Question

2x+5y+3z=2

x+2y+2z=4

x+y+4z=11. Use the elementary row reduction to solve for the linear system.

Answer 1

$\begin{pmatrix} 2 & 5&3&|&2\\ 1 & 2&2&|&4\\1&1&4&|&11 \end{pmatrix}$

$R_{1}\leftrightarrow R_{3}\begin{pmatrix} 1&1&4&|&11\\ 1 & 2&2&|&4\\2 & 5&3&|&2 \end{pmatrix}$

$-R_{1}+R_{2}\rightarrow R_{2}$

$-2R_{1}+R_{3}\rightarrow R_{3}$ $\begin{pmatrix} 1&1&4&|&11\\ 0 & 1&-2&|&-7\\0 & 3&-5&|&-20 \end{pmatrix}$

$-3R_{2}+R_{3}\rightarrow R_{3}$ $\begin{pmatrix} 1&1&4&|&11\\ 0 & 1&-2&|&-7\\0 & 0&1&|&1 \end{pmatrix}$

Hence, we have;

$x+y+4z=11$

$y-2z=-7$

$z=1$

$=> y=-7+2z=-7+2(1)=-5$

$=>x=11-4z-y=11-4(1)-(-5)=12$

$∴ x=12, y=-5, z=1$