Chapter 7.5, Problem 57E

a.

To determine

To determine: Using matrix multiplication to determine whether matrix

  $\left[\begin{array}{c}2\\ 1\end{array}\right]$

is a solution of the system of equations .

Answer to Problem 57E

It is verified that

  $X=\left[\begin{array}{c}2\\ 1\end{array}\right]$

is not the solution of system of equation

Explanation of Solution

Given:

The matrix which needs to verify is

  $\left[\begin{array}{c}2\\ 1\end{array}\right]$

The system of equation is

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\end{array}$

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

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\\ \\ A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right],B=\left[\begin{array}{c}4\\ 0\end{array}\right]\\ \text{Taking }X=\left[\begin{array}{c}2\\ 1\end{array}\right]\\ AX=B\\ \text{Left hand side:}\\ AX=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]\times \left[\begin{array}{c}2\\ 1\end{array}\right]=\left[\begin{array}{c}4\\ 8\end{array}\right]\\ \text{Right hand side:}\\ B=\left[\begin{array}{c}4\\ 0\end{array}\right]\end{array}$

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

Hence

  $\left[\begin{array}{c}2\\ 1\end{array}\right]$

is not the solution of given system of equation.

Graph

  

From graph it is clear that

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\end{array}$

Intersect at point $\left(-2,3\right)$ , therefore

  $\begin{array}{l}x=-2\\ y=3\end{array}$

is the solution of the given system of equation.

Conclusion:

Hence ,it is verified that $X=\left[\begin{array}{c}2\\ 1\end{array}\right]$ is not the solution of system of equation

ii.

To determine

To determine: Using matrix multiplication to determine whether matrix

  $\left[\begin{array}{c}-2\\ 3\end{array}\right]$

is a solution of the system of equations .

Answer to Problem 57E

It is verified that

  $X=\left[\begin{array}{c}-2\\ 3\end{array}\right]$

is the solution of given system of equation

Explanation of Solution

Given:

The matrix which needs to verify is

  $\left[\begin{array}{c}-2\\ 3\end{array}\right]$

The system of equation is

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\end{array}$

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

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\\ \\ A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right],B=\left[\begin{array}{c}4\\ 0\end{array}\right]\\ \text{Taking }X=\left[\begin{array}{c}-2\\ 3\end{array}\right]\\ AX=B\\ \text{Left hand side:}\\ AX=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]\times \left[\begin{array}{c}-2\\ 3\end{array}\right]=\left[\begin{array}{c}4\\ 0\end{array}\right]\\ \text{Right hand side:}\\ B=\left[\begin{array}{c}4\\ 0\end{array}\right]\end{array}$

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

Hence

  $\left[\begin{array}{c}-2\\ 3\end{array}\right]$

is the solution of given system of equation.

Graph:

  

From graph it is clear that

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\end{array}$

Intersect at point $\left(-2,3\right)$ , therefore

  $\begin{array}{l}x=-2\\ y=3\end{array}$

is the solution of the given system of equation.

Conclusion:

Hence ,it is verified that

  $X=\left[\begin{array}{c}-2\\ 3\end{array}\right]$

is the solution of given system of equation

c.

To determine

To determine: Using matrix multiplication to determine whether matrix

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

is a solution of the system of equations .

Answer to Problem 57E

It is verified that

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

is the not the solution of given system of equation.

Explanation of Solution

Given:

The matrix which needs to verify is

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

The system of equation is

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\end{array}$

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

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\\ \\ A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right],B=\left[\begin{array}{c}4\\ 0\end{array}\right]\\ \text{Taking }X=\left[\begin{array}{c}-4\\ 4\end{array}\right]\\ AX=B\\ \text{Left hand side:}\\ AX=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]\times \left[\begin{array}{c}-4\\ 4\end{array}\right]=\left[\begin{array}{c}4\\ -4\end{array}\right]\\ \text{Right hand side:}\\ B=\left[\begin{array}{c}4\\ 0\end{array}\right]\end{array}$

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

Hence

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

is the not the solution of given system of equation.

Graph:

  

From graph it is clear that

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\end{array}$

Intersect at point $\left(-2,3\right)$ , therefore

  $\begin{array}{l}x=-2\\ y=3\end{array}$

is the solution of the given system of equation.

Conclusion:

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

d.

To determine

To determine: Using matrix multiplication to determine whether matrix

  $\left[\begin{array}{c}2\\ -3\end{array}\right]$

is a solution of the system of equations.

Answer to Problem 57E

It is verified that

  $X=\left[\begin{array}{c}2\\ -3\end{array}\right]$

is the not the solution of given system of equation.

Explanation of Solution

Given:

The matrix which needs to verify is

  $\left[\begin{array}{c}2\\ -3\end{array}\right]$

The system of equation is

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\end{array}$

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

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\\ \\ A=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right],B=\left[\begin{array}{c}4\\ 0\end{array}\right]\\ \text{Taking }X=\left[\begin{array}{c}2\\ -3\end{array}\right]\\ AX=B\\ \text{Left hand side:}\\ AX=\left[\begin{array}{cc}1& 2\\ 3& 2\end{array}\right]\times \left[\begin{array}{c}2\\ -3\end{array}\right]=\left[\begin{array}{c}-4\\ 0\end{array}\right]\\ \text{Right hand side:}\\ B=\left[\begin{array}{c}4\\ 0\end{array}\right]\end{array}$

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

Hence

  $\left[\begin{array}{c}2\\ -3\end{array}\right]$

is the not the solution of given system of equation.

Graph:

  

From graph it is clear that

  $\begin{array}{l}x+2y=4\\ 3x+2y=0\end{array}$

Intersect at point $\left(-2,3\right)$ , therefore

  $\begin{array}{l}x=-2\\ y=3\end{array}$

is the solution of the given system of equation.

Conclusion:

Hence ,it is verified that

  $X=\left[\begin{array}{c}2\\ -3\end{array}\right]$

is the not the solution of given system of equation.