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 infinite number of solutions if k $\ne$ $-$2.
Which of the following statements are correct?
Solution
$x - 2y + 0.z = 1$<br><br>$x - y + kz = - 2$<br><br>$0.x + ky + 4z = 6$<br><br>$$\Delta = \left| {\matrix{
1 & { - 2} & 0 \cr
1 & { - 1} & k \cr
0 & k & 4 \cr
} } \right| = 4 - {k^2}$$<br><br>For unique solution $4 - {k^2} \ne 0$<br><br>$\Rightarrow$ k $\ne$ $\pm$ 2<br><br><b>For k = 2 :</b><br><br>$x - 2y + 0.z = 1$<br><br>$x - y + 2z = - 2$<br><br>$0.x + 2y + 4z = 6$<br><br>$$\Delta x = \left| {\matrix{
1 & { - 2} & 0 \cr
2 & { - 1} & 2 \cr
6 & 2 & 4 \cr
} } \right| = ( - 8) + 2[ - 20]$$<br><br>$\Delta x = - 48 \ne 0$<br><br>For k = 2, $\Delta x \ne 0$<br><br>$\therefore$ For K = 2; The system has no solution.
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
This question is part of PrepWiser's free JEE Main question bank. 274 more solved questions on Matrices and Determinants are available — start with the harder ones if your accuracy is >70%.