Chapter 11.4, Problem 56AYU

To determine

To solve: System of equation using matrix inverse.

Answer to Problem 56AYU

Solution:

  $x=1,y=-1$ and $z=\frac{1}{2}$

Explanation of Solution

Given:

  $x+2z=2$

  $-x+2y+3z=-3/2$

  $x-y=2$

Calculation:

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

  $AX=B$

Where,

  $A=\left[\begin{array}{ccc}1& 0& 2\\ -1& 2& 3\\ 1& -1& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $B=\left[\begin{array}{c}2\\ -\frac{3}{2}\\ 2\end{array}\right]$

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

Determining $A^{-1}$

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

  

Step 2: Transform the matrix [A|I ${}_n$ ] into reduced row echelon form.

  

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

  $A^{-1}=\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]$

Solving x, y and $z$

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

  $=\left[\begin{array}{ccc}3& -2& -4\\ 3& -2& -5\\ -1& 1& 2\end{array}\right]\left[\begin{array}{c}2\\ -\frac{3}{2}\end{array}\right]$

  $\left[\begin{array}{cc}6+3-8& 7\\ 6+3-10& \\ -2-\frac{3}{2}+4& \end{array}\right]$

  $=\left[\begin{array}{c}1\\ -1\\ \frac{1}{2}\end{array}\right]$

The solution for the given system of equations is $x=1,y=-1$ and $z=1/2$ .