If the system of simultaneous linear equation | vedclass

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

Question

(C) system of linear equations $AX=B$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero,i.e.,$|A| 
eq 0$

Vedclass · Accepted Answer

$\mu 
eq 23$ and $\lambda \in R$

Vedclass · Answer

(C) system of linear equations $AX=B$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero,i.e.,$|A| 
eq 0$.The coefficient matrix $A$ is given by:$A = \begin{bmatrix} 1 & 1 & 1 \ 5 & -1 & \mu \ 2 & 3 & -1 \end{bmatrix}$We calculate the determinant $|A|$:$|A| = 1((-1)(-1) - (3)(\mu)) - 1((5)(-1) - (2)(\mu)) + 1((5)(3) - (2)(-1))$$|A| = 1(1 - 3\mu) - 1(-5 - 2\mu) + 1(15 + 2)$$|A| = 1 - 3\mu + 5 + 2\mu + 17$$|A| = 23 - \mu$For a unique solution,we require $|A| 
eq 0$.$23 - \mu 
eq 0 \implies \mu 
eq 23$.Since the determinant is independent of $\lambda$,the condition for a unique solution depends only on $\mu$. Thus,$\mu 
eq 23$ and $\lambda$ can be any real number $(\lambda \in R)$.

If the system of simultaneous linear equations $x+y+z=\lambda$,$5x-y+\mu z=10$,and $2x+3y-z=6$ has a unique solution,then:

Similar Questions

If the homogeneous system of linear equations $x-2y+3z=0, 2x+4y-5z=0, 3x+\lambda y+\mu z=0$ has a non-trivial solution,then $8\mu+11\lambda=$

The value of $\lambda$ for which the system of equations $2x - y - z = 12,$ $x - 2y + z = -4,$ and $x + y + \lambda z = 4$ has no solution is

If the system of equations
$ 11 x+y+\lambda z=-5 $
$ 2 x+3 y+5 z=3 $
$ 8 x-19 y-39 z=\mu $
has infinitely many solutions,then $ \lambda^4-\mu $ is equal to :

The system of linear equations $x + 2y + z = -3$,$3x + 3y - 2z = -1$,and $2x + 7y + 7z = -4$ has:

Let $A$ be a matrix such that $AB$ is a scalar matrix where $B = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ and $\det(3A) = 27$. Then $3A^{-1} + A^2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module