Download App

STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
The solution $(s)$ of the equation $\left| {\begin{array}{*{20}{c}}x&a&b\\ a&x&a\\b&b&x\end{array}} \right|$ $= 0$ is/are :
  • A
    $x = - (a + b)$
  • B
    $x = a$
  • C
    $x = b$
  • ✓
    All of the above
✓

Answer

Correct option: D.
All of the above
d
Use $c_1 \rightarrow c_1 -c_2$ and then $R_1 \rightarrow R_1 + R_2 $ to get $\left| {\,\begin{array}{*{20}{c}}0&{a + x}&{b + a}\\{ - (x - a)}&x&a\\0&b&x \end{array}\,} \right|$ $= 0$. Now open by $c_1$ and foctorize

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

${d \over {dx}}\left( {{{\tan }^{ - 1}}{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$ is equal to
→
If $u = {\tan ^{ - 1}}{y \over x}$, then by Euler’s Theorem the value of $x$ ${{\partial u} \over {\partial x}} + y{{\partial u} \over {\partial y}} = $
→
If ${s_n} = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + ........ + \frac{1}{{{2^{n - 1}}}}$ , then the least integral value of $n$ such that $2 - {s_n} < \frac{1}{{100}}$ is
→
$\int {\frac{{dx}}{{\sin (x - a)\sin (x - b)}}} $ is
→
The integral $\int \limits_{0}^{2} \| x-1|-x| d x$ is equal to
→
If $x = \cos \theta + i\sin \theta $ and $y = \cos \varphi + i\sin \phi $, then ${x^m}{y^n} + {x^{ - m}}{y^{ - n}}$ is equal to
→
Sum of all distinct integral value $(s)$ of $\alpha$ such that equation $x^2 -\alpha x + \alpha + 1 = 0$ has integral roots, is equal to-
→
General solution of $\tan 5\theta = \cot 2\theta $ is  $($ where $n \in Z )$
→
If $y = {\sin ^{ - 1}}\sqrt {1 - {x^2}} $, then $dy/dx = $
→
Let $\alpha=\sum_{k=0}^n\left(\frac{\left({ }^n C_k\right)^2}{k+1}\right)$ and $\beta=\sum_{k=0}^{n-1}\left(\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\right)$. If $5 \alpha=6 \beta$, then $n$ equals
→