Chapter 11.4, Problem 47AYU

In Problems 45-64, use the inverses found in Problems 35-44 to solve each system of equations.

$\{\begin{array}{l}6x+5y=13\\ 2x+2y=5\end{array}$

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:

$6x+\text{ 5}y=\text{ 13}$

$2x+\text{ 2}y=\text{ 5}$

Calculation:

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

$AX=B$

Where,

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

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

$AX=B$

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

$A^{-1}\left(AX\right)=A^{-1}B$

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

$(A^{-1}A)X=A^{-1}B$

By the definition of inverse matrix $(A^{-1}A)= I_{n.}$.

$I_{n}X=A^{-1}B$

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

Hence,

$X=A^{-1}B$

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

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

$\left[A|I_n\right] = [\begin{array}{cc}6& 5\\ 2& 2\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}]$

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

$= [\begin{array}{cc}6& 5\\ 2& 2\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}]$

$\underset{R_1= \left(\frac{1}{6}\right)r_1}{\to } [\begin{array}{cc}1& \frac{5}{6}\\ 2& 2\end{array}|\begin{array}{cc}\frac{1}{6}& 0\\ 0& 1\end{array}]$

$\underset{R_2= r_2-2r_1}{\to } [\begin{array}{cc}1& \frac{5}{6}\\ 0& \frac{1}{3}\end{array}|\begin{array}{cc}\frac{1}{6}& 0\\ -\frac{1}{3}& 1\end{array}]$

$\underset{R_1= r_1-\left(\frac{5}{2}\right) r_2}{\to } [\begin{array}{cc}1& 0\\ 0& \frac{1}{3}\end{array}|\begin{array}{cc}1& -\frac{5}{2}\\ -\frac{1}{3}& 1\end{array}]$

$\underset{R_2= 3r_2}{\to } [\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{cc}1& -\frac{5}{2}\\ -1& 3\end{array}]$

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

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

Solving $x$ and $y$,

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

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

$=\left[\begin{array}{c}13-\frac{25}{2}\\ -13+15\end{array}\right]$

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

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