# Finding coordinate vector

The coordinate vector of:

$$\begin{pmatrix} 1&2&-1\\0 & 0 & 6 \end{pmatrix}$$

with respect to the standard basis of $M2x3$ would be:

Are you just finding the inverse of this matrix and multiplying it with $(100)(010)$? I have a feeling that would be wrong.

b)THen Polynomials $-x,2,x^{2}$ form a basis of $p2$. Find the coordinate vector of the polynomial $x-8$ with respect to this basis.

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That matrix is not invertible. What they mean is write that matrix as a linear combination of the basis. The standard basis consists of the matrices $E_{11}, E_{12}, E_{13}, E_{21}, E_{22}, E_{23}$ where $E_{ij}$ is the $2 \times 3$ matrix with a $1$ in entry $(i, j)$ and $0$'s elsewhere. For example,

$E_{21} = \begin{bmatrix}0&0&0\\1&0&0\end{bmatrix}$

$\begin{bmatrix}1&2&-1\\0&0&6\end{bmatrix} = E_{11} + 2E_{12} - E_{13} + 6E_{23}$
Hence it's coordinate vector is $(1, 2, -1, 0, 0, 6)$.