If A= [ cc & - & ] and A adj A= [ ll k & 0 0 & k ], then k is equ… - Vidyadip

MCQ

If $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ and $A \cdot \operatorname{adj} A=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]$, then $k$ is equal to

A
$0$
✓
1
C
$\sin \alpha \cos \alpha$
D
$\cos 2 \alpha$

✓

Answer

Correct option: B.

1

(B) Using Shortcut 4(i),
$A (\operatorname{adj} A )=| A | \cdot I$
$\therefore \quad\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]=\left(\cos ^2 \alpha+\sin ^2 \alpha\right)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow k=1$

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 "MCQ" from Matrics→Matrics — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

If the vectors $x \hat{i}-3 \hat{j}+7 \hat{k}$ and $\hat{i}+y \hat{j}-z \hat{k}$ are collinear then the value of $\frac{x y^2}{z}$ is equal to

If $\bar{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k}), \bar{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$, then the value of $(2 \bar{a}-\bar{b}) \cdot\{(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})\}$ is

The position vector of a point $R$ which divides the line joining two points P and Q whose position vectors are $\hat{i}+2\hat{j}- \hat{k}$ and -$\hat{i}+\hat{j}- \hat{k}$ respectively, in the ratio $2: 1$ externally is

If $X$ is a random variable with probability mass function
$P (x)=k x \quad, \quad$ for $x=1,2,3$
$=0$. otherwise then $k=\ldots$

If $\log 10\left(\frac{x^3-y^3}{x^3+y^3}\right)=2$, then $\frac{d y}{d x}=$

$\int_1^e \frac{1+\log x}{x} d x=$

For any three vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ the expression $\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big\}$ equals:

If $A=\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]$ and $A(\operatorname{adj} A)=K I$, then the value of $K$ is (where $I$ is unit matrix of order 3 )

If the slope of one of the lines represented by $a x^2+2 h x y+b y^2=0$ be the square of the other, then

At $\text{x}=\frac{5\pi}{6}, $ $\text{f}(\text{x})=2\sin3\text{x}+3 \cos3\text{x}$ is: