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3 and 4 .Determinants and Matrices
normal
Consider system of equations in $x$ , $y$ and $z$
$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is
A
$ - \frac{1}{3}$
B
$ - \frac{1}{{12}}$
C
$ - \frac{1}{{6}}$
D
$ - \frac{1}{{4}}$
Solution
$\left|\begin{array}{ccc}{12} & {b} & {c} \\ {a} & {24} & {c} \\ {a} & {b} & {36}\end{array}\right|=0$
$(12) (24)(36)-12 b c-36 b a+2 a b c-24 a c=0$ …….$(i)$
$\frac{1}{a-12}+\frac{2}{b-24}+\frac{3}{c-36}$
$=\frac{(b-24)(c-36)+2(a-12)(c-36)+3(a-12)(b-24)}{(a-12)(b-24)(c-36)}$
on solve
$=-\frac{1}{6}$
Standard 12
Mathematics