Download App

Algebra of Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices1 Mark
MCQ
If $\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}^\text{k}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then the least positive integral value of k is:
  • A
    3
  • B
    4
  • C
    6
  • ✓
    7
✓

Answer

Correct option: D.
7
$\text{A}=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$$\Rightarrow\text{A}^2=\text{A}\times\text{A}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos^2\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}&\Big(-2\cos\frac{2\pi}{7}-\sin\frac{2\pi}{7}\Big)\\2\cos\frac{2\pi}{7}\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos^2\theta-\sin^2\theta=\cos2\theta\\2\sin\theta\cos\theta=\sin2\theta\end{bmatrix}$
$\Rightarrow\text{A}^3=\text{A}^2\times\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\Big(\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}-\sin\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}-\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\\\Big(\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos\text{(A+B)}=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}\\\sin\text{(A+B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\end{bmatrix}$
Now we check if the pattern is same for k = 6.
Here,
$\text{A}^6=\text{A}^3.\text{A}^3$
$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{12\pi}{7}&-\sin\frac{12\pi}{7}\\\sin\frac{12\pi}{7}&\cos\frac{12\pi}{7}\end{bmatrix}$
Now, we check if the pattern is same for k = 7.
Here,
$\text{A}^7=\text{A}^6\times\text{A}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{14\pi}{7}&-\sin\frac{14\pi}{7}\\\sin\frac{14\pi}{7}&\cos\frac{14\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos2\pi&-\sin2\pi\\\sin2\pi&\cos2\pi\end{bmatrix}$$\begin{bmatrix}\because\ \frac{14\pi}{7}=2\pi\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
So, the least positive integral value of k is 7.

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 Algebra of Matrices→Algebra of Matrices — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The area of a trapezium is defined by function $f$ and given by $f(x)=(10+x) \sqrt{100-x^2}$, then the area when it is maximised is
→
The angle between the lines $\frac{{x + 4}}{1} = \frac{{y - 3}}{2} = \frac{{z + 2}}{3}$ and $\frac{x}{3} = \frac{{y - 1}}{{ - 2}} = \frac{z}{1}$ is
→
In a sphere the rate of change of surface area is:
→
The differential equation of all circles having centre lies on $x$ -axis and touches the $y$ -axis at origin is-
→
The projection of the line segment joining the points $(1,-1,3)$ and $(2,-4,11)$ on the line joining the points $(-1,2,3)$ and $(3,-2,10)$ is
→
Choose the correct answer from the given four options : The points at which the tangents to the curve $y = x^3 - 12x + 18$ are parallel to $x-$ axis are : $z$
→
Let $R$ be a relation on the set $N$ be defined by $\{(x, y)| x, y \in N, 2x + y = 41\}$. Then $R$ is
→
When the tangent to the curve $\text{y}=\text{x}\log\text{x}$ is parallel to the chord joining the points $(1, 0)$ and $(e, e),$ the value of $x$ is :
→
Let $a-2 b+c=1$

If $f(x)=\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|,$ then

→
The solution of the differential equation $y - x\frac{{dy}}{{dx}} = a\left( {{y^2} + \frac{{dy}}{{dx}}} \right)$ is
→