Solve the system of linear equations using th | vedclass

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

Question

(C) The given system of equations can be written in the matrix form $AX = B$,where:$A = \begin{bmatrix} 2 & -1 \ 3 & 4 \end{bmatrix}$,$X = \begin{bma

Vedclass · Accepted Answer

$x = \frac{-5}{11}, y = \frac{12}{11}$

Vedclass · Answer

(C) The given system of equations can be written in the matrix form $AX = B$,where:$A = \begin{bmatrix} 2 & -1 \ 3 & 4 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \end{bmatrix}$,and $B = \begin{bmatrix} -2 \ 3 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (2)(4) - (-1)(3) = 8 + 3 = 11$.Since $|A| 
eq 0$,the matrix $A$ is non-singular and invertible.The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$:$A^{-1} = \frac{1}{11} \begin{bmatrix} 4 & 1 \ -3 & 2 \end{bmatrix}$.Now,solve for $X$ using $X = A^{-1}B$:$X = \frac{1}{11} \begin{bmatrix} 4 & 1 \ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 \ 3 \end{bmatrix}$$X = \frac{1}{11} \begin{bmatrix} (4)(-2) + (1)(3) \ (-3)(-2) + (2)(3) \end{bmatrix}$$X = \frac{1}{11} \begin{bmatrix} -8 + 3 \ 6 + 6 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} -5 \ 12 \end{bmatrix} = \begin{bmatrix} -\frac{5}{11} \ \frac{12}{11} \end{bmatrix}$.Thus,$x = -\frac{5}{11}$ and $y = \frac{12}{11}$.

Solve the system of linear equations using the matrix method: $2x - y = -2$ and $3x + 4y = 3$.

Similar Questions

Let the system of equations: $2x + 3y + 5z = 9$,$7x + 3y - 2z = 8$,$12x + 3y - (4 + \lambda)z = 16 - \mu$ have infinitely many solutions. Then the radius of the circle centred at $(\lambda, \mu)$ and touching the line $4x = 3y$ is

If the system of equations $kx + (k+1)y + (k-1)z = 0$,$(k-1)x + (k+2)y + kz = 0$,and $(k+1)x + ky + (k+2)z = 0$ has a non-trivial solution,then the sum of all possible values of $k$ is:

Let $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$ such that $AX = B$,then $X =$

The value of $\lambda$ such that the system of equations $2x-y-2z=2$, $x-2y+z=-4$, and $x+y+\lambda z=4$ has no solution, is:

Consider two systems of $3$ linear equations in $3$ unknowns $AX=B$ and $CX=D$. If $AX=B$ has a unique solution $D$ and $CX=D$ has a unique solution $B$,then the solution of $(A-C^{-1})X=O$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module