The matrix begin{bmatrix} lambda & -1 & 4 -3 | vedclass

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

Question

(B) square matrix $A$ is invertible if and only if its determinant is non-zero,i.e.,$|A| 
eq 0$.Given the matrix $A = \begin{bmatrix} \lambda & -1 &

Vedclass · Accepted Answer

$\lambda 
eq -17$

Vedclass · Answer

(B) square matrix $A$ is invertible if and only if its determinant is non-zero,i.e.,$|A| 
eq 0$.Given the matrix $A = \begin{bmatrix} \lambda & -1 & 4 \ -3 & 0 & 1 \ -1 & 1 & 2 \end{bmatrix}$.Calculating the determinant $|A|$:$|A| = \lambda(0 	imes 2 - 1 	imes 1) - (-1)(-3 	imes 2 - 1 	imes (-1)) + 4(-3 	imes 1 - 0 	imes (-1))$$|A| = \lambda(0 - 1) + 1(-6 + 1) + 4(-3 - 0)$$|A| = -\lambda - 5 - 12$$|A| = -\lambda - 17$For the matrix to be invertible,$|A| 
eq 0$:$-\lambda - 17 
eq 0$$-\lambda 
eq 17$$\lambda 
eq -17$.

The matrix $\begin{bmatrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is invertible,if

Similar Questions

If $A = \begin{bmatrix} 2-k & 2 \\ 1 & 3-k \end{bmatrix}$ is a singular matrix,then the value of $5k - k^2$ is equal to

If the system of equations
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
$(k+1) x + (k+2) y = k+3$
$x + y = 1$
is consistent,then the value of $k$ is

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

If $\left|\begin{array}{ccc}2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1\end{array}\right|=Ak^2+Bk+C$, then $A+B+C=$

If $\left| \begin{array}{ccc} x+1 & 1 & 1 \\ 2 & x+2 & 2 \\ 3 & 3 & x+3 \end{array} \right| = 0$,then $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module