The system of linear equations
$\lambda$x + 2y + 2z = 5
2$\lambda$x + 3y + 5z = 8
4x + $\lambda$y + 6z = 10 has
Solution
$\Delta$ = $$\left| {\matrix{
\lambda & 2 & 2 \cr
{2\lambda } & 3 & 5 \cr
4 & \lambda & 6 \cr
} } \right|$$
<br><br>= $\lambda$ ( 18 – 5$\lambda$) – 2(12$\lambda$ – 20) + 2(2$\lambda$<sup>2</sup>
– 12)
<br><br>= 18$\lambda$ – 5$\lambda$<sup>2</sup>
– 24$\lambda$ + 40 + 4$\lambda$<sup>2</sup>
– 24
<br><br>= – $\lambda$<sup>2</sup>
– 6$\lambda$ + 16
<br><br>= – ($\lambda$ + 8)($\lambda$ – 2)
<br><br>For no solutions $\lambda$ = 0 $\Rightarrow$ $\lambda$ = – 8, $\lambda$ = 2
<br><br>when $\lambda$ = 2
<br><br>$\Delta$<sub>x</sub> = $$\left| {\matrix{
5 & 2 & 2 \cr
8 & 3 & 5 \cr
{10} & 2 & 6 \cr
} } \right|$$
<br><br>= 5 (18 – 10) – 2 (48 – 50) + 2 (16 – 30)
<br><br>= 40 + 4 – 28 $\ne$ 0
<br><br>So no solution for $\lambda$ = 2
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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