Matrix A = [ * 20 c 1&0& - k 2&1&3 &0&1 ] is invertible for - Vidyadip

MCQ

Matrix $A = \left[ {\begin{array}{*{20}{c}}1&0&{ - k}\\2&1&3\\k&0&1\end{array}} \right]$ is invertible for

A
$k = 1$
B
$k = - 1$
C
$k = 0$
✓
All real $k$

✓

Answer

Correct option: D.

All real $k$

d
(d) On expansion, $|A| = {k^2} + 1$, which can be never zero.

.Hence matrix $A$ is invertible for all real $k.$

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 STD 12 - 3 and 4 . determinant and metrices→STD 12 - 3 and 4 . determinant and metrices — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Which of the following is correct:

Let $A \equiv (\lambda + 2, 1 - 2\lambda , \lambda + 2)$ and $B \equiv (2k + 1, k, k +1)$ and $ \lambda , k \in R.$ Then minimum distance between $A$ and $B$ is -

If $y = \sqrt {\sin \sqrt x } $, then ${{dy} \over {dx}} = $

An iso$-$profit line represents:

Area bounded by the curve $\text{y}=\cos\text{x}$ between $\text{x}=0$ and $\text{x}=3\frac{\pi}{2}$ is :

The solution of the differential equation $({x^2} + {y^2})dx = 2xydy$ is

Match List $I$ with List $II$ and select the correct answer using the code given below the lists :

List $I$	List $II$
$P.\quad$ Volume of parallelopiped determined by vectors $\vec{a}, \vec{b}$ and $\overrightarrow{ c }$ is $2$ . Then the volume of the parallelepiped determined by vectors $2(\vec{a} \times \vec{b}), 3(\vec{b} \times \vec{c})$ and $(\vec{c} \times \vec{a})$ is	$1.\quad$ $100$
$Q.\quad$ Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $5$ . Then the volume of the parallelepiped determined by vectors $3(\overrightarrow{ a }+\overrightarrow{ b }),(\overrightarrow{ b }+\overrightarrow{ c })$ and $2(\overrightarrow{ c }+\overrightarrow{ a })$ is	$2.\quad$ $30$
$R.\quad$ Area of a triangle with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $20$ . Then the area of the triangle with adjacent sides determined by vectors $(2 \vec{a}+3 \vec{b})$ and $(\vec{a}-\vec{b})$ is	$3.\quad$ $24$
$S.\quad$ Area of a paralelogram with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $30$ . Then the area of the parallelogram with adjacent sides determined by vectors $(\vec{a}+\vec{b})$ and $\vec{a}$ is	$4.\quad$ $60$

Codes: $ \quad P \quad Q \quad R \quad S $

Let $\vec{x}, \vec{y}$ and $\vec{z}$ be three vectors each of magnitude $\sqrt{2}$ and the angle between each pair of them is $\frac{\pi}{3}$. If $\vec{a}$ is a nonzero vector perpendicular to $\vec{x}$ and $\vec{y} \times \vec{z}$ and $\vec{b}$ is a nonzero vector perpendicular to $\vec{y}$ and $\overrightarrow{ z } \times \overrightarrow{ x }$, then

$(A)$ $\vec{b}=(\vec{b} \cdot \vec{z})(\vec{z}-\vec{x})$

$(B)$ $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{y}-\vec{z})$

$(C)$ $\vec{a} \cdot \vec{b}=-(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$

$(D)$ $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{z}-\vec{y})$

If abscissa of vertex of parabola $y = a{x^2} + bx + c$ is $1\left( {a,b,c > 0} \right)$ and $f(x) = \int\limits_0^x {\left( {3a{x^2} + bx + c} \right)dx} $ is strictly increasing function $\forall \,\,\,x\, \in \,R$ , then maximum possible value of $\left[ {\frac{a}{c}} \right]$ is (where [.] denotes greatest integer function)

$\int_{}^{} {\frac{1}{{({x^2} - 1)\sqrt {{x^2} + 1} }}} \;dx = $