Chapter 11.4, Problem 47AYU

To determine

To solve: system of equation using matrix inverse.

Answer to Problem 47AYU

Solution: $x=\frac{1}{2}$ and $y=2$

Explanation of Solution

Given:

  $6x+5y=13$

  $2x+2y=5$

Calculation:

The original system of equations can be written compactly as a matrix equation

  $AX=B$

Where,

  $A=\left[\begin{array}{ll}6\hfill & 5\hfill \\ 2\hfill & 2\hfill \end{array}\right],X=\left[\begin{array}{l}x\hfill \\ y\hfill \end{array}\right]$ and $B=\left[\begin{array}{c}13\\ 5\end{array}\right]$

To solve the for $\text{x}$ and $\text{y}$ , the solution of $\text{X}$ has to be found as given below:

  $AX=B$

Multiply both sides by $A^{-1}$

  $A^{-1}(AX)=A^{-1}B$

By association property of matrix multiplication, the above equation can be rewritten as

  $\left(A^{-1}A\right)X=A^{-1}B$

By the definition of inverse matrix $\left(A^{-1}A\right)=I_n$ .

  $I_{n}X=A^{-1}B$

By the property of Identity matrix $,I_{n}X=X$

Hence,

  $X=A^{-1}B$

i.e. $\left[\begin{array}{l}x\hfill \\ y\hfill \end{array}\right]=A^{-1}B$

Step 1: Construction of $\left[A\mid I_n\right]$ .

  

Step 2: Transform the matrix $\left[A\mid I_n\right]$ into reduced row echelon form.

  

Step- 3 ; Separate the matrix on the right of the vertical bar which is the inverse of $A$ .

  $A^{-1}=\left[\begin{array}{rr}\hfill 1& \hfill -\frac{5}{2}\\ \hfill -1& \hfill 3\end{array}\right]$

Solving $x$ and $y$ .

  $\left[\begin{array}{l}x\hfill \\ y\hfill \end{array}\right]=A^{-1}B$

  $\begin{array}{l}=\left[\begin{array}{cc}1& -\frac{5}{2}\\ -1& 3\end{array}\right]\left[\begin{array}{c}13\\ 5\end{array}\right]\hfill \\ =\left[\begin{array}{c}13-\frac{25}{2}\\ -13+15\end{array}\right]\hfill \\ =\left[\begin{array}{l}\frac{1}{2}\hfill \\ \frac{2}{2}\hfill \end{array}\right]\hfill \end{array}$

The solution for the given system of equations is $x=\frac{1}{2}$ and $y=2$ .