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:

  1. 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] $



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