(A) To find the determinant of matrix $A$,we expand along the first row: $|A| = \begin{vmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{vmatrix}$ $|A

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

(A) To find the determinant of matrix $A$,we expand along the first row: $|A| = \begin{vmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{vmatrix}$ $|A| = 1 \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} - 0 \begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix}$ $|A| = 1(1 \times 1 - 0 \times 2) - 0 + 1(2 \times 2 - 1 \times 3)$ $|A| = 1(1 - 0) + 1(4 - 3)$ $|A| = 1(1) + 1(1) = 1 + 1 = 2$ Thus,$\det A = 2$.

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

Easy

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det A$ =

A
$2$
B
$3$
C
$4$
D
$5$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

3 and 4 .Determinants and Matrices

Subtopic

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

The set of all values of $\lambda$ for which the system of linear equations $x - 2y - 2z = \lambda x$,$x + 2y + z = \lambda y$,and $-x - y = \lambda z$ has non-zero solutions.

DifficultJEE MAIN 2019

View Solution

The sum of the real roots of the equation $\left| \begin{matrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{matrix} \right| = 0$ is equal to

DifficultJEE MAIN 2019

View Solution

The value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$ is

Medium

View Solution

If $\alpha+\beta+\gamma=2 \pi$,then the system of equations
$x+(\cos \gamma) y+(\cos \beta) z=0$
$(\cos \gamma) x+y+(\cos \alpha) z=0$
$(\cos \beta) x+(\cos \alpha) y+z=0$
has :

DifficultJEE MAIN 2021

View Solution

If $C = 2 \cos \theta$,then the value of the determinant $\Delta = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$ is

Medium

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 $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det A$ =

Similar Questions

The set of all values of $\lambda$ for which the system of linear equations $x - 2y - 2z = \lambda x$,$x + 2y + z = \lambda y$,and $-x - y = \lambda z$ has non-zero solutions.

The sum of the real roots of the equation $\left| \begin{matrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{matrix} \right| = 0$ is equal to

The value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$ is

If $\alpha+\beta+\gamma=2 \pi$,then the system of equations$x+(\cos \gamma) y+(\cos \beta) z=0$$(\cos \gamma) x+y+(\cos \alpha) z=0$$(\cos \beta) x+(\cos \alpha) y+z=0$has :

If $C = 2 \cos \theta$,then the value of the determinant $\Delta = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If $\alpha+\beta+\gamma=2 \pi$,then the system of equations
$x+(\cos \gamma) y+(\cos \beta) z=0$
$(\cos \gamma) x+y+(\cos \alpha) z=0$
$(\cos \beta) x+(\cos \alpha) y+z=0$
has :