MCQ
In order that the matrix $\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\\3&\lambda &5\end{array}} \right]$ be non-singular, $\lambda $ should not be equal to
- A$1$
- B$2$
- C$3$
- ✓$4$
only if $\left| {\,\begin{array}{*{20}{c}}1&2&3\\4&5&6\\3&\lambda &5\end{array}\,} \right| \ne 0$
==> $\,1(25 - 6\lambda ) - 2(20 - 18) + 3(4\lambda - 15) \ne 0$
==> $25 - 6\lambda - 4 + 12\lambda - 45 \ne 0$
==> $6\lambda - 24 \ne 0$ ==> $\lambda \ne 4$.
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$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(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