trouble 703

Using the an interpretation of the selection of a matrix, define the selection of the matrix\

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by definition, the variety $\calR(A)$ of the procession $A$ is offered by\<\calR(A)=\left \ \quad A\mathbfx=\mathbfb \text for part \mathbfx \in \R^4 \right \.\>Thus, a vector $\mathbfb=\beginbmatrix b_1 \\ b_2 \\ b_3 \endbmatrix$ in $\R^3$ is in the selection $\calR(A)$ if and only if the system $A\mathbfx=\mathbfb$ is consistent. So, let us discover the conditions on $\mathbfb$ so the the mechanism is consistent.

To execute this, we take into consideration the augmented matrix of the system and also reduce it together follows. \beginalign*\left<\beginarrayrrrr 2 & 4 & 1 & -5 & b_1\\ 1 &2 & 1 & -2 & b_2 \\ 1 & 2 & 0 & -3 &b_3 \endarray\right> \xrightarrowR_1 \leftrightarrow R_2 \left<\beginarrayrrrr 1 &2 & 1 & -2 & b_2 \\ 2 & 4 & 1 & -5 & b_1\\ 1 & 2 & 0 & -3 &b_3 \endarray\right> \xrightarrowR_2-2R_1\\<6pt> \left<\beginarrayr 1 &2 & 1 & -2 & b_2 \\ 0 & 0 & -1 & -1 & b_1-2b_2\\ 0 & 0 & 0 & -1 & b_3-b_2 \endarray\right> \xrightarrow-R_2 \left<\beginarrayrrrr 1 &2 & 1 & -2 & b_2 \\ 0 & 0 & 1 & 1 & -b_1+2b_2\\ 0 & 0 & -1 & -1 & b_3-b_2 \endarray\right>\\<6pt> \xrightarrowR_1-R_2 \left<\beginarrayrrrr 1 &2 & 0 & -3 & b_1-b_2 \\ 0 & 0 & 1 & 1 & -b_1+2b_2\\ 0 & 0 & 0 & 0 & -b_1 +b_2 +b_3 \endarray\right>.\endalign*

Note that if the $(3, 5)$-entry $-b_1+b_2+b_3$ is not zero, climate the mechanism $A\mathbfx=\mathbf0$ is inconsistent due to the fact that this implies $0=1$.On the other hand, if $-b_1+b_2+b_3=0$, then we check out that the mechanism is consistent.Hence, the vector $\mathbfb$ is in the selection $\calR(A)$ if and only if $-b_1+b_2+b_3=0$.

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In summary, us have\<\calR(A)=\left\ \quad -b_1+b_2+b_3=0 \right \.\>

Spanning collection for the range

With a little bit extr computation, we can uncover the spanning collection for the variety as follows.

Thus, $\mathbfb \in \calR(A)$ if and only if\beginalign*\mathbfb=\beginbmatrix b_1 \\ b_2 \\ b_3 \endbmatrix=\beginbmatrix b_2+b_3 \\ b_2 \\ b_3 \endbmatrix=b_2\beginbmatrix 1 \\ 1 \\ 0 \endbmatrix+b_3\beginbmatrix 1 \\ 0 \\ 1 \endbmatrix.\endalign*

In summary, us have\beginalign*\calR(A)&=\left\ \mathbfb \in \R^3 \quad \middle \\<6pt> &=\Span\left\ \beginbmatrix 1 \\ 1 \\ 0 \endbmatrix, \beginbmatrix 1 \\ 0 \\ 1 \endbmatrix \right \. \endalign* Hence, the spanning set is \< \left\ \beginbmatrix 1 \\ 1 \\ 0 \endbmatrix, \beginbmatrix 1 \\ 0 \\ 1 \endbmatrix \right \.\>This spanning set is linearly independent, therefore it’s a basis for the range.