Chapter 10.5, Problem 35E

To determine

To calculate: Solution of the equations given.

Answer to Problem 35E

Solution of the equations is $x=3,y=-2,z=2$

Explanation of Solution

Given information:

  $\{\begin{array}{l}12x+\frac{1}{2}y-7z=21\\ 11x-2y+3z=43\\ 13x+y-4z=29\end{array}$

Calculation:

Given equations are,

  $\{\begin{array}{l}12x+\frac{1}{2}y-7z=21\\ 11x-2y+3z=43\\ 13x+y-4z=29\end{array}$

Rewrite the system as a matrix equation of form $AX=B$

  $\left[\begin{array}{ccc}24& 1& -14\\ 11& -2& 3\\ 13& 1& -4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}42\\ 43\\ 29\end{array}\right]$

Where,

  $A=\left[\begin{array}{ccc}24& 1& -14\\ 11& -2& 3\\ 13& 1& -4\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}42\\ 43\\ 29\end{array}\right]$

Using rule for finding the inverse of $3\times 3$ matrix, we get $B=adjmatrixofA$

  $\begin{array}{l}c=adjmatrixofA\\ c=\left[\begin{array}{ccc}5& 83& 37\\ -10& 86& -11\\ -25& -226& -59\end{array}\right]\end{array}$

Then,

  $\begin{array}{l}adjA=C^r=\left[\begin{array}{ccc}5& 83& 37\\ -10& 86& -11\\ -25& -226& -59\end{array}\right]\\ \left|A\right|=24.5+1.83-14.37\\ =120+83-518\\ =-315\end{array}$

Then,

  $\begin{array}{l}X=A^{-1}B\\ =\frac{-1}{315}\left[\begin{array}{ccc}5& 83& 37\\ -10& 86& -11\\ -25& -226& -59\end{array}\right]\left[\begin{array}{c}42\\ 43\\ 29\end{array}\right]\\ X=\frac{-1}{315}\left[\begin{array}{ccc}5& 83& 37\\ -10& 86& -11\\ -25& -226& -59\end{array}\right]\left[\begin{array}{c}42\\ 43\\ 29\end{array}\right]\\ =\frac{-1}{315}\left[\begin{array}{c}-945\\ 630\\ -630\end{array}\right]\\ =\left[\begin{array}{c}3\\ -2\\ 2\end{array}\right]\end{array}$

Thus, $x=3,y=-2,z=2$ is unique solution of given system.