The following system of equations x+y+z=9, 2x | vedclass

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

Question

(D) The given system of equations is:$x+y+z=9$$2x+5y+7z=52$$x+7y+11z=77$We represent this system as an augmented matrix $[A|B]$:$[A|B] = \begin{bmatri

Vedclass · Accepted Answer

infinitely many solutions

Vedclass · Answer

(D) The given system of equations is:$x+y+z=9$$2x+5y+7z=52$$x+7y+11z=77$We represent this system as an augmented matrix $[A|B]$:$[A|B] = \begin{bmatrix} 1 & 1 & 1 & 9 \ 2 & 5 & 7 & 52 \ 1 & 7 & 11 & 77 \end{bmatrix}$Applying row operations $R_2 ightarrow R_2-2R_1$ and $R_3 ightarrow R_3-R_1$:$\begin{bmatrix} 1 & 1 & 1 & 9 \ 0 & 3 & 5 & 34 \ 0 & 6 & 10 & 68 \end{bmatrix}$Applying row operation $R_3 ightarrow R_3-2R_2$:$\begin{bmatrix} 1 & 1 & 1 & 9 \ 0 & 3 & 5 & 34 \ 0 & 0 & 0 & 0 \end{bmatrix}$Here, the rank of the coefficient matrix $ho(A) = 2$ and the rank of the augmented matrix $ho(A|B) = 2$. Since $ho(A) = ho(A|B) < 3$ (where $3$ is the number of variables), the system has infinitely many solutions.

The following system of equations $x+y+z=9$, $2x+5y+7z=52$, $x+7y+11z=77$ has

Similar Questions

The system of equations $\begin{cases} \alpha x + y + z = \alpha - 1 \\ x + \alpha y + z = \alpha - 1 \\ x + y + \alpha z = \alpha - 1 \end{cases}$ has no solution,if $\alpha$ is

If the system of equations $x + 2y - 3z = 1$,$(k + 3)z = 3$,and $(2k + 1)x + z = 0$ is inconsistent,then the value of $k$ is

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 $x + y + z = 2$,$3x - y + 2z = 6$ and $3x + y + z = -18$ has

If the system of linear equations: $x + y + z = 6, x + 2y + 5z = 10, 2x + 3y + \lambda z = \mu$ has infinitely many solutions, then the value of $\lambda + \mu$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module