Question

Solve the following non-homogeneous system of linear equations by the determinant method.
\[2x+y-z=4\] ,
\[x+y-2z=0\] ,
\[3x+2y-3z=4\] .

Answer 1

Hint: First of all make a matrix \[\Delta \] having coefficients of x in the \[{{1}^{st}}\] row, coefficient of y in the \[{{2}^{nd}}\] row, and coefficient of z in the \[{{3}^{rd}}\] row. Now, get the determinant value of \[{{\Delta }_{1}}\] by replacing the \[{{1}^{st}}\] column of the matrix \[\Delta \] by the constant terms and the constant terms are 4, 0, and 4. Similarly, get the determinant value of \[{{\Delta }_{2}}\] by replacing the \[{{2}^{nd}}\] column of the matrix \[\Delta \] by the constant terms and the constant terms are 4, 0, and 4. Similarly, get the determinant value of \[{{\Delta }_{3}}\] by replacing the \[{{3}^{rd}}\] column of the matrix \[\Delta \] by the constant terms and the constant terms are 4, 0, and 4. Now, get the values of x, y, and z by using the formula \[x=\dfrac{{{\Delta }_{1}}}{\Delta }\] , \[y=\dfrac{{{\Delta }_{2}}}{\Delta }\] , and \[z=\dfrac{{{\Delta }_{3}}}{\Delta }\] and then, solve it further.

Complete step-by-step answer:

