If the system of linear equations $$x + 2ay + az = 0 ; x + 3by + bz = 0 ; x + 4cy + cz = 0$$ has a non - zero solution, then $$a, b, c.$$
A.
satisfy $$a + 2b + 3c = 0$$
B.
are in A.P.
C.
are in G.P.
D.
are in H.P.
Answer :
are in H.P.
Solution :
For homogeneous system of equations to have non zero solution, $$\Delta = 0$$
\[\begin{array}{l}
\left| {\begin{array}{*{20}{c}}
1&{2a}&a\\
1&{3b}&b\\
1&{4c}&c
\end{array}} \right| = 0\left[ {\therefore {C_2} \to {C_2} - 2{C_3}} \right]\\
\left| {\begin{array}{*{20}{c}}
1&0&a\\
1&b&b\\
1&{2c}&c
\end{array}} \right| = 0\left[ {{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}} \right]
\end{array}\]
On simplification, $$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$
∴ $$a, b, c$$ are in Harmonic Progression.
Releted MCQ Question on Algebra >> Matrices and Determinants
Releted Question 1
Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then
A.
$$C$$ is empty
B.
$$B$$ has as many elements as $$C$$
C.
$$A = B \cup C$$
D.
$$B$$ has twice as many elements as elements as $$C$$
Let $$a, b, c$$ be the real numbers. Then following system of equations in $$x, y$$ and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ has