(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

If $a+x=b+y=c+z+1,$ where $a, b, c, x, y, z$ are non-zero distinct real numbers,then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to

DifficultJEE MAIN 2020

View Solution

For any $a, b, c \in R$,the determinant $\left|\begin{array}{lll}bc & b+c & 1 \\ ca & c+a & 1 \\ ab & a+b & 1\end{array}\right|$ is equal to

EasyAP EAMCET 2020

View Solution

If $a$ is a non-real complex number for which the system of equations $ax - a^2y + a^3z = 0$,$-a^2x + a^3y + az = 0$,and $a^3x + ay - a^2z = 0$ has non-trivial solutions,then $|a|$ is:

View Solution

If $a, b, c$ are positive and unequal,show that the value of the determinant $\Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$ is negative.

Difficult

View Solution

The value of the determinant $\left| \begin{array}{ccc} 4 & -6 & 1 \\ -1 & -1 & 1 \\ -4 & 11 & -1 \end{array} \right|$ is

Easy

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

If $a+x=b+y=c+z+1,$ where $a, b, c, x, y, z$ are non-zero distinct real numbers,then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to

For any $a, b, c \in R$,the determinant $\left|\begin{array}{lll}bc & b+c & 1 \\ ca & c+a & 1 \\ ab & a+b & 1\end{array}\right|$ is equal to

If $a$ is a non-real complex number for which the system of equations $ax - a^2y + a^3z = 0$,$-a^2x + a^3y + az = 0$,and $a^3x + ay - a^2z = 0$ has non-trivial solutions,then $|a|$ is:

If $a, b, c$ are positive and unequal,show that the value of the determinant $\Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$ is negative.

The value of the determinant $\left| \begin{array}{ccc} 4 & -6 & 1 \\ -1 & -1 & 1 \\ -4 & 11 & -1 \end{array} \right|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module