If the system of equations x+y+z=5, x+2y+2z=6 | vedclass

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

Question

(A) The given system of linear equations is:$x+y+z=5$$x+2y+2z=6$$x+3y+\lambda z=\mu$The system is solvable by the Matrix Inversion Method if and only

Vedclass · Accepted Answer

$\lambda 
eq 3, \mu \in R$

Vedclass · Answer

(A) The given system of linear equations is:$x+y+z=5$$x+2y+2z=6$$x+3y+\lambda z=\mu$The system is solvable by the Matrix Inversion Method if and only if the determinant of the coefficient matrix $A$ is non-zero $(|A| 
eq 0)$.The coefficient matrix $A$ is given by:$A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 2 \ 1 & 3 & \lambda \end{bmatrix}$Calculating the determinant $|A|$:$|A| = 1(2\lambda - 6) - 1(\lambda - 2) + 1(3 - 2)$$|A| = 2\lambda - 6 - \lambda + 2 + 1$$|A| = \lambda - 3$For the system to be solvable by the Matrix Inversion Method, we require $|A| 
eq 0$, which implies $\lambda - 3 
eq 0$, or $\lambda 
eq 3$.Since the matrix is invertible for any $\lambda 
eq 3$, the system has a unique solution for any value of $\mu \in R$.Therefore, the condition is $\lambda 
eq 3, \mu \in R$.

If the system of equations $x+y+z=5$, $x+2y+2z=6$, and $x+3y+\lambda z=\mu$ (where $\lambda, \mu \in R$) is solvable by the Matrix Inversion Method, then:

Similar Questions

Let $A = \begin{bmatrix} 1 & -4 & 7 \\ 0 & 3 & -5 \\ -2 & 5 & -9 \end{bmatrix}$ and $B = \begin{bmatrix} a \\ -b \\ -c \end{bmatrix}$. If $A$ and $[A: B]$ have the same rank, then:

If $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4 \end{bmatrix}$,$B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$ and $X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ such that $AX = B$,then the value of $x_1 + x_2 + x_3$ is:

Given the system of linear equations: $2x + 3y + 4z = 9$,$4x + 9y + 3z = 10$,and $5x + 10y + 5z = 11$. The value of $x$ is given by:

The solution of equations $x + y = 10$,$2x + y = 18$,and $4x - 3y = 26$ is:

If $\begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ 0 & -6 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix}$,then $(x, y, z) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module