Solve each equation.52

$\left[\begin{array}{c}\text{x}+3\text{z}\\ -2\text{x}+\text{y}-\text{z}\\ 5\text{y}-7\text{z}\end{array}\right]=\left[\begin{array}{c}-19\\ -2\\ 24\end{array}\right]$

A rectangular array of numbers arranged in rows and columns is called a matrix

Using the definition of equal matices solve the equations.

If the dimension of the matrices are same and corresponding elements are equal then the matrices are equal.

Here the matrices are:$\left[\begin{array}{c}x+3z\\ -2x+y-z\\ 5y-7z\end{array}\right]=\left[\begin{array}{c}-19\\ -2\\ 24\end{array}\right]$

Since, the matrices are equal then corresponding elements are also equal.

Then, the three linear equations formed are:$x+3z=-19$ (1)

$-2x+y-z=-2$ (2)

$5y-7z=24$ (3)

Solve the equations using substitution.

Solve equation 1 as:

$\begin{array}{c}x+3z=-19\\ x=-19-3z\end{array}$

Solve equation 2 as:

$\begin{array}{c}-2x+y-z=-2\\ y=-2+2x+z\end{array}$

Substitute y with $y=-2+2x+z$ in equation 3:

$\begin{array}{c}5y-7z=24\\ 5(-2+2x+z)-7z=24\\ -10+10x+5z-7z=24\end{array}$

That implies,$-10+10x-2z=24$ (4)

To obtain the value of z , substitue x with $x=-19-3z$ in equation equation 4,

$\begin{array}{c}-10+10x-2z=24\\ -10+10(-19-3z)-2z=24\\ -10-190-30z-2z=24\\ -200-32z=24\end{array}$

Further simplify as,

$\begin{array}{c}-200-32z+200=24+200\\ -32z=224\\ z=-7\end{array}$

To obatin value of x , put the value of z in equation 1,

$\begin{array}{c}x+3z=-19\\ x+(3\times -7)=-19\\ x-21=-19\\ x-21+21=-19+21\\ x=2\end{array}$

To obatin value of y , put the value of x and z in equation 2,

$\begin{array}{c}-2x+y-z=-2\\ (-2\times 2)+y-(-7)=-2\\ -4+y+7=-2\\ y+3=-2\\ y+3-3=-2-3\\ y=-5\end{array}$

Therefore, $x=2$, $y=-5$ and $z=-7$. The solution is $(2,-5,-7)$.