Question

In Exercise 3, find the vector x determined by the given coordinate vector ${\left(x\right)}_{\mathrm{B}}$ and the given basis $\mathrm{B}$.

3. $\mathrm{B}=\left\{\left(\begin{array}{c}\mathbf{1}\\ -\mathbf{4}\\ \mathbf{3}\end{array}\right),\left(\begin{array}{c}\mathbf{5}\\ \mathbf{2}\\ -\mathbf{2}\end{array}\right),\left(\begin{array}{c}\mathbf{4}\\ -\mathbf{7}\\ \mathbf{0}\end{array}\right)\right\},{\left(x\right)}_{\mathrm{B}}=\left(\begin{array}{c}\mathbf{3}\\ \mathbf{0}\\ -\mathbf{1}\end{array}\right)$

Short answer

Vector $x=\left(\begin{array}{c}-1\\ -5\\ 9\end{array}\right)$

Step-by-step solution

Use the definition

The coordinates of x relative to basis$\mathrm{B}=\left\{b_{\mathbf{1}},b_{\mathbf{2}},...,b_n\right\}$are the weights$c_{\mathbf{1}},c_{\mathbf{2}},...,c_n$,such that $x=c_{\mathbf{1}}{b}_{\mathbf{1}}+c_{\mathbf{2}}{b}_{\mathbf{2}}+...+c_n{b}_n$. Then,${\left(x\right)}_{\mathrm{B}}=\left(\begin{array}{c}{c}_1\\ c_2\\ ⋮\\ c_n\end{array}\right)$.

Find x

By the above definition, you get

$$\begin{array}{c}x=3\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right]+0\left[\begin{array}{c}5\\ 2\\ -2\end{array}\right]+\left(-1\right)\left[\begin{array}{c}4\\ -7\\ 0\end{array}\right]\\ =\left[\begin{array}{c}3\\ -12\\ 9\end{array}\right]+0+\left[\begin{array}{c}-4\\ 7\\ 0\end{array}\right]\\ =\left[\begin{array}{c}3-4\\ -12+7\\ 9\end{array}\right]\\ x=\left[\begin{array}{c}-1\\ -5\\ 9\end{array}\right].\end{array}$$

Draw a conclusion

Hence, vector $x=\left(\begin{array}{c}-1\\ -5\\ 9\end{array}\right)$.