Question

For the system of linear equations: $x - 2y = 1,x - y + kz = - 2,ky + 4z = 6,k \ in R$
Consider the following statements:
(A) The system has unique solution if $k \ne 2,k \ne - 2$
(B) The system has unique solution if $k = - 2$
(C) The system has unique solution if $k = 2$
(D) The system has no-solution if $k = 2$
(E) The system has an infinite number of solutions if $k \ne - 2$
Which of the following statements is correct?
1. (B) and (E) only
2. (C) and (D) only
3. (A) and (D) only
4. (A) and (E) only

Answer 1

Hint: In this question, we are given a system of linear equations $x - 2y = 1,x - y + kz = - 2,ky + 4z = 6,k \ in R$ and we have to find at which value of $k$, the system has unique solution or no solution. The first step is to write the equation in the form of $AX = B$. Then find the determinant of $A$.To check at which value the equation has a unique value put the determinant equal to zero. Now, at $k = 2$ where change the first element with the elements of matrix B. If the determinant is zero it means at $k = 2$ linear equations have a unique solution but if not, then it has no solutions.

Formula Used:
Determinant –
$\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right| = a\left( {ei - fh} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)$

Complete step by step Solution:
A set of two or more linear equations with the same variables is referred to as a system of linear equations.
Given that,
System of linear equations,
$x - 2y = 1$
$x - y + kz = - 2$
$ky + 4z = 6,k \ in R$
It can also be written as,
$AX = B$
$\left[ {\begin{array}{*{20}{c}}
  1&{ - 2}&0 \\
  1&{ - 1}&k \\
  0&k&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1 \\
  { - 2} \\
  6
\end{array}} \right]$
Now, the Determinant of $A$
$\left| A \right| = \left| {\begin{array}{*{20}{c}}
  1&{ - 2}&0 \\
  1&{ - 1}&k \\
  0&k&4
\end{array}} \right|$
$\left| A \right| = 1\left( { - 4 + {k^2}} \right) - \left( { - 2} \right)\left( {4 - 0} \right) + 0\left( {k - 0} \right)$
$\left| A \right| = 4 + {k^2}$
Now for unique solutions,
$\left| A \right| \ne 0$
$4 + {k^2} \ne 0$
Which implies that, $k \ne \pm 2$
$ \Rightarrow $ The system has unique solution if $k \ne 2,k \ne - 2$
Now let’s check at $k = 2$,
The first step is to replace the elements of column $1$ by the matrix $B$ and find its determinant,
$\left| {A'} \right| = \left| {\begin{array}{*{20}{c}}
  1&{ - 2}&0 \\
  { - 2}&{ - 1}&2 \\
  6&2&4
\end{array}} \right|$
$\left| {A'} \right| = 1\left( { - 4 - 4} \right) - \left( { - 2} \right)\left( { - 8 - 12} \right) + 0\left( { - 4 - \left( { - 6} \right)} \right)$
$\left| {A'} \right| = - 8 - 40$
$\left| {A'} \right| = - 48 \ne 0$
Therefore, at $k = 2$ system has no solutions

Hence, the correct option is 3.

Note: The key concept involved in solving this problem is a good knowledge of determinants. Students must know that the Determinant of $3 \times 3$ formula is derived from the $2 \times 2$ matrix formula only.
Let, $P = \left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right]$be the $3 \times 3$ matrix
Then, $\left| P \right| = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|$
$ = a\left| {\begin{array}{*{20}{c}}
  e&f \\
  h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
  d&f \\
  g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
  d&e \\
  g&h
\end{array}} \right|$
$ = a\left( {ei - fh} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)$