Let S be the set of all $\lambda$ $\in$ R for which the system
of linear equations
2x – y + 2z = 2
x – 2y +
$\lambda$z = –4
x +
$\lambda$y + z = 4
has no solution. Then the set S :
Solution
For no solution :
<br><br>$\Delta$ = 0 and $\Delta$<sub>1</sub>/$\Delta$<sub>2</sub>/$\Delta$<sub>3</sub> $\ne$ 0
<br><br>$\Delta$ = $$\left| {\matrix{
2 & { - 1} & 2 \cr
1 & { - 2} & \lambda \cr
1 & \lambda & 1 \cr
} } \right|$$ = 0
<br><br>$\Rightarrow$ 2(–2 – $\lambda$<sup>2</sup>) + 1 (1 – $\lambda$) + 2($\lambda$ + 2) = 0
<br><br>$\Rightarrow$ –2$\lambda$<sup>2</sup>
+ $\lambda$ + 1 = 0
<br><br>$\Rightarrow$ $\lambda$ = 1, $- {1 \over 2}$
<br><br>When <b>$\lambda$ = 1</b>
<br><br>2x – y + 2z = 2 ...(1)
<br><br>x – 2y + z = –4 ...(2)
<br><br>x + y + z = 4 ...(3)
<br><br>Adding (2) and (3), we get
<br><br>2x – y + 2z = 0 (contradiction) hence no solution.
<br><br>$\therefore$ $\lambda$ = 1 belongs to set S.
<br><br>When $\lambda$ = $- {1 \over 2}$
<br><br>2x – y + 2z = 2 ...(1)
<br><br>x – 2y $- {1 \over 2}$z
= –4 ...(2)
<br><br>x $- {1 \over 2}$y
+ z = 4 ...(3)
<br><br>(1) and (3) contradict each other, hence no
solution.
<br><br>$\therefore$ $\lambda$ = $- {1 \over 2}$ belongs to set S.
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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