If the system of linear equations
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0,
where a, b, c $\in$ R are non-zero distinct; has a non-zero solution, then:
Solution
For non-zero solution
<br><br>$$\left| {\matrix{
2 & {2a} & a \cr
2 & {3b} & b \cr
2 & {4c} & c \cr
} } \right| = 0$$
<br><br>$\Rightarrow$ $$\left| {\matrix{
1 & {2a} & a \cr
0 & {3b - 2a} & {b - a} \cr
0 & {4c - 2a} & {c - a} \cr
} } \right| = 0$$
<br><br>$\Rightarrow$ (3b – 2a) (c –a) – (b – a) (4c – 2a) = 0
<br><br>$\Rightarrow$ 2ac = bc + ab
<br><br>$\Rightarrow$ ${2 \over b} = {1 \over a} + {1 \over c}$
<br><br>$\therefore$ ${1 \over a},{1 \over b},{1 \over c}$ are in A.P.
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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