Question

The number of solutions of the equation $3x+3y-z=5$, $x+y+z=3$, $2x+2y-z=3$ is
A. 1
B. 0
C. infinite
D. 2

Answer 1

Hint : Use the Cramer’s rule for finding the number of solutions of the three given rules.
For the equations to have an unique solution, $\Delta \ne 0$.
If $\Delta =0$ and ${{\Delta }_{x}}={{\Delta }_{y}}={{\Delta }_{z}}=0$, then the given equations will have infinitely many solutions.
If $\Delta =0$ and anyone of ${{\Delta }_{x}}$, ${{\Delta }_{y}}$, ${{\Delta }_{z}}$ is equal to zero, then the given equations will have no solution.

Complete step-by-step answer :
Let us first write down the given three equations.
$3x+3y-z=5$
$x+y+z=3$
$2x+2y-z=3$
To find the number of solutions for the three equations we will Cramer’s rule.
Before, we will understand some terms used in Cramer’s rule, that are $\Delta$, ${{\Delta }_{x}}$, ${{\Delta }_{y}}$ and ${{\Delta }_{z}}$.
$\Delta$ is the determinant of the coefficients of the variables present in the three equations. The arrangement of the coefficient follows in the same way as we have written the three equations.
In this case, $\Delta =\left( \begin{matrix} 3 & 3 & -1 \\\ 1 & 1 & 1 \\\ 2 & 2 & -1 \\\ \end{matrix} \right)$.
${{\Delta }_{x}}$ is the determinant by replacing the coefficients of x by the number on the right hand side of each equation.
Therefore, in this case, ${{\Delta }_{x}}=\left( \begin{matrix} 5 & 3 & -1 \\\ 3 & 1 & 1 \\\ 3 & 2 & -1 \\\ \end{matrix} \right)$.
Similarly, the determinants found by replacing the y and z coefficients with the same number as done above are ${{\Delta }_{y}}$ and ${{\Delta }_{z}}$, respectively.
So, in this case,
${{\Delta }_{y}}=\left( \begin{matrix} 3 & 5 & -1 \\\ 1 & 3 & 1 \\\ 2 & 3 & -1 \\\ \end{matrix} \right)$
And
${{\Delta }_{z}}=\left( \begin{matrix} 3 & 3 & 5 \\\ 1 & 1 & 3 \\\ 2 & 2 & 3 \\\ \end{matrix} \right)$.
According to Cramer’s there are three cases for the solution of the equation.
For the equations to have an unique solution, $\Delta \ne 0$.
If $\Delta =0$ and ${{\Delta }_{x}}={{\Delta }_{y}}={{\Delta }_{z}}=0$, then the given equations will have infinitely many solutions.
If $\Delta =0$ and anyone of ${{\Delta }_{x}}$, ${{\Delta }_{y}}$, ${{\Delta }_{z}}$ is equal to zero, then the given equations will have no solution.
Now, let us check which of the conditions is satisfied.
Now, let us solve the four determinants.
$\Delta =\left( \begin{matrix} \overset{+}{\mathop{3}}\, & \overset{-}{\mathop{3}}\, & \overset{+}{\mathop{-1}}\, \\\ 1 & 1 & 1 \\\ 2 & 2 & -1 \\\ \end{matrix} \right)=3\left( (1)(-1)-(1)(2) \right)-3\left( (1)(-1)-(1)(2) \right)-1\left( (1)(2)-(1)(2) \right)$
$\Rightarrow \Delta =3\left( -1-2 \right)-3\left( -1-2 \right)-1\left( 2-2 \right)=0$
${{\Delta }_{x}}=\left( \begin{matrix} 5 & 3 & -1 \\\ 3 & 1 & 1 \\\ 3 & 2 & -1 \\\ \end{matrix} \right)=5\left( (1)(-1)-(1)(2) \right)-3\left( (3)(-1)-(1)(3) \right)-1\left( (3)(2)-(1)(3) \right)$
$\Rightarrow {{\Delta }_{x}}=5\left( -1-2 \right)-3\left( -3-3 \right)-1\left( 6-3 \right)=-15+18-3=0$
${{\Delta }_{y}}=\left( \begin{matrix} 3 & 5 & -1 \\\ 1 & 3 & 1 \\\ 2 & 3 & -1 \\\ \end{matrix} \right)=3\left( (3)(-1)-(1)(3) \right)-5\left( (1)(-1)-(1)(2) \right)-1\left( (3)(1)-(2)(3) \right)$
$\Rightarrow {{\Delta }_{y}}=3\left( -3-3 \right)-5\left( -1-2 \right)-1\left( 3-6 \right)=-18+15+3=0$
And
${{\Delta }_{z}}=\left( \begin{matrix} 3 & 3 & 5 \\\ 1 & 1 & 3 \\\ 2 & 2 & 3 \\\ \end{matrix} \right)=3\left( (3)(1)-(2)(3) \right)-3\left( (1)(3)-(3)(2) \right)+5\left( (2)(1)-(2)(1) \right)$
$\Rightarrow {{\Delta }_{z}}=3\left( 3-6 \right)-3\left( 3-6 \right)+5\left( 2-2 \right)=-9-9+0=0$
Therefore, the three given equations $\Delta =0$ and ${{\Delta }_{x}}={{\Delta }_{y}}={{\Delta }_{z}}=0$. This means that the equations do have any solution, in fact has infinitely many solutions.
Hence, the correct option is C.
So, the correct answer is “Option C”.

Note : The Cramer’s rule can be used from any number of variables. However, the number of equations must be equal to the number of variables and at least one equation must be non-homogenous.
The value of a determinant can be found in several ways. If we know the properties of determinants then it becomes easier. Like, if two rows or two columns of a determinant are the same then the determinant is zero.