The system of linear equations x + y + z = 2, | vedclass

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

Question

(A) 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| 
e 0

Vedclass · Accepted Answer

$k 
e 0$

Vedclass · Answer

(A) 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.,$|A| 
e 0$.The coefficient matrix $A$ is given by:$A = \begin{bmatrix} 1 & 1 & 1 \ 2 & 1 & -1 \ 3 & 2 & k \end{bmatrix}$We calculate the determinant $|A|$:$|A| = 1(1 \cdot k - (-1) \cdot 2) - 1(2 \cdot k - (-1) \cdot 3) + 1(2 \cdot 2 - 1 \cdot 3)$$|A| = 1(k + 2) - 1(2k + 3) + 1(4 - 3)$$|A| = k + 2 - 2k - 3 + 1$$|A| = -k$For a unique solution,we require $|A| 
e 0$,which implies $-k 
e 0$,or $k 
e 0$.

The system of linear equations $x + y + z = 2$,$2x + y - z = 3$,and $3x + 2y + kz = 4$ has a unique solution if

Similar Questions

The existence of a unique solution for the system of equations $x+y+z=\beta$,$5x-y+\alpha z=10$,and $2x+3y-z=6$ depends on:

If the system of linear equations $x-2y+z=-4$; $2x+\alpha y+3z=5$; $3x-y+\beta z=3$ has infinitely many solutions,then $12\alpha+13\beta$ is equal to

Let the system of linear equations $x+y+\alpha z=2$,$3x+y+z=4$,and $x+2z=1$ have a unique solution $(x^{}, y^{}, z^{})$. If $(\alpha, x^{}), (y^{}, \alpha)$ and $(x^{}, -y^{*})$ are collinear points,then the sum of absolute values of all possible values of $\alpha$ is

The number of solutions of the system of equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y - 3z = 5$ is:

If the system of linear equations,$x+y+z = 6$,$x+2y+3z = 10$,and $3x+2y+\lambda z = \mu$ has more than two solutions,then $\mu-\lambda^{2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module