Chapter 7.5, Problem 58E

a.

To determine

To determine: Use matrix multiplication to determine whether matrix

  $\left[\begin{array}{c}-4\\ -5\end{array}\right]$

is a solution of the system of equations .

Answer to Problem 58E

It is verified that

  $X=\left[\begin{array}{c}-4\\ -5\end{array}\right]$

is not the solution of given system of equation.

Explanation of Solution

Given:

The matrix which needs to verify is

  $\left[\begin{array}{c}-4\\ -5\end{array}\right]$

The system of equation is

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\end{array}$

Concept Used:

Matrix Multiplication:

Two matrix of order $m\times n$ and $p\times q$ can be multiplied to get a matrix of order $m\times q$ when , number of columns in the first matrix must be equal to the number of rows in the second matrix

that is $n=q$ .

Calculation:

The system of equation is

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\\ \\ A=\left[\begin{array}{cc}5& -7\\ 3& 1\end{array}\right],B=\left[\begin{array}{c}-15\\ 17\end{array}\right]\\ \text{Taking }X=\left[\begin{array}{c}-4\\ -5\end{array}\right]\\ AX=B\\ \text{Left hand side:}\\ AX=\left[\begin{array}{cc}5& -7\\ 3& 1\end{array}\right]\times \left[\begin{array}{c}-4\\ -5\end{array}\right]=\left[\begin{array}{c}15\\ -17\end{array}\right]\\ \text{Right hand side:}\\ B=\left[\begin{array}{c}-15\\ 17\end{array}\right]\end{array}$

In this case Left hand side is not equal to right hand side

Hence

  $\left[\begin{array}{c}-4\\ -5\end{array}\right]$

is not the solution of given system of equation.

Graph

  

From graph it is clear that

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\end{array}$

Intersect at point $\left(4,5\right)$ , therefore

  $\begin{array}{l}x=4\\ y=5\end{array}$

is the solution of the given system of equation.

Conclusion:

Hence ,it is verified that $X=\left[\begin{array}{c}-4\\ -5\end{array}\right]$ is not the solution of system of equation

b.

To determine

To determine: Use matrix multiplication to determine whether matrix

  $\left[\begin{array}{c}5\\ 2\end{array}\right]$

is a solution of the system of equations .

Answer to Problem 58E

It is verified that

  $X=\left[\begin{array}{c}5\\ 2\end{array}\right]$

is not the solution of given system of equation.

Explanation of Solution

Given:

The matrix which needs to verify is

  $\left[\begin{array}{c}5\\ 2\end{array}\right]$

The system of equation is

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\end{array}$

Concept Used:

Matrix Multiplication:

Two matrix of order $m\times n$ and $p\times q$ can be multiplied to get a matrix of order $m\times q$ when , number of columns in the first matrix must be equal to the number of rows in the second matrix

that is $n=q$ .

Calculation:

The system of equation is

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\\ \\ A=\left[\begin{array}{cc}5& -7\\ 3& 1\end{array}\right],B=\left[\begin{array}{c}-15\\ 17\end{array}\right]\\ \text{Taking }X=\left[\begin{array}{c}5\\ 2\end{array}\right]\\ AX=B\\ \text{Left hand side:}\\ AX=\left[\begin{array}{cc}5& -7\\ 3& 1\end{array}\right]\times \left[\begin{array}{c}5\\ 2\end{array}\right]=\left[\begin{array}{c}11\\ 17\end{array}\right]\\ \text{Right hand side:}\\ B=\left[\begin{array}{c}-15\\ 17\end{array}\right]\end{array}$

In this case Left hand side is not equal to right hand side

Hence

  $\left[\begin{array}{c}5\\ 2\end{array}\right]$

is not the solution of given system of equation.

Graph

  

From graph it is clear that

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\end{array}$

Intersect at point $\left(4,5\right)$ , therefore

  $\begin{array}{l}x=4\\ y=5\end{array}$

is the solution of the given system of equation.

Conclusion:

Hence ,it is verified that $X=\left[\begin{array}{c}5\\ 2\end{array}\right]$ is not the solution of system of equation

c.

