If a,b,c are positive integers, then the determinant =

MCQ

If $a,b,c$ are positive integers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + x}&{ab}&{ac}\\{ab}&{{b^2} + x}&{bc}\\{ac}&{bc}&{{c^2} + x}\end{array}\,} \right|$ is divisible by

A
${x^3}$
✓
${x^2}$
C
$({a^2} + {b^2} + {c^2})$
D
None of these

✓

Answer

Correct option: B.

${x^2}$

b
(b) $\Delta = \frac{1}{{abc}}\,\left| {\,\begin{array}{*{20}{c}}{{a^3} + ax}&{{a^2}b}&{{a^2}c}\\{a{b^2}}&{{b^3} + bx}&{{b^2}c}\\{{c^2}a}&{{c^2}b}&{{c^3} + cx}\end{array}\,} \right|$

=$({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+x)\times \left| \,\begin{matrix}
1 & 1 & 1 \\
{{b}^{2}} & {{b}^{2}}+x & {{b}^{2}} \\
{{c}^{2}} & {{c}^{2}} & {{c}^{2}}+x \\
\end{matrix}\, \right|$

$\{$Applying ${R_1} \to {R_1} + {R_2} + {R_3}\} $

$ = ({a^2} + {b^2} + {c^2} + x)\,\left| {\,\begin{array}{*{20}{c}}1&0&0\\{{b^2}}&x&0\\{{c^2}}&0&x\end{array}\,} \right|$$\left. \begin{array}{l}{C_2} \to {C_2} - {C_1}\\{C_3} \to {C_3} - {C_1}\end{array} \right\}$

= ${x^2}({a^2} + {b^2} + {c^2} + x)$.

Hence $\Delta $ is divisible by ${x^2}$ as well as by $ x.$

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

More "M.C.Q (1 Marks)" from 3 and 4 . determinant and metrices→3 and 4 . determinant and metrices — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int_{\pi /4}^{\pi /2} {{\rm{cose}}{{\rm{c}}^2}xdx = } $

→

The cartesian equation of line which is parallel to $3 \hat{i}+2 \hat{j}-8 \hat{k}$ and passes through the point $(5,2,-4)$ is __________ .

→

If the minimum and the maximum values of the function $f :\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \rightarrow R ,$ defined by :

$f (\theta)=\left|\begin{array}{ccc}-\sin ^{2} \theta & -1-\sin ^{2} \theta & 1 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 1 \\ 12 & 10 & -2\end{array}\right|$ are $m$ and $M$ respectively, then the ordered pair $( m , M )$ is equal to

→

If $\vec a$ and $\vec b$ are perpendicular unit vectors and vector $\vec c$ is such that $\vec c = \vec a + \vec b$ , then $\left( {\vec a \times \vec b} \right).\left( {\vec b \times \vec c} \right) + \left( {\vec b \times \vec c} \right).\left( {\vec c \times \vec a} \right) + \left( {\vec c \times \vec a} \right).\left( {\vec a \times \vec b} \right)$ is

→

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{\left(1+x^2\right)}} ; y(0)=0$. Then the area enclosed by the curve $f(\mathrm{x})=\mathrm{y}(\mathrm{x}) \mathrm{e}^{-\frac{1}{\left(1+\mathrm{x}^2\right)}}$ and the line $\mathrm{y}-\mathrm{x}=4$ is....................

→

The projection of the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ along the vector of $\hat{\text{j}}$ is: