Chapter 11.4, Problem 52AYU

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

$\{\begin{array}{l}bx+3y=14\\ bx+2y=10\end{array}b\ne 0$

To determine

To solve: System of equation using matrix inverse.

Answer to Problem 52AYU

Solution:

$x=\frac{2}{b}\text{ and }y=4$.

Explanation of Solution

Given:

$bx+3y=14,bx+2y=10$ where $b\ne 0$.

Calculation:

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

$AX=B$

Where,

$A=\left[\begin{array}{cc}b& 3\\ b& 2\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]\text{ and }B=\left[\begin{array}{c}14\\ 10\end{array}\right]$

To solve the for $x\text{ and }y$, 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. $\left[\begin{array}{c}x\\ y\end{array}\right]=A^{-1}B$.

Determining $A^{-1}$.

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

$\left[A|I_n\right]=[\begin{array}{cc}b& 3\\ b& 2\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}]$

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

$=[\begin{array}{cc}b& 3\\ b& 2\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}]$

$\underset{R_1=\frac{1}{b}*r_1}{\to } [\begin{array}{cc}1& \frac{3}{b}\\ b& 2\end{array}|\begin{array}{cc}\frac{1}{b}& 0\\ 0& 1\end{array}]$

$\underset{R_2=r_2-\left(b\right)r_1}{\to } [\begin{array}{cc}1& \frac{3}{b}\\ 0& -1\end{array}|\begin{array}{cc}\frac{1}{b}& 0\\ -1& 1\end{array}]$

$\underset{\begin{array}{c}{R}_1=r_1-\left(-\frac{3}{b}\right)r_2\\ R_2\to \left(-1\right)r_2\end{array}}{\to } [\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}]$

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

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

Solving $x\text{ and }y$.

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

$=\left[\begin{array}{cc}-\frac{2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]\left[\begin{array}{c}14\\ 10\end{array}\right]$

$=\left[\begin{array}{c}-\frac{28}{b}+\frac{30}{b}\\ 14-10\end{array}\right]$

$=\left[\begin{array}{c}\frac{2}{b}\\ 4\end{array}\right]$

The solution for the given system of equations is $x=\frac{2}{b}\text{ and }y=4$.