Chapter 11.4, Problem 55AYU

To determine

To solve: System of equation using matrix inverse.

Answer to Problem 55AYU

Solution:

  $x=\frac{1}{2},y=-\frac{1}{2}\mathrm{and}z=1$

Explanation of Solution

Given:

  $x-y+z=2$

  $-2y+z=2$

  $-2x-3y=1/2$

Calculation:

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

  $AX=B$

Where,

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

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

Detemining $A^{-1}$

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.

  

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

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

Solving x, y and $z$

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

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

  $=\left[\begin{array}{c}6-6+\frac{1}{2}\\ -4+4-\frac{1}{2}\\ -8+10-1\end{array}\right]$

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

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