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

& \begin{matrix} 2 & 1 & -1 \\\ \end{matrix} \\\ & \begin{matrix} 1 & 1 & -2 \\\ \end{matrix} \\\ & \begin{matrix} 3 & 2 & -3 \\\ \end{matrix} \\\ \end{aligned} \right|$$ …………………….(4) $$\begin{aligned} & \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{aligned}$$ $$\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{aligned} & \begin{matrix} 4 & 1 & -1 \\\ \end{matrix} \\\ & \begin{matrix} 0 & 1 & -2 \\\ \end{matrix} \\\ & \begin{matrix} 4 & 2 & -3 \\\ \end{matrix} \\\ \end{aligned} \right|$$ $$\begin{aligned} & \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{aligned}$$ $$\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{aligned} & \begin{matrix} 2 & 4 & -1 \\\ \end{matrix} \\\ & \begin{matrix} 1 & 0 & -2 \\\ \end{matrix} \\\ & \begin{matrix} 3 & 4 & -3 \\\ \end{matrix} \\\ \end{aligned} \right|$$ $$\begin{aligned} & \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{aligned}$$ $$\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{aligned} & \begin{matrix} 2 & 1 & 4 \\\ \end{matrix} \\\ & \begin{matrix} 1 & 1 & 0 \\\ \end{matrix} \\\ & \begin{matrix} 3 & 2 & 4 \\\ \end{matrix} \\\ \end{aligned} \right|$$ $$\begin{aligned} & \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{aligned}$$ $$\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{aligned} & \begin{matrix} 4 & 1 & -1 \\\ \end{matrix} \\\ & \begin{matrix} 0 & 1 & -2 \\\ \end{matrix} \\\ & \begin{matrix} 4 & 2 & -3 \\\ \end{matrix} \\\ \end{aligned} \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)\\}$$ .