Question

1. $3\

$$\begin{bmatrix}0 & 6 & 2 \\\\ 3 & -4 & -3\\end{bmatrix}$$

\cdot\

$$\begin{bmatrix}5 & 0 & 6 \\\\ -3 & 6 & -5\\end{bmatrix}$$

$

1. $\

$$\begin{bmatrix}-2 \\\\ 6 \\\\ -4\\end{bmatrix}$$

-\

$$\begin{bmatrix}-5 \\\\ 0 \\\\ -1\\end{bmatrix}$$

$

1. $\

$$\begin{bmatrix}-6 & -1 & -4 & 0 \\\\ -5 & -6 & -4 & 4\\end{bmatrix}$$

\cdot\

$$\begin{bmatrix}-2 & -2 \\\\ -3 & -1 \\\\ -3 & -6 \\\\ 5 & 3\\end{bmatrix}$$

+\

$$\begin{bmatrix}1 & -4 \\\\ 5 & 5\\end{bmatrix}$$

$

1. $\

$$\begin{bmatrix}2 \\\\ 1 \\\\ 0\\end{bmatrix}$$

-\

$$\begin{bmatrix}1 \\\\ 1 \\\\ 0\\end{bmatrix}$$

$

Explanation:

Problem 8

Step 1: Multiply the first matrix by 3

First, multiply the matrix \(

$$\begin{bmatrix}0&6&2\\3&-4&-3\end{bmatrix}$$

\) by 3. We get:
\(3

$$\begin{bmatrix}0&6&2\\3&-4&-3\end{bmatrix}$$

=

$$\begin{bmatrix}0\times3&6\times3&2\times3\\3\times3&-4\times3&-3\times3\end{bmatrix}$$

=

$$\begin{bmatrix}0&18&6\\9&-12&-9\end{bmatrix}$$

\)

Step 2: Multiply the two matrices

Now, multiply \(

$$\begin{bmatrix}0&18&6\\9&-12&-9\end{bmatrix}$$

\) with \(

$$\begin{bmatrix}5&0&6\\-3&6&-5\end{bmatrix}$$

\). The formula for matrix multiplication \(C = AB\) where \(A\) is \(m\times n\) and \(B\) is \(n\times p\) is \(c_{ij}=\sum_{k = 1}^{n}a_{ik}b_{kj}\)

For the first row and first column:
\(0\times5 + 18\times(-3)+6\times0=0 - 54+0=- 54\)

First row and second column:
\(0\times0+18\times6 + 6\times0 = 0 + 108+0 = 108\)

First row and third column:
\(0\times6+18\times(-5)+6\times0=0 - 90 + 0=-90\)

Second row and first column:
\(9\times5+(-12)\times(-3)+(-9)\times0 = 45 + 36+0 = 81\)

Second row and second column:
\(9\times0+(-12)\times6+(-9)\times0=0 - 72+0=-72\)

Second row and third column:
\(9\times6+(-12)\times(-5)+(-9)\times0=54 + 60+0 = 114\)

So the product is \(

$$\begin{bmatrix}-54&108&-90\\81&-72&114\end{bmatrix}$$

\)

To subtract two vectors (column matrices), we subtract the corresponding elements.

For the first element: \(- 2-(-5)=-2 + 5 = 3\)

Second element: \(6 - 0=6\)

Third element: \(-4-(-1)=-4 + 1=-3\)

Step 1: Multiply the two matrices

Let \(A=

$$\begin{bmatrix}-6&-1&-4&0\\-5&-6&-4&4\end{bmatrix}$$

\), \(B=

$$\begin{bmatrix}-2&-2\\-3&-1\\-3&-6\\5&3\end{bmatrix}$$

\)

For the first row and first column of \(AB\):
\((-6)\times(-2)+(-1)\times(-3)+(-4)\times(-3)+0\times5 = 12 + 3+12 + 0=27\)

First row and second column:
\((-6)\times(-2)+(-1)\times(-1)+(-4)\times(-6)+0\times3=12 + 1+24 + 0 = 37\)

Second row and first column:
\((-5)\times(-2)+(-6)\times(-3)+(-4)\times(-3)+4\times5=10 + 18+12 + 20 = 60\)

Second row and second column:
\((-5)\times(-2)+(-6)\times(-1)+(-4)\times(-6)+4\times3=10 + 6+24 + 12 = 52\)

So \(AB=

$$\begin{bmatrix}27&37\\60&52\end{bmatrix}$$

\)

Step 2: Add the matrix \(

$$\begin{bmatrix}1&-4\\5&5\end{bmatrix}$$

\)
\(

$$\begin{bmatrix}27&37\\60&52\end{bmatrix}$$

+

$$\begin{bmatrix}1&-4\\5&5\end{bmatrix}$$

=

$$\begin{bmatrix}27 + 1&37-4\\60 + 5&52 + 5\end{bmatrix}$$

=

$$\begin{bmatrix}28&33\\65&57\end{bmatrix}$$

\)

Answer:

\(

$$\begin{bmatrix}-54&108&-90\\81&-72&114\end{bmatrix}$$

\)

Problem 9