Question

1. if $b = \

$$\begin{bmatrix} 2 & -3 & 1 \\\\ 4 & 2 & -6 \\end{bmatrix}$$

$ and $e = \

$$\begin{bmatrix} -4 & 2 \\\\ 0 & 1 \\\\ 4 & 2 \\end{bmatrix}$$

$, find the following:
a. $be$
b. $2b - e^t$

Explanation:

Part a: Calculate \( BE \)

Step 1: Recall matrix multiplication rule

For two matrices \( B \) (size \( m \times n \)) and \( E \) (size \( n \times p \)), the product \( BE \) has size \( m \times p \), and the element in row \( i \), column \( j \) is \( \sum_{k = 1}^{n} B_{ik}E_{kj} \). Here, \( B=

$$\begin{bmatrix}2&-3&1\\4&2&-6\end{bmatrix}$$

\) (size \( 2 \times 3 \)) and \( E=

$$\begin{bmatrix}-4&2\\0&1\\4&2\end{bmatrix}$$

\) (size \( 3 \times 2 \)), so \( BE \) will be \( 2 \times 2 \).

Step 2: Calculate first row, first column of \( BE \)

\( (2)(-4)+(-3)(0)+(1)(4)= -8 + 0 + 4=-4 \)

Step 3: Calculate first row, second column of \( BE \)

\( (2)(2)+(-3)(1)+(1)(2)=4 - 3 + 2 = 3 \)

Step 4: Calculate second row, first column of \( BE \)

\( (4)(-4)+(2)(0)+(-6)(4)=-16 + 0 - 24=-40 \)

Step 5: Calculate second row, second column of \( BE \)

\( (4)(2)+(2)(1)+(-6)(2)=8 + 2 - 12=-2 \)

Step 1: Find \( 2B \)

Multiply each element of \( B \) by 2. \( B=

$$\begin{bmatrix}2&-3&1\\4&2&-6\end{bmatrix}$$

\), so \( 2B=

$$\begin{bmatrix}2\times2&2\times(-3)&2\times1\\2\times4&2\times2&2\times(-6)\end{bmatrix}$$

=

$$\begin{bmatrix}4&-6&2\\8&4&-12\end{bmatrix}$$

\)

Step 2: Find \( E^T \) (transpose of \( E \))

The transpose of \( E=

$$\begin{bmatrix}-4&2\\0&1\\4&2\end{bmatrix}$$

\) is \( E^T=

$$\begin{bmatrix}-4&0&4\\2&1&2\end{bmatrix}$$

\) (rows become columns).

Step 3: Subtract \( E^T \) from \( 2B \)

Subtract corresponding elements:
First row: \( 4 - (-4)=8 \), \( -6 - 0=-6 \), \( 2 - 4=-2 \)
Second row: \( 8 - 2 = 6 \), \( 4 - 1 = 3 \), \( -12 - 2=-14 \)
So \( 2B - E^T=

$$\begin{bmatrix}8&-6&-2\\6&3&-14\end{bmatrix}$$

\)

Answer:

\( BE=

$$\begin{bmatrix}-4&3\\-40&-2\end{bmatrix}$$

\)

Part b: Calculate \( 2B - E^T \)