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Suppose $Q$ is any matrix such that$[\mathbf{v}]_{\mathcal{C}}=Q[\mathbf{v}]_{\mathcal{B}}$ for each $\mathbf{v}$ in $V$Set $\mathbf{v}=\mathbf{b}_{1}$ in $(9) .$ Then $(9)$ shows that $\left[\mathbf{b}_{1}\right] c$ is the first column of $Q$ because (a) ______. Similarly, for $k=2, \ldots, n,$ the $k$th column of $Q$ is (b) ______ because (c) ______. This shows that the matrix $_{c \leftarrow \mathcal{B}}$ defined by $(5)$ in Theorem 15 is the only matrix that satisfies condition $(4)$ .

A. $\left[\mathbf{b}_{1}\right]_{c}=Q \mathbf{e}_{1}$B. $\left[\mathbf{b}_{k}\right]_{c}$C. $Q \mathbf{e}_{k}$

Calculus 3

Chapter 4

Vector Spaces

Section 7

Change of Basis

Vectors

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gate with make sure to keep is equal to B Times eight on We want to find the first row. First column element of Q on the second row. Second column Element off key. So to do this, remember Mood Image publication. It's the Rose. The first column against Sorry Rose, the first major COMSEC matrix so fast against first G. A job against a C E here. G a to us hate C plus he I This would be equal to Q and one on bank UTC. Well, that second wrote against Second Column said JK elegant B d F. Said have JB um plus, uh, DK don't keep come the drift king and finally f out on this is a key t t.

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