Question

1. determine if the matrices are inverses of each other by multiplying the matrices.

$x = \

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

$ and $y = \

$$\begin{bmatrix}\\frac{1}{4} & 0 \\\\ -\\frac{1}{8} & \\frac{1}{2}\\end{bmatrix}$$

$
\bigcirc yes
\bigcirc no

1. select yes if the determinant shows that the matrix will reduce to an identity matrix; select no if it does not.

$\

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

$
\bigcirc no
\bigcirc yes

1. find the result of applying the given elementary row operations.

$\

$$\begin{bmatrix}1 & 3 & \\bigm| & 1 & 0 \\\\ 5 & 2 & \\bigm| & 0 & 1\\end{bmatrix}$$

r_2 + (-5r_1) \to r_2 \to$

Explanation:

Question 8

Step1: Recall matrix multiplication rule

For two matrices \( A =

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

\) and \( B =

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

\), the product \( AB \) is \(

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

\). We need to multiply \( X \) and \( Y \) and check if the result is the identity matrix \( I =

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

\).

Step2: Multiply \( X \) and \( Y \)

Given \( X =

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

\) and \( Y =

$$\begin{bmatrix} \frac{1}{4} & 0 \\ -\frac{1}{8} & \frac{1}{2} \end{bmatrix}$$

\)

First element (1,1): \( 4\times\frac{1}{4}+0\times(-\frac{1}{8}) = 1 + 0 = 1 \)

First element (1,2): \( 4\times0 + 0\times\frac{1}{2}=0 + 0 = 0 \)

Second element (2,1): \( 1\times\frac{1}{4}+2\times(-\frac{1}{8})=\frac{1}{4}-\frac{2}{8}=\frac{1}{4}-\frac{1}{4}=0 \)

Second element (2,2): \( 1\times0 + 2\times\frac{1}{2}=0 + 1 = 1 \)

So \( XY=

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

\), which is the identity matrix.

Step1: Recall determinant formula for \( 2\times2 \) matrix

For matrix \( A=

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

\), determinant \( \det(A)=ad - bc \). If \( \det(A)
eq0 \), the matrix is invertible (can reduce to identity matrix), else not.

Step2: Calculate determinant of given matrix

Given matrix \(

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

\), \( a = 2, b = 3, c = 4, d = 6 \)

\( \det(A)=2\times6 - 3\times4 = 12 - 12 = 0 \)

Since determinant is 0, the matrix is not invertible and cannot reduce to identity matrix.

Step1: Recall elementary row operation

The operation is \( R_2 + (-5R_1) \to R_2 \), which means we take row 2, subtract 5 times row 1, and replace row 2 with the result.

Given augmented matrix \(

$$\begin{bmatrix} 1 & 3 &\bigm| & 1 & 0 \\ 5 & 2 &\bigm| & 0 & 1 \end{bmatrix}$$

\) (assuming the matrix is \(

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

\) as per the format, with first two columns as coefficient matrix and last two as identity part for inverse maybe, but the operation is on row 2)

Step2: Apply the operation to each element of row 2

Row 1: \( [1, 3, 1, 0] \)

Row 2 original: \( [5, 2, 0, 1] \)

New row 2: \( [5 - 5\times1, 2 - 5\times3, 0 - 5\times1, 1 - 5\times0] \)

Calculate each element:

First element: \( 5 - 5 = 0 \)

Second element: \( 2 - 15 = -13 \)

Third element: \( 0 - 5 = -5 \)

Fourth element: \( 1 - 0 = 1 \)

So the new matrix is \(

$$\begin{bmatrix} 1 & 3 & 1 & 0 \\ 0 & -13 & -5 & 1 \end{bmatrix}$$

\)

Answer:

yes

Question 9