Question

1. use the formula for finding the inverse of the 2 x 2 square matrix.

$m = \

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

$
$m^{-1} = \frac{1}{\text{det}m}\

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

$
options:
$\

$$\begin{bmatrix} \\frac{1}{7} & \\frac{3}{14} \\\\ -\\frac{2}{7} & \\frac{1}{14} \\end{bmatrix}$$

$
$\

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

$
$\

$$\begin{bmatrix} 7 & 14 \\\\ 7 & 14 \\end{bmatrix}$$

$
$\

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

$

1. determine if the matrices are inverses of each other by multiplying $a \times b$.

$a = \

$$\begin{bmatrix}2 & 3 \\\\ 4 & 7 \\end{bmatrix}$$

$ and $b = \

$$\begin{bmatrix} \\frac{7}{2} & -\\frac{3}{2} \\\\ -2 & 1 \\end{bmatrix}$$

$
options: no, yes

Explanation:

Question 7

Step1: Calculate determinant of \( M \)

For a matrix \( M =

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

\), \( \det M = ad - bc \). Here, \( a = 1 \), \( b = -3 \), \( c = 4 \), \( d = 2 \). So \( \det M = (1)(2) - (-3)(4) = 2 + 12 = 14 \).

Step2: Apply inverse formula

Using \( M^{-1} = \frac{1}{\det M}

$$\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

\), substitute \( a = 1 \), \( b = -3 \), \( c = 4 \), \( d = 2 \), \( \det M = 14 \). So \( M^{-1} = \frac{1}{14}

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

=

$$\begin{bmatrix} \frac{2}{14} & \frac{3}{14} \\ \frac{-4}{14} & \frac{1}{14} \end{bmatrix}$$

=

$$\begin{bmatrix} \frac{1}{7} & \frac{3}{14} \\ -\frac{2}{7} & \frac{1}{14} \end{bmatrix}$$

\).

Step1: Multiply matrices \( A \) and \( B \)

Matrix multiplication: For \( A =

$$\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$

\) and \( B =

$$\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$$

\), \( A \times B =

$$\begin{bmatrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \end{bmatrix}$$

\). Here, \( A =

$$\begin{bmatrix} 2 & 3 \\ 4 & 7 \end{bmatrix}$$

\), \( B =

$$\begin{bmatrix} \frac{7}{2} & -\frac{3}{2} \\ -2 & 1 \end{bmatrix}$$

\).

Step2: Compute each element

First row first column: \( 2\times\frac{7}{2} + 3\times(-2) = 7 - 6 = 1 \).
First row second column: \( 2\times(-\frac{3}{2}) + 3\times1 = -3 + 3 = 0 \).
Second row first column: \( 4\times\frac{7}{2} + 7\times(-2) = 14 - 14 = 0 \).
Second row second column: \( 4\times(-\frac{3}{2}) + 7\times1 = -6 + 7 = 1 \).
So \( A \times B =

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

\), which is the identity matrix. Thus, \( A \) and \( B \) are inverses.

Answer:

\(

$$\begin{bmatrix} \frac{1}{7} & \frac{3}{14} \\ -\frac{2}{7} & \frac{1}{14} \end{bmatrix}$$

\) (the first option)

Question 8