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# Solve the following system of linear equations by Cramer’s rulex + y = 0, y + z = 1 and z + x = 3.

Last updated date: 20th Jun 2024
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Hint: To solve this question firstly we will write the system of linear equations in determinant form. Then, we will find the determinants${{D}_{1}}$, ${{D}_{2}}$ and${{D}_{3}}$. And then using formula $x=\dfrac{D}{{{D}_{1}}}$ , $y=\dfrac{D}{{{D}_{2}}}$ and $z=\dfrac{D}{{{D}_{3}}}$, we will evaluate the variables x, y and z.

Now , if we want to calculate the determinant of matrix A of order $3\times 3$, then determinant of matrix A of $3\times 3$ is evaluated as,
$\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$
Also, we know that if we have linear equation as, px + qy + rz = u, lx + my + nz = v and ax + by + cz = w, then we can represent coefficients of system of linear equation in determinant as $D=\left| \begin{matrix} p & q & r \\ l & m & n \\ a & b & c \\ \end{matrix} \right|$
Then in Cramer’s rule, we find three more determinants as ${{D}_{1}}=\left| \begin{matrix} u & q & r \\ v & m & n \\ w & b & c \\ \end{matrix} \right|$, ${{D}_{2}}=\left| \begin{matrix} p & u & r \\ l & v & n \\ a & w & c \\ \end{matrix} \right|$ and ${{D}_{3}}=\left| \begin{matrix} p & q & u \\ l & m & v \\ a & b & w \\ \end{matrix} \right|$ and we evaluate $x=\dfrac{D}{{{D}_{1}}}$ , $y=\dfrac{D}{{{D}_{2}}}$ and $z=\dfrac{D}{{{D}_{3}}}$.

Now, we can re – write the system of linear equations as,
x + y + 0.z = 0,
0.x + y + z = 1,
x + 0.y + z = 3 and in determinant form as
$D=\left| \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ \end{matrix} \right|$
Expanding determinant along ${{R}_{1}}$, we get
$D=1(1)-1(-1)+0$
$\Rightarrow$ D = 2
Now, ${{D}_{1}}=\left| \begin{matrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 3 & 0 & 1 \\ \end{matrix} \right|$
Expanding determinant along ${{R}_{1}}$, we get
${{D}_{1}}=0-1(1-3)+0$
$\Rightarrow {{D}_{1}}=2$
Now, ${{D}_{2}}=\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 3 & 1 \\ \end{matrix} \right|$
Expanding determinant along ${{R}_{1}}$, we get
${{D}_{2}}=1(1-3)-0+0$
$\Rightarrow {{D}_{2}}=-2$
Now, ${{D}_{3}}=\left| \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 3 \\ \end{matrix} \right|$
Expanding determinant along ${{R}_{1}}$, we get
${{D}_{3}}=1(3-1)-1+0$
$\Rightarrow {{D}_{3}}=1$
So, we know that according to Cramer’s rule
$x=\dfrac{D}{{{D}_{1}}}$ , $y=\dfrac{D}{{{D}_{2}}}$ and $z=\dfrac{D}{{{D}_{3}}}$.
So, $x=\dfrac{2}{2}$,$y=\dfrac{2}{-2}$ and $z=\dfrac{2}{1}$
We get, x = 1, y = -1 and z = 2.

Note: To solve this question, one must know how we expand the determinants and also one must know the concept of Cramer’s rule. Also, this rule works for any number of variable linear equations. While solving determinant and evaluating the values of variable x, y and z try not to make any calculation mistakes as this may give you wrong values of variables.