Matrices and Determinants 4 Question 22
Assertion and Reason
For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
23. Consider the system of equations $x-2 y+3 z=-1$, $x-3 y+4 z=1$ and $-x+y-2 z=k$
Statement I The system of equations has no solution for $k \neq 3$ and
Statement II The determinant $\left|\begin{array}{rrr}1 & 3 & -1 \\ -1 & -2 & k \\ 1 & 4 & 1\end{array}\right| \neq 0$, for
$k \neq 0$.
(2008, 3M)
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Answer:
Correct Answer: 23. (a)
Solution:
- The given system of equations can be expressed as
$ \left[\begin{array}{rrr} 1 & -2 & 3 \\ 1 & -3 & 4 \\ -1 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} -1 \\ 1 \\ k \end{array}\right] $
Applying $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3+R_1$
$ \sim\left[\begin{array}{lll} 1 & -2 & 3 \\ 0 & -1 & 1 \\ 0 & -1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ k-1 \end{array}\right] $
Applying $R_3 \rightarrow R_3-R_2$
$ \sim\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ k-3 \end{array}\right] $
When $k \neq 3$, the given system of equations has no solution.
$\Rightarrow$ Statement I is true. Clearly, Statement II is also true as it is rearrangement of rows and columns of
$ \left[\begin{array}{rrc} 1 & -2 & 3 \\ 1 & -3 & 4 \\ -1 & 1 & -2 \end{array}\right] $
