(C) Given the matrix $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.We are given that $|A^3| = 27$.Using the property of determinants,$|

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) Given the matrix $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$.We are given that $|A^3| = 27$.Using the property of determinants,$|A^n| = |A|^n$,we have $|A|^3 = 27$.Taking the cube root on both sides,we get $|A| = 3$.Now,calculate the determinant of matrix $A$:$|A| = (\alpha \times \alpha) - (2 \times 2) = \alpha^2 - 4$.Equating this to $3$:$\alpha^2 - 4 = 3$$\alpha^2 = 7$$\alpha = \pm \sqrt{7}$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

Easy

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 27$,then $\alpha = $

KCET 2015 All KCET

A
$\pm 1$
B
$\pm 2$
C
$\pm \sqrt{7}$
D
$\pm \sqrt{5}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

3-4. Matrices and Determinants

Subtopic

Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

Previous Year

KCET 2015 Paper All KCET years →

If $P, Q$ and $R$ are angles of $\Delta PQR$, then the value of $\left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$ is equal to

MediumWBJEE 2012

View Solution

If $\Delta_k=\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & k & k-1 \\ 0 & k-1 & k\end{array}\right|$,then $\Delta_1+\Delta_2+\ldots+\Delta_{20}$ is equal to

EasyAP EAMCET 2021

View Solution

If $\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{array}\right|=0$, then $B$ is equal to

MediumTS EAMCET 2003

View Solution

Let $A(a, 0)$,$B(b, 2b+1)$,and $C(0, b)$,where $b \neq 0$ and $|b| \neq 1$,be points such that the area of triangle $ABC$ is $1 \, \text{sq. unit}$. Then,the sum of all possible values of $a$ is:

DifficultJEE MAIN 2021

View Solution

If $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}\right|$, then $f(2012)$ is equal to:

MediumAP EAMCET 2012

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 27$,then $\alpha = $

Similar Questions

If $P, Q$ and $R$ are angles of $\Delta PQR$, then the value of $\left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$ is equal to

If $\Delta_k=\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & k & k-1 \\ 0 & k-1 & k\end{array}\right|$,then $\Delta_1+\Delta_2+\ldots+\Delta_{20}$ is equal to

If $\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{array}\right|=0$, then $B$ is equal to

Let $A(a, 0)$,$B(b, 2b+1)$,and $C(0, b)$,where $b \neq 0$ and $|b| \neq 1$,be points such that the area of triangle $ABC$ is $1 \, \text{sq. unit}$. Then,the sum of all possible values of $a$ is:

If $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}\right|$, then $f(2012)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module