Question

The system of linear equations:
$\begin{align}
  & x+y+z=2 \\
 & 2x+3y+2z=5 \\
 & 2x+3y+\left( {{a}^{2}}-1 \right)z=a+1 \\
\end{align}$
(A) Has infinitely many solutions for a=4
(B) Is consistent when $\left| a \right|=\sqrt{3}$
(C) Is consistent when $\left| a \right|=4$
(D) Has a unique solution for $\left| a \right|=\sqrt{3}$

Answer 1

Hint: We start solving this problem by considering all the three given equations and then we apply Cramer’s rule to find $D,{{D}_{1}}$. We start solving the determinant $D$ by converting the column ${{C}_{2}}$ as ${{C}_{2}}-{{C}_{1}}$ and the column ${{C}_{3}}$as ${{C}_{3}}-{{C}_{1}}$ and then we solve the obtained determinant to get $D$. Then we start solving the determinant ${{D}_{1}}$ by converting the column ${{C}_{3}}$as ${{C}_{3}}-{{C}_{2}}$ and the column ${{C}_{2}}$ as ${{C}_{2}}-\dfrac{1}{2}{{C}_{1}}$ and then we solve the obtained determinant to get ${{D}_{1}}$. Then we substitute the value 4 for $a$ in the determinant ${{D}_{1}}$ to know whether it has infinitely many solutions for $a=4$ as mentioned in option (A). Then we check whether it is consistent or inconsistent.

Complete step-by-step solution:
Let us consider the given equations,
$\begin{align}
  & x+y+z=2 \\
 & 2x+3y+2z=5 \\
 & 2x+3y+\left( {{a}^{2}}-1 \right)z=a+1 \\
\end{align}$
Let us consider the Cramer’s method, that is, if the system of linear equations are
$\begin{align}
  & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \\
 & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \\
 & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}} \\
\end{align}$ , then
$\begin{align}
  & D=\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|, \\
 & {{D}_{1}}=\left| \begin{matrix}
   {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|, \\
 & {{D}_{2}}=\left| \begin{matrix}
   {{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\
\end{matrix} \right|, \\
 & {{D}_{3}}=\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\
\end{matrix} \right| \\
\end{align}$
Let us consider the different conditions for consistent and inconsistent solutions.
(i) If $D\ne 0$ and at least one of the ${{D}_{1}},{{D}_{2}},{{D}_{3}}$ is not equal to zero, then the system of equations has unique solution. (Consistent)
(ii) If $D=0,{{D}_{1}}=0,{{D}_{2}}=0,{{D}_{3}}=0$ , then the system of equations has infinitely many solutions. (Consistent)
(iii) If $D=0$ and at least one of the ${{D}_{1}},{{D}_{2}},{{D}_{3}}$ is not equal to zero, then the system of equations has no solution. (Inconsistent)
Let us now find the determinant $D$ , we get,
$D=\left| \begin{matrix}
   1 & 1 & 1 \\
   2 & 3 & 2 \\
   2 & 3 & {{a}^{2}}-1 \\
\end{matrix} \right|$
Now, let us convert the column ${{C}_{2}}$ as ${{C}_{2}}-{{C}_{1}}$ and the column ${{C}_{3}}$as ${{C}_{3}}-{{C}_{1}}$, we get,
$\begin{align}
  & D=\left| \begin{matrix}
   1 & 1-1 & 1-1 \\
   2 & 3-2 & 2-2 \\
   2 & 3-2 & {{a}^{2}}-1-2 \\
\end{matrix} \right| \\
 & D=\left| \begin{matrix}
   1 & 0 & 0 \\
   2 & 1 & 0 \\
   2 & 1 & {{a}^{2}}-3 \\
\end{matrix} \right| \\
\end{align}$
Let us expand the above determinant along ${{R}_{1}}$ , we get,
$\begin{align}
  & D=1\left( {{a}^{2}}-3-0 \right)-0+0 \\
 & D={{a}^{2}}-3 \\
\end{align}$
Now, we find the determinant ${{D}_{1}}$.
${{D}_{1}}=\left| \begin{matrix}
   2 & 1 & 1 \\
   5 & 3 & 2 \\
   a+1 & 3 & {{a}^{2}}-1 \\
\end{matrix} \right|$
Let us now convert the column ${{C}_{3}}$ as ${{C}_{3}}-{{C}_{2}}$ and the column ${{C}_{2}}$ as ${{C}_{2}}-\dfrac{1}{2}{{C}_{1}}$ , we get,
$\begin{align}
  & {{D}_{1}}=\left| \begin{matrix}
   2 & 1-\dfrac{2}{2} & 1-1 \\
   5 & 3-\dfrac{5}{2} & 2-3 \\
   a+1 & 3-\dfrac{a+1}{2} & {{a}^{2}}-1-3 \\
\end{matrix} \right| \\
 & {{D}_{1}}=\left| \begin{matrix}
   2 & 0 & 0 \\
   5 & \dfrac{1}{2} & -1 \\
   a+1 & \dfrac{5}{2}-\dfrac{a}{2} & {{a}^{2}}-4 \\
\end{matrix} \right| \\
\end{align}$
Now, we expand the determinant ${{D}_{1}}$ along ${{R}_{1}}$ , we get,
$\begin{align}
  & {{D}_{1}}=2\left[ \dfrac{1}{2}\left( {{a}^{2}}-4 \right)+1\left( \dfrac{5}{2}-\dfrac{a}{2} \right) \right] \\
 & {{D}_{1}}={{a}^{2}}-4+5-a \\
 & {{D}_{1}}={{a}^{2}}-a+1 \\
\end{align}$
Now, let us substitute $\sqrt{3}$ in $D$ and ${{D}_{1}}$ , we get,
$\begin{align}
  & D={{\left( \sqrt{3} \right)}^{2}}-3 \\
 & D=3-3 \\
 & D=0 \\
\end{align}$
And
$\begin{align}
  & {{D}_{1}}=3-\sqrt{3}+1 \\
 & {{D}_{1}}=4-\sqrt{3} \\
 & {{D}_{1}}\ne 0 \\
\end{align}$
Therefore, the given system of linear equations has no solution when $\left| a \right|=\sqrt{3}$.
Now, let us substitute $4$ in $D$ and ${{D}_{1}}$ , we get,
$\begin{align}
  & D={{\left( 4 \right)}^{2}}-3 \\
 & D=16-3 \\
 & D=13\Rightarrow D\ne 0 \\
\end{align}$
And
$\begin{align}
  & {{D}_{1}}=16-4+1 \\
 & {{D}_{1}}=13 \\
 & {{D}_{1}}=13\Rightarrow {{D}_{1}}\ne 0 \\
\end{align}$
So, the given system of linear equations has unique solution when $\left| a \right|=4$.
Therefore, the given system of linear equations is consistent when $\left| a \right|=4$.
Hence, the answer is an option (C).

Note: The possibilities for making mistakes in this type of problems are one may make a mistake by converting the column or row while finding out the determinant, in a wrong way. One must know which column or row should be converted by which process and one must also know how to expand the determinant through any row or column for easy calculation.