(A) matrix $A$ is not invertible if and only if its determinant is zero,i.e.,$|A| = 0$. Given the matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1

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

(A) matrix $A$ is not invertible if and only if its determinant is zero,i.e.,$|A| = 0$. Given the matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$. Calculating the determinant $|A|$: $|A| = a(0 \times 2 - 1 \times 1) - (-1)(-3 \times 2 - 1 \times (-1)) + 4(-3 \times 1 - 0 \times (-1)) = 0$ $|A| = a(0 - 1) + 1(-6 + 1) + 4(-3 - 0) = 0$ $|A| = a(-1) + 1(-5) + 4(-3) = 0$ $-a - 5 - 12 = 0$ $-a - 17 = 0$ $a = -17$. Thus,the matrix is not invertible when $a = -17$.

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

The matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is not invertible only if $a =$

MHT CET 2020 All MHT CET

A
-$17$
B
-$16$
C
$16$
D
$17$

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

Previous Year

MHT CET 2020 Paper All MHT CET years →

If $\omega$ is a cube root of unity,then a root of the equation $\left| \begin{array}{ccc} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{array} \right| = 0$ is

Easy

View Solution

If the system of equations $2x + 3y - z = 0$,$x + ky - 2z = 0$ and $2x - y + z = 0$ has a non-trivial solution $(x, y, z)$,then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to

DifficultJEE MAIN 2019

View Solution

If $\omega$ is a complex cube root of unity,then the value of the determinant $\left| \begin{array}{ccc} 2 & 2\omega & -\omega^2 \\ 1 & 1 & 1 \\ 1 & -1 & 0 \end{array} \right|$ is:

Easy

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 $\Delta = \begin{vmatrix} 1 & \cos \theta & 1 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{vmatrix}$,then $\Delta$ lies in the interval

EasyAP EAMCET 2017

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

The matrix $A = \begin{bmatrix} a & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is not invertible only if $a =$

Similar Questions

If $\omega$ is a cube root of unity,then a root of the equation $\left| \begin{array}{ccc} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{array} \right| = 0$ is

If the system of equations $2x + 3y - z = 0$,$x + ky - 2z = 0$ and $2x - y + z = 0$ has a non-trivial solution $(x, y, z)$,then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to

If $\omega$ is a complex cube root of unity,then the value of the determinant $\left| \begin{array}{ccc} 2 & 2\omega & -\omega^2 \\ 1 & 1 & 1 \\ 1 & -1 & 0 \end{array} \right|$ is:

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

If $\Delta = \begin{vmatrix} 1 & \cos \theta & 1 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{vmatrix}$,then $\Delta$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module