Question

How do you solve the system of equations: $2x+6y+3z=5$, $-2x+2y+z=15$ and $-3y+4z=9$ ?

Answer 1

Hint: We have been given three equations of three planes which have to be solved simultaneously. Each equation consists of three variables, x, y and z. Thus, these equations cannot be solved by simple substitution or elimination. In order to solve three plane equations in 3 variables simultaneously, we shall use the cramer’s rule.

Complete step by step answer:
Given a system of equations: $2x+6y+3z=5$, $-2x+2y+z=15$ and $-3y+4z=9$
We shall form four different $3\times 3$ matrices from the given system of equations.
In the first matrix, $D$ we shall write the coefficients of x, y and z in each column from each of the three equations.
$D=\left[ \begin{matrix}
   2 & 6 & 3 \\
   -2 & 2 & 1 \\
   0 & -3 & 4 \\
\end{matrix} \right]$
We will calculate the determinant, $\left| D \right|$ of this matrix by expanding it along the column.
$\Rightarrow \left| D \right|=2\left( 2\times 4-\left( -3 \right)\times 1 \right)-6\left( \left( -2\times 4 \right)-0\times 1 \right)+3\left( \left( -2\times -3 \right)-0\times 2 \right)$
$\Rightarrow \left| D \right|=2\left( 8+3 \right)-6\left( -8 \right)+3\left( 6 \right)$
$\Rightarrow \left| D \right|=22+48+18$
$\Rightarrow \left| D \right|=88$
In the second matrix, in column 1, we will write the constant terms of the given equations, in column 2, we will write the coefficients of variable-y of all the equations and in column 3, we shall write the coefficients of variable-z of all the given equations.
${{D}_{x}}=\left[ \begin{matrix}
   5 & 6 & 3 \\
   15 & 2 & 1 \\
   9 & -3 & 4 \\
\end{matrix} \right]$
We will calculate the determinant, $\left| {{D}_{x}} \right|$ of this matrix by expanding it along the column.
\[\Rightarrow \left| {{D}_{x}} \right|=5\left( 2\times 4-\left( -3 \right)\times 1 \right)-6\left( \left( 15\times 4 \right)-9\times 1 \right)+3\left( \left( 15\times -3 \right)-9\times 2 \right)\]
$\Rightarrow \left| {{D}_{x}} \right|=5\left( 8+3 \right)-6\left( 60-9 \right)+3\left( -45-18 \right)$
$\Rightarrow \left| {{D}_{x}} \right|=5\left( 11 \right)-6\left( 51 \right)+3\left( -63 \right)$
$\Rightarrow \left| {{D}_{x}} \right|=55-306-189$
$\Rightarrow \left| {{D}_{x}} \right|=-440$
In the third matrix, in column 1, we will write the coefficients of variable-x of all the equations in column 2, we will write the constant terms of the given equations, and in column 3, we shall write the coefficients of variable-z of all the given equations.
${{D}_{y}}=\left[ \begin{matrix}
   2 & 5 & 3 \\
   -2 & 15 & 1 \\
   0 & 9 & 4 \\
\end{matrix} \right]$
We will calculate the determinant, $\left| {{D}_{y}} \right|$ of this matrix by expanding it along the column.
\[\Rightarrow \left| {{D}_{y}} \right|=2\left( 15\times 4-9\times 1 \right)-5\left( \left( -2\times 4 \right)-0\times 1 \right)+3\left( \left( -2\times 9 \right)-0\times 15 \right)\]
$\Rightarrow \left| {{D}_{y}} \right|=2\left( 60-9 \right)-5\left( -8-0 \right)+3\left( -18-0 \right)$
$\Rightarrow \left| {{D}_{y}} \right|=2\left( 51 \right)-5\left( -8 \right)+3\left( -18 \right)$
$\Rightarrow \left| {{D}_{y}} \right|=102+40-54$
$\Rightarrow \left| {{D}_{y}} \right|=88$
In the fourth matrix, in column 1, we will write the coefficients of variable-x of all the equations in column 2, we shall write the coefficients of variable-y of all the given equations and in column 3, we will write the constant terms of the given equations.
${{D}_{z}}=\left[ \begin{matrix}
   2 & 6 & 5 \\
   -2 & 2 & 15 \\
   0 & -3 & 9 \\
\end{matrix} \right]$
We will calculate the determinant, $\left| {{D}_{z}} \right|$ of this matrix by expanding it along the column.
\[\Rightarrow \left| {{D}_{z}} \right|=2\left( 2\times 9-\left( -3 \right)\times 15 \right)-6\left( \left( -2\times 9 \right)-0\times 15 \right)+5\left( \left( -2\times -3 \right)-0\times 2 \right)\]
$\Rightarrow \left| {{D}_{z}} \right|=2\left( 18+45 \right)-6\left( -18-0 \right)+5\left( 6-0 \right)$
$\Rightarrow \left| {{D}_{z}} \right|=2\left( 63 \right)-6\left( -18 \right)+5\left( 6 \right)$
$\Rightarrow \left| {{D}_{z}} \right|=126+108+30$
$\Rightarrow \left| {{D}_{z}} \right|=264$
Now, the x, y and z-coordinates of intersection of this system of equations are given as
$x=\dfrac{\left| {{D}_{x}} \right|}{\left| D \right|}$ , $y=\dfrac{\left| {{D}_{y}} \right|}{\left| D \right|}$ and $z=\dfrac{\left| {{D}_{z}} \right|}{\left| D \right|}$.
Substituting the values of these determinants, we get
$\Rightarrow x=\dfrac{-440}{88}$
$\Rightarrow x=-5$
And $y=\dfrac{88}{88}$
$\Rightarrow y=1$
And $z=\dfrac{264}{88}$
$\Rightarrow z=3$

Therefore, the solution of the system of equations: $2x+6y+3z=5$, $-2x+2y+z=15$ and $-3y+4z=9$ is $x=-5,y=1$ and $z=3$.

Note: We must be careful while performing the calculations of the determinants of the matrices as these calculations are very complex and little ignorance can lead to big calculation errors as a result of which the entire solution might go wrong.