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 system $x+2y+3z=4$,$4x+5y+3z=5$,$3x+4y+3z=\lambda$ is consistent and $3\lambda=n+100$,then $n=$

If the system of equations $x + y + z = 5$,$x + 2y + 3z = 9$,and $x + 3y + \alpha z = \beta$ has infinitely many solutions,then $\beta - \alpha$ equals:

The system of equations $x + 3y + 7 = 0$, $3x + 10y - 3z + 18 = 0$ and $3y - 9z + 2 = 0$ has

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

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 5 \\ 0 & 2 & 1 \end{bmatrix}$ and $b = \begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$. Which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module