Question

which is the transpose of the matrix?
4.13)
$a=\

$$\begin{bmatrix}0 & -1 & 3\\\\4 & -2 & 5\\\\0 & 7 & 1\\end{bmatrix}$$

$
$a^t = ?$
a. $\

$$\begin{bmatrix}3 & -1 & 3\\\\5 & 2 & 4\\\\1 & 7 & 0\\end{bmatrix}$$

$
b. $\

$$\begin{bmatrix}0 & 4 & 0\\\\-1 & -2 & 7\\\\3 & 5 & 1\\end{bmatrix}$$

$
c. $\

$$\begin{bmatrix}1 & 7 & 0\\\\5 & -2 & 4\\\\3 & 1 & 0\\end{bmatrix}$$

$
d. $\

$$\begin{bmatrix}3 & -1 & 0\\\\5 & -2 & 4\\\\1 & 7 & 0\\end{bmatrix}$$

$

which is the sum of the matrices?
4.14)
$\

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

+\

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

$
a. $\

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

$
b. $\

$$\begin{bmatrix}-2 & -3\\\\0 & -8\\end{bmatrix}$$

$
c. $\

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

$
d. $\

$$\begin{bmatrix}1 & 2\\\\5 & 2\\end{bmatrix}$$

$

which is the product of the matrices?
4.15)
$3\

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

$
a. $\

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

$
b. $\

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

$
c. $\

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

$
d. $\

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

$

4.16)
$\

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

\times\

$$\begin{bmatrix}1\\\\3\\\\2\\end{bmatrix}$$

$
a. $\

$$\begin{bmatrix}6\\end{bmatrix}$$

$
b. $\

$$\begin{bmatrix}13\\end{bmatrix}$$

$
c. $\

$$\begin{bmatrix}3 & 4 & 8\\end{bmatrix}$$

$
d. $\

$$\begin{bmatrix}8\\end{bmatrix}$$

$

4.17)
$\

$$\begin{bmatrix}1 & 0 & 3\\end{bmatrix}$$

\times\

$$\begin{bmatrix}1 & 3 & 2\\\\0 & -1 & 4\\\\-2 & 0 & 1\\end{bmatrix}$$

$
a. $\

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

$
b. $\

$$\begin{bmatrix}3 & 0 & 0\\end{bmatrix}$$

$
c. $\

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

$
d. $\

$$\begin{bmatrix}5 & 3 & 1\\end{bmatrix}$$

$

Explanation:

4.13)

Step1: Recall transpose definition

The transpose of a matrix \( A \), denoted \( A^T \), swaps rows and columns. So for \( A =

$$\begin{bmatrix} 0 & -1 & 3 \\ 4 & -2 & 5 \\ 0 & 7 & 1 \end{bmatrix}$$

\), the first row of \( A \) becomes the first column of \( A^T \), second row becomes second column, third row becomes third column.

Step2: Construct \( A^T \)

First row of \( A \): \( [0, -1, 3] \) → first column of \( A^T \): \(

$$\begin{bmatrix} 0 \\ -1 \\ 3 \end{bmatrix}$$

\)
Second row of \( A \): \( [4, -2, 5] \) → second column of \( A^T \): \(

$$\begin{bmatrix} 4 \\ -2 \\ 5 \end{bmatrix}$$

\)
Third row of \( A \): \( [0, 7, 1] \) → third column of \( A^T \): \(

$$\begin{bmatrix} 0 \\ 7 \\ 1 \end{bmatrix}$$

\)
So \( A^T =

$$\begin{bmatrix} 0 & 4 & 0 \\ -1 & -2 & 7 \\ 3 & 5 & 1 \end{bmatrix}$$

\), which matches option B.

Step1: Matrix addition rule

To add two matrices, add corresponding elements. For \(

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

+

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

\), add element-wise:
First row, first column: \( -1 + 2 = 1 \)
First row, second column: \( 3 + (-1) = 2 \)
Second row, first column: \( 5 + 0 = 5 \)
Second row, second column: \( -2 + 4 = 2 \)

Step2: Resulting matrix

The sum is \(

$$\begin{bmatrix} 1 & 2 \\ 5 & 2 \end{bmatrix}$$

\)? Wait, no, wait: Wait, first row second column: 3 + (-1) = 2? Wait, no, the options: Wait, let's recalculate. Wait, the first matrix is \(

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

\), second is \(

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

\). So:
Row 1, Col 1: -1 + 2 = 1
Row 1, Col 2: 3 + (-1) = 2
Row 2, Col 1: 5 + 0 = 5
Row 2, Col 2: -2 + 4 = 2
Wait, but option D is \(

$$\begin{bmatrix} 1 & 2 \\ 5 & 2 \end{bmatrix}$$

\). Wait, let's check options:
A: \(

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

\) – no. B: \(

$$\begin{bmatrix} -2 & -3 \\ 0 & -8 \end{bmatrix}$$

\) – no. C: \(

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

\) – no. D: \(

$$\begin{bmatrix} 1 & 2 \\ 5 & 2 \end{bmatrix}$$

\) – yes.

Step1: Scalar multiplication rule

To multiply a matrix by a scalar, multiply each element by the scalar. For \( 3

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

\), multiply each element by 3:
First row, first column: \( 3 \times (-1) = -3 \)
First row, second column: \( 3 \times 2 = 6 \)
Second row, first column: \( 3 \times (-2) = -6 \)
Second row, second column: \( 3 \times 1 = 3 \)

Step2: Resulting matrix

The product is \(

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

\), which matches option A.

Answer:

B. \(

$$\begin{bmatrix} 0 & 4 & 0 \\ -1 & -2 & 7 \\ 3 & 5 & 1 \end{bmatrix}$$

\)

4.14)