MCQ
Choose the correct answer from given four options in each of the Exercise:
The determinant $\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$ equals to:
  • A
    $abc(b - c)(c - a)(a - b)$
  • B
    $(b - c)(c - a)(a - b)$
  • C
    $(a + b + c)(b - c)(c - a)(a - b)$
  • D
    None of these

Answer

$\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}=\begin{vmatrix}\text{b}(\text{b}-\text{a})&\text{b}-\text{c}&\text{c}(\text{b}-\text{a})\\\text{a}(\text{b}-\text{a})&\text{a}-\text{b}&\text{b}(\text{b}-\text{a})\\\text{c}(\text{b}-\text{a})&\text{c}-\text{a}&\text{a}(\text{b}-\text{a})\end{vmatrix}$
$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}&\text{b}-\text{c}&\text{c}\\\text{a}&\text{a}-\text{b}&\text{b}\\\text{c}&\text{c}-\text{a}&\text{a}\end{vmatrix}$
$[$on taking $(b - a)$ common from $C_1$ and $C_3$ each$]$
$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}-\text{c}&\text{b}-\text{c}&\text{c}\\\text{a}-\text{b}&\text{a}-\text{b}&\text{b}\\\text{c}-\text{a}&\text{c}-\text{a}&\text{a}\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_1-\text{C}_3\big]$
$=0$
$[$Since, two columns $C_1$ and $C_2$ are identical, so the value of determinant is zero$]$

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

Similar questions

If $A$ is a non$-$singular square matrix of order $3$ such that $A^2=3 A,$ then value of $|A|$ is
Choose the correct answer from the given four options.
If $|\vec{\text{a}}|=4$ and $-3\leq\lambda\leq2,$ then the range of $|\lambda\vec{\text{a}}|$ is:
  1. [0,8]
  2. [-12,8]
  3. [0,12]
  4. [8,12]
If $m$ and $n$, respectively, are the order and the degree of the differential equation $\frac{d}{d x}\left[\left(\frac{d y}{d x}\right)\right]^4=0$, then $m+n=$
A point P lies on the line segment joining the points (-1, 3, 2) and (5, 0, 6). If x-coordinate of P is 2, then its z-coordinate is:
Find the value of x, if $\begin{bmatrix}1&-1\\3&-5\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}^2\\3&-5\end{bmatrix}$ is:
  1. $\text{x}=2,-\frac{1}{3}$
  2. $\text{x}=-1,-\frac{1}{3}$
  3. $\text{x}=-2,-\frac{1}{3}$
  4. $\text{x}=0,-\frac{1}{3}$
$\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)+4 \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is equal to
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are any three mutualy perpendicular vectors of equal magnitude a, then $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|$ is equal to
  1. $\text{a}$
  2. $\sqrt{2}\text{a}$
  3. $\sqrt{3}\text{a}$
  4. $2\text{a}$
  5. $\text{None of these}$
Find the value of $\lambda$ so that the vectors $2 \hat{i}-4 \hat{j}+\hat{k}$ and $4 \hat{i}-8 \hat{j}+\lambda \hat{k}$ are perpendicular.
If the corner points of the feasible region of an LPP are $(0,3),(3,2)$ and $(0,5)$, then the minimum value of $z=11 x+7 y$ is
The general solution of the differntial equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$ is:
  1. $\log\text{y}=\text{kx}$
  2. $\text{y}=\text{kx}$
  3. $\text{xy}=\text{k}$
  4. $\text{y}=\text{k}\log\text{x}$