According to the question, we have three equations that are linear in terms of x, y, and z.
\[2x+y-z=4\] ………………………(1)
\[x+y-2z=0\] …………………..(2)
\[3x+2y-3z=4\] ……………………..(3)
First of all, we have to get the value of \[\Delta \] and we know that \[\Delta \] is the determinant value of coefficients of x, y, and z.
Let us make a matrix and in that matrix, we have to put the values of coefficients of x in the \[{{1}^{st}}\] row, coefficient of y in the \[{{2}^{nd}}\] row, and coefficient of z in the \[{{3}^{rd}}\] row.
For the \[{{1}^{st}}\] , we have 2, 1, and -1 as the coefficients of x, y, and z.
For the \[{{2}^{nd}}\] , we have 1, 1, and -2 as the coefficients of x, y, and z.
For the \[{{3}^{rd}}\] , we have 3, 3, and -3 as the coefficients of x, y, and z.
Putting the value of coefficients of x, y, and z in the rows of the matrix, we get
\[\Delta =\left| \begin{align}
  & \begin{matrix}
   2 & 1 & -1 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   3 & 2 & -3 \\
\end{matrix} \\
\end{align} \right|\] …………………….(4)
\[\begin{align}
  & \Rightarrow \Delta =2\{1(-3)-2(-2)\}-1\left\{ 1(-3)-3(-2) \right\}+(-1)\{1(2)-1(3)\} \\
 & \Rightarrow \Delta =2\left( -3+4 \right)-1\left( -3+6 \right)-1\left( 2-3 \right) \\
 & \Rightarrow \Delta =2(1)-1(3)-1(-1) \\
 & \Rightarrow \Delta =2-3+1 \\
\end{align}\]
\[\Rightarrow \Delta =0\] ………………….(5)
Now, we need to find the determinant value of \[{{\Delta }_{1}}\] . But for the determinant value, we need the matrix \[{{\Delta }_{1}}\] . Replace the \[{{1}^{st}}\] column of the matrix \[\Delta \] by the constant terms and the constant terms are 4, 0, and 4.
On replacing we get,
\[{{\Delta }_{1}}=\left| \begin{align}
  & \begin{matrix}
   4 & 1 & -1 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   4 & 2 & -3 \\
\end{matrix} \\
\end{align} \right|\]
 \[\begin{align}
  & \Rightarrow {{\Delta }_{1}}=4\{1(-3)-2(-2)\}-1\left\{ 0(-3)-4(-2) \right\}+(-1)\{0(2)-1(4)\} \\
 & \Rightarrow {{\Delta }_{1}}=4\left( -3+4 \right)-1\left( 0+8 \right)-1\left( 0-4 \right) \\
 & \Rightarrow {{\Delta }_{1}}=4(1)-1(8)-1(-4) \\
 & \Rightarrow {{\Delta }_{1}}=4-8+4 \\
\end{align}\]
\[\Rightarrow {{\Delta }_{1}}=0\]
Now, we need to find the determinant value of \[{{\Delta }_{2}}\] . But for the determinant value, we need the matrix \[{{\Delta }_{2}}\] . Replace the \[{{2}^{nd}}\] column of the matrix \[\Delta \] by the constant terms and the constant terms are 4, 0, and 4.
\[{{\Delta }_{2}}=\left| \begin{align}
  & \begin{matrix}
   2 & 4 & -1 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 0 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   3 & 4 & -3 \\
\end{matrix} \\
\end{align} \right|\]
\[\begin{align}
  & \Rightarrow {{\Delta }_{2}}=2\{0(-3)-4(-2)\}-4\left\{ 1(-3)-3(-2) \right\}+(-1)\{1(4)-0(3)\} \\
 & \Rightarrow {{\Delta }_{2}}=2\left( 0+8 \right)-4\left( -3+6 \right)-1\left( 4-0 \right) \\
 & \Rightarrow {{\Delta }_{2}}=2(8)-4(3)-1(4) \\
 & \Rightarrow {{\Delta }_{2}}=16-12-4 \\
\end{align}\]
\[\Rightarrow {{\Delta }_{2}}=0\]
Now, we need to find the determinant value of \[{{\Delta }_{3}}\] . But for the determinant value, we need the matrix \[{{\Delta }_{3}}\] . Replace the \[{{3}^{rd}}\] column of the matrix \[\Delta \] by the constant terms and the constant terms are 4, 0, and 4.
\[{{\Delta }_{3}}=\left| \begin{align}
  & \begin{matrix}
   2 & 1 & 4 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   3 & 2 & 4 \\
\end{matrix} \\
\end{align} \right|\]
\[\begin{align}
  & \Rightarrow {{\Delta }_{3}}=2\{1(4)-2(0)\}-1\left\{ 1(4)-3(0) \right\}+(4)\{1(2)-1(3)\} \\
 & \Rightarrow {{\Delta }_{3}}=2\left( 4-0 \right)-1\left( 4-0 \right)+4\left( 2-3 \right) \\
 & \Rightarrow {{\Delta }_{3}}=2(4)-1(4)+4(-1) \\
 & \Rightarrow {{\Delta }_{3}}=8-4-4 \\
\end{align}\]
\[\Rightarrow {{\Delta }_{3}}=0\]
We know that,
\[x=\dfrac{{{\Delta }_{1}}}{\Delta }=\dfrac{0}{0}=Not\,defined\]
\[y=\dfrac{{{\Delta }_{2}}}{\Delta }=\dfrac{0}{0}=Not\,defined\]
\[z=\dfrac{{{\Delta }_{3}}}{\Delta }=\dfrac{0}{0}=Not\,defined\]
We don’t have any defined values of x, y, and z.
Hence, there is no solution of x, y, and z.

Note: In this question, one may do a calculation mistake in getting the determinant value of the matrix. Here, we are finding the determinant value of the matrices \[\Delta \] , \[{{\Delta }_{1}}\] , \[{{\Delta }_{2}}\] , and \[{{\Delta }_{3}}\] .
\[{{\Delta }_{1}}=\left| \begin{align}
  & \begin{matrix}
   4 & 1 & -1 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   4 & 2 & -3 \\
\end{matrix} \\
\end{align} \right|\]
While finding the determinant value of the matrix \[{{\Delta }_{1}}\] , one may miss the negative sign while expanding the matrix with respect to the \[{{2}^{nd}}\] row and write it as,
\[{{\Delta }_{1}}=4\{1(-3)-2(-2)\}+1\left\{ 0(-3)-4(-2) \right\}+(-1)\{0(2)-1(4)\}\] which is wrong. The correct way to expand the matrix is \[{{\Delta }_{1}}=4\{1(-3)-2(-2)\}-1\left\{ 0(-3)-4(-2) \right\}+(-1)\{0(2)-1(4)\}\] .