Chapter 11.4, Problem 43AYU

To determine

To solve: system of equation using matrix inverse.

Answer to Problem 43AYU

Solution:

  $x=-5$ and $y=10$

Explanation of Solution

Given:

  $2x+y=0,x+y=5$

Calculation:

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

  $AX=B$

Where,

  $A=\left[\begin{array}{ll}2\hfill & 1\hfill \\ 1\hfill & 1\hfill \end{array}\right],X=\left[\begin{array}{l}x\hfill \\ y\hfill \end{array}\right]$ and $B=\left[\begin{array}{l}0\hfill \\ 5\hfill \end{array}\right]$

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 $\mathrm{matrix}\left(A^{-1}A\right)=I_n$

  $I_{n}X=A^{-1}B$

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

Hence,

  $X=A^{-1}B$

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

Finding $A^{-1}$

  

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

  

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

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

Solving $x$ and $y$ ,

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

  $=\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]\left[\begin{array}{l}0\hfill \\ 5\hfill \end{array}\right]$

  $=\left[\begin{array}{c}0-5\\ 0+10\end{array}\right]$

  $=\left[\begin{array}{l}-5\hfill \\ 10\hfill \end{array}\right]$

The solution for the given system of equations is $x=-5$ and $y=10$ .