Chapter 11.4, Problem 42AYU

To determine

To solve: system of equation using matrix inverse.

Answer to Problem 42AYU

Solution:

  $x=12$ and $y=28$

Explanation of Solution

Given:

  $3x-y=8,-2x+y=4$

Calculation:

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

  $AX=B$

Where,

  $A=\left[\begin{array}{cc}3& -1\\ -2& 1\end{array}\right],X=\left[\begin{array}{l}x\hfill \\ y\hfill \end{array}\right]$ and $B=\left[\begin{array}{l}8\hfill \\ 4\hfill \end{array}\right]$

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 re written as,

  $\left(A^{-1}A\right)X=A^{-1}B$

By the definition of inverse $ma\mathrm{trix}\left(A^{-1}A\right)=I_n$

  $I_{n}X=A^{-1}B$

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

Hence,

  $X=A^{-1}B$

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

Finding $A^{-1}$

Step-1: Construction $\rho f\left[A\mid I_n\right]$ .

  

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

  

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

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

Solving $x$ and $y$

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

  $\begin{array}{l}=\left[\begin{array}{ll}1\hfill & 1\hfill \\ 2\hfill & 3\hfill \end{array}\right]\left[\begin{array}{l}8\hfill \\ 4\hfill \end{array}\right]\hfill \\ =\left[\begin{array}{c}8+4\\ 16+12\end{array}\right]\hfill \\ =\left[\begin{array}{c}12\\ 28\end{array}\right]\hfill \end{array}$

The solution for the given system of equations is $x=12$ and $y=28$ .