In matrix notation,if the system of equations | vedclass

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(B) Given the matrix equation: $\begin{bmatrix} 1 \ -1 \ 2 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{b

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a plane not parallel to any of the coordinate planes.

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(B) Given the matrix equation: $\begin{bmatrix} 1 \ -1 \ 2 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 5 \ -5 \ 10 \end{bmatrix}$ Multiplying the first two matrices: $\begin{bmatrix} 1(1) & 1(-1) & 1(2) \ -1(1) & -1(-1) & -1(2) \ 2(1) & 2(-1) & 2(2) \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 5 \ -5 \ 10 \end{bmatrix}$ $\Rightarrow \begin{bmatrix} 1 & -1 & 2 \ -1 & 1 & -2 \ 2 & -2 & 4 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 5 \ -5 \ 10 \end{bmatrix}$ This results in the system of linear equations: $x - y + 2z = 5$ $-x + y - 2z = -5$ $2x - 2y + 4z = 10$ All three equations are equivalent to the single plane equation $x - y + 2z = 5$. Since the coefficients of $x, y, z$ are all non-zero,the plane is not parallel to any of the coordinate planes $(XY, YZ, ZX)$. Thus,the solutions lie on a plane not parallel to any of the coordinate planes.

In matrix notation,if the system of equations $\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ -5 \\ 10 \end{bmatrix}$ has an infinite number of solutions,then all these solutions lie on

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