If {a_1}x + {b_1}y + {c_1}z = 0, {a_2}x + {b_ | vedclass

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

Question

(D) The given system of equations is a homogeneous system of linear equations of the form $AX = 0$,where $A$ is the coefficient matrix and $X = [x, y,

Vedclass · Accepted Answer

Infinite solution

Vedclass · Answer

(D) The given system of equations is a homogeneous system of linear equations of the form $AX = 0$,where $A$ is the coefficient matrix and $X = [x, y, z]^T$.For a homogeneous system $AX = 0$,the nature of the solution depends on the determinant of the coefficient matrix,denoted by $|A|$.If $|A| 
eq 0$,the system has only the trivial solution $(x = 0, y = 0, z = 0)$.If $|A| = 0$,the system has infinitely many non-trivial solutions in addition to the trivial solution.Since the problem states that $\left| \begin{matrix} {a_1} & {b_1} & {c_1} \ {a_2} & {b_2} & {c_2} \ {a_3} & {b_3} & {c_3} \end{matrix} ight| = 0$,the determinant of the coefficient matrix is zero.Therefore,the system has infinitely many solutions.

If ${a_1}x + {b_1}y + {c_1}z = 0, {a_2}x + {b_2}y + {c_2}z = 0, {a_3}x + {b_3}y + {c_3}z = 0$ and $\left| \begin{matrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{matrix} \right| = 0$,then the given system has

Similar Questions

If the system of linear equations $8x + y + 4z = -2$,$x + y + z = 0$,and $\lambda x - 3y = \mu$ has infinitely many solutions,then the distance of the point $\left(\lambda, \mu, -\frac{1}{2}\right)$ from the plane $8x + y + 4z + 2 = 0$ is:

If $x=\alpha, y=\beta, z=\gamma$ is the solution for the system of equations:
$\begin{aligned} 2x-y+8z &= 13 \\ 3x+4y+5z &= 18 \\ 5x-2y+7z &= 20 \end{aligned}$
then $\alpha\beta+\beta\gamma+\gamma\alpha=$

The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ is equal to:

The system of equations $\lambda x - y + (\cos\theta) z = 0$,$3x + y + 2z = 0$,and $(\cos\theta) x + y + 2z = 0$ for $0 < \theta < 2\pi$ has non-trivial solution$(s)$:

If the solution of the system of simultaneous linear equations $x+y-z=6$,$3x+2y-z=5$ and $2x-y-2z+3=0$ is $x=\alpha, y=\beta, z=\gamma$,then $\alpha+\beta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module