Chapter 11.4, Problem 52AYU

To determine

To solve: System of equation using matrix inverse.

Answer to Problem 52AYU

Solution:

  $x=\frac{2}{b}$ and $=4$

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,

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

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

  $AX=B$

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

  $A^{-1}(AX)=A^{-1}B$

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

Determining $A^{-1}$

Step 1: Construction $\mathrm{of}\left[A\mid I_n\right]$

  

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

  

  

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

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

Solving $x$ and $y$ .

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

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

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

  $=\left[\frac{2}{b}\right]$

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