Question

Use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, use t for the parameter.) x1 − x2 − x3 = −1 2x1 + 3x2 + 5x3 = −5 x1 − 2x2 + 3x3 = −22

Answer 1

Answer:

$x_{1} = 0\\ x_{2} = 5 \\x_{3} =-4$

Step-by-step explanation:

$\left[\begin{array}{cccc}1&-1&-1&-1\\2&3&5&-5\\1&-2&3&-22\end{array}\right]$

new $R_{2}$=$R_{2} -2R_{3}$

= $\left[\begin{array}{cccc}1&-1&-1&-1\\0&7&-1&39\\1&-2&3&-22\end{array}\right]$

(first )new $R_{3} = R_{3} -R_{1}$

=$\left[\begin{array}{cccc}1&-1&-1&-1\\0&7&-1&39\\0&1&-4&21\end{array}\right]$

to make $R_{32}$ = 0 from the new matrix;

(second ) new $R_{3}$ = new $R_{2}$ - (first) new $R_{3}$

= $\left[\begin{array}{cccc}1&-1&-1&-1\\0&7&-1&39\\0&0&27&-108\end{array}\right]$ ------------------------- eqn k

therefore $27x_{3}= -108\\x_{3} = -4$

$7x_{2}-x_{3}=39$

$x_{2} = 5$

And $x_{1}- x_{2}- x_{3} =-1$ from first line of equation k

$x_{1} = 0$

Hence

$x_{1} = 0\\ x_{2} = 5 \\x_{3} =-4$