][/tex]
5. Multiply [tex]\( 5 \)[/tex] by [tex]\( 3 \)[/tex]:
[tex]\[
3 \times 5 = 15
\][/tex]
6. Multiply [tex]\( 2 \)[/tex] by [tex]\( 3 \)[/tex]:
[tex]\[
3 \times 2 = 6
\][/tex]
Combining these results, the resulting matrix is:
[tex]\[
\left[\begin{array}{ccc}
12 & 0 & 3 \\
-3 & 15 & 6
\end{array}\right]
\][/tex]
So, after multiplying the given matrix by the scalar 3, we get:
[tex]\[
\left[\begin{array}{ccc}
12 & 0 & 3 \\
-3 & 15 & 6
\end{array}\right]
\][/tex]
