Question

Solve the following simultaneous equation using Cramer's rule: $2x+3y-z=1,4x+y-3z=11$ and $3x-2y+5z=21$.

Answer 1

To solve this question we need to know about Cramer’s Rule. This rule is used to solve a linear system with three equations having three unknowns. The first step to solve for the variable is to find the determinant of the coefficient of the variable $x,y,z$. The next step is to find ${{D}_{x}},{{D}_{y}},{{D}_{z}}$. Final step is to divide the values of ${{D}_{x}},{{D}_{y}},{{D}_{z}}$ with $D$ to find $x,y,z$ respectively, which in mathematical form is written as$x=\dfrac{{{D}_{x}}}{D},y=\dfrac{{{D}_{y}}}{D},z=\dfrac{{{D}_{z}}}{D}.$

Complete step-by-step solution:
The question asks us to find the value of $x,y$ and $z$ for the three equations given which are $2x+3y-z=1,4x+y-3z=11$ and $3x-2y+5z=21$ by using Cramer's rule. The Cramer’s Rule is represented as:
$x=\dfrac{{{D}_{x}}}{D},y=\dfrac{{{D}_{y}}}{D},z=\dfrac{{{D}_{z}}}{D},$
Where ${{D}_{x}},{{D}_{y}},{{D}_{z}},D$ are the determinants. Starting to solve the problem by calculating the four determinants. Firstly we will calculate for $D$
$\Rightarrow D=\left| \begin{matrix} 2 & 3 & -1 \\\ 4 & 1 & -3 \\\ 3 & -2 & 5 \\\ \end{matrix} \right|$
$\Rightarrow D=2\left| \begin{matrix} 1 & -3 \\\ -2 & 5 \\\ \end{matrix} \right|-4\left| \begin{matrix} 3 & -1 \\\ -2 & 5 \\\ \end{matrix} \right|+3\left| \begin{matrix} 3 & -1 \\\ 1 & -3 \\\ \end{matrix} \right|$
$\Rightarrow D=2\left( 5-6 \right)-4\left( 15-2 \right)+3\left( -9+1 \right)$
$\Rightarrow D=-2-52-24$
$\Rightarrow D=-78$
Now we will find the value of ${{D}_{x}}$ which means the $x-values$ in the first column will be replaced by the constant term after the equal sign, leaving the$y-values$ and $z-values$ unchanged. On doing this the determinant we get is:

1 & 3 & -1 \\\ 11 & 1 & -3 \\\ 21 & -2 & 5 \\\ \end{matrix} \right|$$ $\Rightarrow {{D}_{x}}=1\left| \begin{matrix} 1 & -3 \\\ -2 & 5 \\\ \end{matrix} \right|-11\left| \begin{matrix} 3 & -1 \\\ -2 & 5 \\\ \end{matrix} \right|+21\left| \begin{matrix} 3 & -1 \\\ 1 & -3 \\\ \end{matrix} \right|$ $\Rightarrow {{D}_{x}}=1\left( 5-6 \right)-11\left( 15-2 \right)+21\left( -9+1 \right)$ $\Rightarrow {{D}_{x}}=-1-143-168$ $\Rightarrow {{D}_{x}}=-312$ Similarly, we will find the value of ${{D}_{y}}$ which means the $y-values$ in the second column will be replaced by the constant term after the equal sign, leaving the$x-values$ and $z-values$ unchanged. On doing this the determinant we get is: $$\Rightarrow {{D}_{y}}=\left| \begin{matrix} 2 & 1 & -1 \\\ 4 & 11 & -3 \\\ 3 & 21 & 5 \\\ \end{matrix} \right|$$ $\Rightarrow {{D}_{y}}=2\left| \begin{matrix} 11 & -3 \\\ 21 & 5 \\\ \end{matrix} \right|-4\left| \begin{matrix} 1 & -1 \\\ 21 & 5 \\\ \end{matrix} \right|+3\left| \begin{matrix} 1 & -1 \\\ 11 & -3 \\\ \end{matrix} \right|$ $\Rightarrow {{D}_{y}}=2\left( 55+63 \right)-4\left( 5+21 \right)+3\left( -3+11 \right)$ $\Rightarrow {{D}_{y}}=236-104+24$ $\Rightarrow {{D}_{y}}=156$ Similarly, we will find the value of ${{D}_{z}}$ which means the $z-values$ in the third column will be replaced by the constant term after the equal sign, leaving the $x-values$ and $y-values$ unchanged. On doing this the determinant we get is: $\Rightarrow {{D}_{z}}=\left| \begin{matrix} 2 & 3 & 1 \\\ 4 & 1 & 11 \\\ 3 & -2 & 21 \\\ \end{matrix} \right|$ $\Rightarrow {{D}_{z}}=2\left| \begin{matrix} 1 & 11 \\\ -2 & 21 \\\ \end{matrix} \right|-4\left| \begin{matrix} 3 & 1 \\\ -2 & 21 \\\ \end{matrix} \right|+3\left| \begin{matrix} 3 & 1 \\\ 1 & 11 \\\ \end{matrix} \right|$ $$\Rightarrow {{D}_{z}}=2\left( 21+22 \right)-4\left( 63+2 \right)+3\left( 33-1 \right)$$ $\Rightarrow {{D}_{z}}=86-260+96$ $\Rightarrow {{D}_{z}}=-78$ The last step is to find x, y and z values using the formula $x=\dfrac{{{D}_{x}}}{D},y=\dfrac{{{D}_{y}}}{D},z=\dfrac{{{D}_{z}}}{D},$ firstly finding the value of $x$ , substituting the value we get: $\Rightarrow x=\dfrac{{{D}_{x}}}{D}$ $\Rightarrow x=\dfrac{-312}{-78}$ $\Rightarrow x=4$ Now finding the value of $y$, substituting the value we get: $\Rightarrow y=\dfrac{{{D}_{y}}}{D}$ $\Rightarrow y=\dfrac{156}{-78}$ $\Rightarrow y=-2$ Now finding the value of $z$, substituting the value we get: $\Rightarrow z=\dfrac{{{D}_{z}}}{D}$ $\Rightarrow y=\dfrac{-78}{-78}$ $\Rightarrow y=1$ **$\therefore $ The value of $x,y,z$ are $4,-2,1$ respectively for the following simultaneous equation using Cramer's rule: $2x+3y-z=1,4x+y-3z=11$ and $3x-2y+5z=21$.** **Note:** Cramer’s Rule takes the help of the determinant for solving the variable. We can check whether the values we got are correct or not. For doing this we need to substitute the values of $x,y,z$ on any one of the equations and check whether LHS is the same as that of RHS. We will be checking by putting the values in the first equation which is $2x+3y-z=1$, we get: $\Rightarrow 2x+3y-z=1$ $\Rightarrow 2\times 4+3\times \left( -2 \right)-1=1$ $\Rightarrow 8-6-1=1$ $\Rightarrow 1=1$ Since both RHS are equal to LHS, the values of $x,y,z$ is correct.