-2
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Express the column matrix $b$ as a linear combination of the columns of $A$. (Use $A_1$, $A_2$, and $A_3$ respectively for the columns of A.)

$$\begin{align} A&=\begin{pmatrix}-2 &-5 &4\\3 &-3&5\end{pmatrix}\\b&=\begin{pmatrix}11\\28\end{pmatrix}\end{align}$$

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    $\begingroup$ (-1) "This question does not show any research effort." $\endgroup$ – TMM Apr 15 '13 at 9:06
1
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\begin{align} A&=\begin{pmatrix}-2 &-5 &4\\3 &-3&5\end{pmatrix}\\b&=\begin{pmatrix}11\\28\end{pmatrix}\\ Ax&=b\\ \begin{pmatrix}-2 &-5 &4\\3 &-3&5\end{pmatrix}x&=\begin{pmatrix}11\\28\end{pmatrix}\\ \begin{pmatrix}-2 &-5 &4\\0 &-10.5&11\end{pmatrix}x&=\begin{pmatrix}11\\44.5\end{pmatrix}\\ \begin{pmatrix}-2 &0 &1.2381\\0 &-10.5&11\end{pmatrix}x&=\begin{pmatrix}-10.1905\\44.5\end{pmatrix}\\ x&=(a\quad b\quad c)^T\\ \text{Let, }b&=0\\ \therefore c&=4.0455\\ \therefore a&=2.5909\\ \blacksquare \end{align}

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  • $\begingroup$ Hi, actually the answer is supposed to come out in the format similar to: b = -A1 - 3A2 + A3 $\endgroup$ – user2153126 Apr 15 '13 at 5:52
  • $\begingroup$ @user2153126. There can be multiple answers. $\endgroup$ – Inquest Apr 15 '13 at 5:54
  • $\begingroup$ Yeah, but I'm looking for the general form - similar to what I indicated: b = ___________ $\endgroup$ – user2153126 Apr 15 '13 at 6:11
  • $\begingroup$ @user2153126, if you are suggesting that [-1 -3 1] is an answer then it's wrong. You can verify this yourself by multiplying. Also, to get an answer in a different form, simply replace the $b=0$ step with your choice of pivot. $\endgroup$ – Inquest Apr 15 '13 at 6:26
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Hint: Try to formulate and to solve the respective system of linear equations.

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    $\begingroup$ yeah, I understand the concept but I can't get the correct answer $\endgroup$ – user2153126 Apr 15 '13 at 5:35
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    $\begingroup$ I don't see how one could understand this concept and not be able to get the answer. $\endgroup$ – jim Apr 15 '13 at 6:13

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