Solve the system of equations 2x + 5y = 1 and | vedclass

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

Question

(A) The system of equations can be written in the form $AX = B$,where$A = \begin{bmatrix} 2 & 5 \ 3 & 2 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \e

Vedclass · Accepted Answer

$x = 3, y = -1$

Vedclass · Answer

(A) The system of equations can be written in the form $AX = B$,where$A = \begin{bmatrix} 2 & 5 \ 3 & 2 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \end{bmatrix}$,and $B = \begin{bmatrix} 1 \ 7 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = (2)(2) - (5)(3) = 4 - 15 = -11$.Since $|A| 
eq 0$,the system has a unique solution given by $X = A^{-1}B$.The inverse of $A$ is $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = -\frac{1}{11} \begin{bmatrix} 2 & -5 \ -3 & 2 \end{bmatrix}$.Now,$X = -\frac{1}{11} \begin{bmatrix} 2 & -5 \ -3 & 2 \end{bmatrix} \begin{bmatrix} 1 \ 7 \end{bmatrix}$.$X = -\frac{1}{11} \begin{bmatrix} (2)(1) + (-5)(7) \ (-3)(1) + (2)(7) \end{bmatrix} = -\frac{1}{11} \begin{bmatrix} 2 - 35 \ -3 + 14 \end{bmatrix} = -\frac{1}{11} \begin{bmatrix} -33 \ 11 \end{bmatrix}$.$X = \begin{bmatrix} 3 \ -1 \end{bmatrix}$.Thus,$x = 3$ and $y = -1$.

Solve the system of equations $2x + 5y = 1$ and $3x + 2y = 7$.

Similar Questions

If the system of simultaneous linear equations $x+y+z=a$,$x-y+bz=2$,and $2x+3y-z=1$ has infinitely many solutions,then $b-5a=$

The system $2x + 3y + z = 5$,$3x + y + 5z = 7$ and $x + 4y - 2z = 3$ has

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be a $2 \times 2$ real matrix with $\det(A) = 1$. If the equation $\det(A - \lambda I_2) = 0$ has imaginary roots (where $I_2$ is the identity matrix of order $2$), then:

If $x = \alpha, y = \beta, z = \gamma$ is the unique solution of the system of equations $5x - 2y + 3z = 0$, $7x + 10y - 8z = 3$ and $2x + 3y - 4z = -4$, then $\beta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module