The solution of the linear system of equation | vedclass

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(A) Given the matrix equation: $\begin{bmatrix} 2 & 2 & 3 \ 7 & 1 & 1 \ 0 & 6 & 5 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{b

Vedclass · Accepted Answer

$x = 4, y = -3, z = 2$

Vedclass · Answer

(A) Given the matrix equation: $\begin{bmatrix} 2 & 2 & 3 \ 7 & 1 & 1 \ 0 & 6 & 5 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 3 y + 11 \ 6 z - 1 \ 5 y + 11 \end{bmatrix} + \begin{bmatrix} x \ x \ 4 z \end{bmatrix} + \begin{bmatrix} z \ 3 x \ 4 y \end{bmatrix}$ Multiplying the left side: $\begin{bmatrix} 2x + 2y + 3z \ 7x + y + z \ 6y + 5z \end{bmatrix} = \begin{bmatrix} x + 3y + z + 11 \ 4x + 6z - 1 \ 9y + 4z + 11 \end{bmatrix}$ Comparing the components,we get: $1) 2x + 2y + 3z = x + 3y + z + 11 \Rightarrow x - y + 2z = 11$ $2) 7x + y + z = 4x + 6z - 1 \Rightarrow 3x + y - 5z = -1$ $3) 6y + 5z = 9y + 4z + 11 \Rightarrow -3y + z = 11$ From equation $(3)$,$z = 3y + 11$. Substituting this into $(1)$ and $(2)$: $x - y + 2(3y + 11) = 11 \Rightarrow x + 5y = -11$ $3x + y - 5(3y + 11) = -1 \Rightarrow 3x - 14y = 54$ Solving the system $x + 5y = -11$ and $3x - 14y = 54$: $x = -11 - 5y \Rightarrow 3(-11 - 5y) - 14y = 54 \Rightarrow -33 - 15y - 14y = 54 \Rightarrow -29y = 87 \Rightarrow y = -3$ Then $x = -11 - 5(-3) = 4$ and $z = 3(-3) + 11 = 2$. Thus,the solution is $x = 4, y = -3, z = 2$.

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