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Let $A$ be a $3 \times 3$ matrix such that $$A \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \\ -13 \end{pmatrix}, \quad A \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} -6 \\ 0 \\ 4 \end{pmatrix}, \quad A \begin{pmatrix} 5 \\ -9 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -11 \end{pmatrix}.$$ Find $$A \begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}.$$


I'm sure there's a way to arrange the equations so that we get the result, but I've tried adding and subtracting but none leads to $\begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}$.

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  • $\begingroup$ Different ways how this can be done. One straight forward method is to set up a 3 by 3 matrix with entries $a,b,c,d,...h,i$ and perform three times a matrix multiplication with the given vectors. That gives you three systems of three equations which can be easily solved $\endgroup$ – imranfat May 11 '17 at 0:31
  • $\begingroup$ You want to write $(3, -11, -1)$ as a linear combination of $(3,4,5)$, $(4,5,6)$, and $(5,-9,1)$. That is, you want to solve the system $(3,-11,-1)=x(3,4,5)+y(4,5,6)+z(5,-9,1)$, i.e., $\begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}= \begin{pmatrix} 3 & 4 & 5 \\ 4 & 5 & -9\\ 5 & 6 & 1\end{pmatrix} \begin{pmatrix} x \\ y \\z \end{pmatrix}$. Does that make it more clear how to proceed? (Pardon the horizontal vectors - not much space in the comment to make them all vertical.) $\endgroup$ – kccu May 11 '17 at 0:31
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You're thinking the right thing. There is a way to do it systematically. Essentially you want to find $x,y,z$ such that $$x\begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix}+ y\begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} + z\begin{pmatrix} 5 \\ -9 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}$$

This amounts to solving this system of equations for $x,y,z$: \begin{align*} 3x + 4y + 5z &= 3\\ 4x + 5y - 9z &= -11\\ 5x + 6y + z &= -1 \end{align*}

Once you have $x,y,z$ you'll be able to express $\begin{pmatrix} 3 \\ -11 \\ -1\end{pmatrix}$ as a linear combination of those other $3$ vectors above. Then use linearity of matrix-vector multiplication to get the desired result. Specifically, you'll want to use the fact that if $A$ is a matrix, $a,b,c$ are scalars, and $\vec u, \vec v, \vec w$ are vectors, then $$A \left[a\vec u + b\vec v + c\vec w\right] = a\cdot A\vec u + b\cdot A\vec v + c\cdot A\vec w.$$

Let me know if you require further assistance.

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  • $\begingroup$ So far, I have $x=2, y=-2, z=1$, but I'm not sure what to do from here. $\endgroup$ – Yuna Kun May 13 '17 at 22:05
  • $\begingroup$ @YunaKun I added another sentence near the end of the post clarifying what you'll want to do with linearity. $\endgroup$ – tilper May 14 '17 at 22:52

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