To determine

To determine: Use matrix multiplication to determine whether matrix

  $\left[\begin{array}{c}4\\ 5\end{array}\right]$

is a solution of the system of equations .

Answer to Problem 58E

It is verified that

  $X=\left[\begin{array}{c}4\\ 5\end{array}\right]$

is the solution of given system of equation.

Explanation of Solution

Given:

The matrix which needs to verify is

  $\left[\begin{array}{c}4\\ 5\end{array}\right]$

The system of equation is

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\end{array}$

Concept Used:

Matrix Multiplication:

Two matrix of order $m\times n$ and $p\times q$ can be multiplied to get a matrix of order $m\times q$ when , number of columns in the first matrix must be equal to the number of rows in the second matrix

that is $n=q$ .

Calculation:

The system of equation is

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\\ \\ A=\left[\begin{array}{cc}5& -7\\ 3& 1\end{array}\right],B=\left[\begin{array}{c}-15\\ 17\end{array}\right]\\ \text{Taking }X=\left[\begin{array}{c}4\\ 5\end{array}\right]\\ AX=B\\ \text{Left hand side:}\\ AX=\left[\begin{array}{cc}5& -7\\ 3& 1\end{array}\right]\times \left[\begin{array}{c}4\\ 5\end{array}\right]=\left[\begin{array}{c}-15\\ 17\end{array}\right]\\ \text{Right hand side:}\\ B=\left[\begin{array}{c}-15\\ 17\end{array}\right]\end{array}$

In this case Left hand side is equal to right hand side

Hence

  $\left[\begin{array}{c}4\\ 5\end{array}\right]$

is the solution of given system of equation.

Graph

  

From graph it is clear that

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\end{array}$

Intersect at point $\left(4,5\right)$ , therefore

  $\begin{array}{l}x=4\\ y=5\end{array}$

is the solution of the given system of equation.

Conclusion:

Hence ,it is verified that $X=\left[\begin{array}{c}4\\ 5\end{array}\right]$ is the solution of system of equation

d.

To determine

To determine: Use matrix multiplication to determine whether matrix

  $\left[\begin{array}{c}2\\ 11\end{array}\right]$

is a solution of the system of equations .

Answer to Problem 58E

It is verified that

  $X=\left[\begin{array}{c}2\\ 11\end{array}\right]$

is not the solution of given system of equation.

Explanation of Solution

Given:

The matrix which needs to verify is

  $\left[\begin{array}{c}2\\ 11\end{array}\right]$

The system of equation is

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\end{array}$

Concept Used:

Matrix Multiplication:

Two matrix of order $m\times n$ and $p\times q$ can be multiplied to get a matrix of order $m\times q$ when , number of columns in the first matrix must be equal to the number of rows in the second matrix

that is $n=q$ .

Calculation:

The system of equation is

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\\ \\ A=\left[\begin{array}{cc}5& -7\\ 3& 1\end{array}\right],B=\left[\begin{array}{c}-15\\ 17\end{array}\right]\\ \text{Taking }X=\left[\begin{array}{c}2\\ 11\end{array}\right]\\ AX=B\\ \text{Left hand side:}\\ AX=\left[\begin{array}{cc}5& -7\\ 3& 1\end{array}\right]\times \left[\begin{array}{c}2\\ 11\end{array}\right]=\left[\begin{array}{c}-67\\ 17\end{array}\right]\\ \text{Right hand side:}\\ B=\left[\begin{array}{c}-15\\ 17\end{array}\right]\end{array}$

In this case Left hand side is equal to right hand side

Hence

  $\left[\begin{array}{c}2\\ 11\end{array}\right]$

is not the solution of given system of equation.

Graph

  

From graph it is clear that

  $\begin{array}{l}5x-7y=-15\\ 3x+y=17\end{array}$

Intersect at point $\left(4,5\right)$ , therefore

  $\begin{array}{l}x=4\\ y=5\end{array}$

is the solution of the given system of equation.

Conclusion:

Hence ,it is verified that $X=\left[\begin{array}{c}2\\ 11\end{array}\right]$ is not the solution of system of equation