Given the matrices [tex]A[/tex] and [tex]B[/tex], compute [tex]4A[/tex].

[tex]\[
A = \begin{bmatrix}
1 & 2 & -4 \\
0 & 3 & -1
\end{bmatrix} \quad B = \begin{bmatrix}
-1 & 0 \\
2 & 4 \\
7 & 2
\end{bmatrix}
\][/tex]

Enter the element in row 2, column 3 of [tex]4A[/tex].



Answer :

To solve this problem, follow these steps:

1. Matrix [tex]\( A \)[/tex] Representation:
We need to work with the matrix [tex]\( A \)[/tex]:
[tex]\[ A = \left[\begin{array}{ccc} 1 & 2 & -4 \\ 0 & 3 & -1 \end{array}\right] \][/tex]

2. Scaling the Matrix [tex]\( A \)[/tex] by 4:
To compute [tex]\( 4A \)[/tex], we multiply each element of the matrix [tex]\( A \)[/tex] by 4. So, we calculate each element as follows:

[tex]\[ 4A = 4 \times \left[\begin{array}{ccc} 1 & 2 & -4 \\ 0 & 3 & -1 \end{array}\right] = \left[\begin{array}{ccc} 4 \times 1 & 4 \times 2 & 4 \times -4 \\ 4 \times 0 & 4 \times 3 & 4 \times -1 \end{array}\right] \][/tex]

Performing the multiplications:

[tex]\[ 4A = \left[\begin{array}{ccc} 4 & 8 & -16 \\ 0 & 12 & -4 \end{array}\right] \][/tex]

3. Extracting the Particular Element:
We are asked to find the element in row 2, column 3 of the matrix [tex]\( 4A \)[/tex]. From the matrix computed:

[tex]\[ 4A = \left[\begin{array}{ccc} 4 & 8 & -16 \\ 0 & 12 & -4 \end{array}\right] \][/tex]

Look at the second row:

[tex]\[ \left[0, 12, -4\right] \][/tex]

The element in column 3 of this row is [tex]\( -4 \)[/tex].

So, the element in row 2, column 3 of [tex]\( 4A \)[/tex] is:
[tex]\[ -4 \][/tex]