Solve the following system of equations by ma | vedclass

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

Question

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

Vedclass · Accepted Answer

$x = 1, y = 2, z = 3$

Vedclass · Answer

(B) The system of equations can be written in the form $AX = B$,where$A = \begin{bmatrix} 3 & -2 & 3 \ 2 & 1 & -1 \ 4 & -3 & 2 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \ z \end{bmatrix}$,and $B = \begin{bmatrix} 8 \ 1 \ 4 \end{bmatrix}$.First,we calculate the determinant $|A| = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4) = 3(-1) + 2(8) + 3(-10) = -3 + 16 - 30 = -17$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.Next,we find the cofactor matrix of $A$:$C_{11} = (2 - 3) = -1, C_{12} = -(4 + 4) = -8, C_{13} = (-6 - 4) = -10$$C_{21} = -(-4 + 9) = -5, C_{22} = (6 - 12) = -6, C_{23} = -(-9 + 8) = 1$$C_{31} = (2 - 3) = -1, C_{32} = -(-3 - 6) = 9, C_{33} = (3 + 4) = 7$Thus,$adj(A) = \begin{bmatrix} -1 & -5 & -1 \ -8 & -6 & 9 \ -10 & 1 & 7 \end{bmatrix}$.$A^{-1} = \frac{1}{|A|} adj(A) = -\frac{1}{17} \begin{bmatrix} -1 & -5 & -1 \ -8 & -6 & 9 \ -10 & 1 & 7 \end{bmatrix}$.Now,$X = A^{-1}B = -\frac{1}{17} \begin{bmatrix} -1 & -5 & -1 \ -8 & -6 & 9 \ -10 & 1 & 7 \end{bmatrix} \begin{bmatrix} 8 \ 1 \ 4 \end{bmatrix} = -\frac{1}{17} \begin{bmatrix} -8 - 5 - 4 \ -64 - 6 + 36 \ -80 + 1 + 28 \end{bmatrix} = -\frac{1}{17} \begin{bmatrix} -17 \ -34 \ -51 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}$.Therefore,$x = 1, y = 2, z = 3$.

Solve the following system of equations by matrix method: $3x - 2y + 3z = 8$,$2x + y - z = 1$,$4x - 3y + 2z = 4$.

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

The values of $p$ and $q$ so that the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$ and $x + y + 3z = 4$ may have no solution are

If $A$ and $B$ are the two real values of $k$ for which the system of equations $x+2y+z=1$,$x+3y+4z=k$,and $x+5y+10z=k^2$ is consistent,then $A+B=$

If $3X + 2Y = I$ and $2X - Y = O$,where $I$ and $O$ are unit and null matrices of order $3$ respectively,then

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module