Chapter 11.4, Problem 76AYU

To determine

To solve: The system of equations algebraically by any method.

Answer to Problem 76AYU

Solution:

  $x=8,y=-3$

Explanation of Solution

Given:

  $2x+8y=-8$

  $x+7y=-13$

Calculation:

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

  $AX=B$

Where,

  $A=\left[\begin{array}{ll}2\hfill & 8\hfill \\ 1\hfill & 7\hfill \end{array}\right],X=\left[\begin{array}{l}x\hfill \\ y\hfill \end{array}\right]$ and $B=\left[\begin{array}{c}-8\\ -13\end{array}\right]$

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 defmition 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 of $\left[A\mid I_n\right].$

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

  $\text{Interchanging Rows}$ ,

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

  $A^{-1}=\left[\begin{array}{cc}\frac{7}{6}& -\frac{4}{3}\\ -\frac{1}{6}& \frac{1}{3}\end{array}\right]$

Solving $x$ and $y$ ?

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

  $=\left[\begin{array}{cc}\frac{7}{6}& -\frac{4}{3}\\ -\frac{1}{6}& \frac{1}{3}\end{array}\right]\left[\begin{array}{c}-8\\ -13\end{array}\right]$

  $=\left[\begin{array}{c}-\frac{56}{6}+\frac{52}{3}\\ \frac{8}{6}-\frac{13}{3}\end{array}\right]$

  $=\left[\begin{array}{c}8\\ -3\end{array}\right]$

The solution for the given system of equations is $x=8,y=-3$ .