20. Let \( \boldsymbol{a}=\left(\begin{array}{c}4 \\ -1 \\ 1\end{array}\right), \boldsymbol{b}=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right), \boldsymbol{c}=\left(\begin{array}{c}-1 \\ 3 \\ 2\end{array}\right), \boldsymbol{d}=\left(\begin{array}{c}-3 \\ 0 \\ 1\end{array}\right) \) Find the scalars \( \alpha, \beta \), and \( \mu \) (or show that they cannot exist) such that \( \boldsymbol{a}=\alpha \boldsymbol{b}+\beta \boldsymbol{c}+\mu \boldsymbol{d} \). \( \begin{array}{l}\text { Figure } 9.15 \text { Diagram for } \\ \text { question } 18\end{array} \)

We are given \[ \boldsymbol{a}=\begin{pmatrix}4\\-1\\1\end{pmatrix},\quad \boldsymbol{b}=\begin{pmatrix}1\\1\\1\end{pmatrix},\quad \boldsymbol{c}=\begin{pmatrix}-1\\3\\2\end{pmatrix},\quad \boldsymbol{d}=\begin{pmatrix}-3\\0\\1\end{pmatrix}, \] and we wish to find scalars \(\alpha\), \(\beta\), and \(\mu\) such that \[ \boldsymbol{a} = \alpha \boldsymbol{b} + \beta \boldsymbol{c} + \mu \boldsymbol{d}. \] This equation expands to \[ \begin{pmatrix}4\\-1\\1\end{pmatrix} = \alpha\begin{pmatrix}1\\1\\1\end{pmatrix} + \beta \begin{pmatrix}-1\\3\\2\end{pmatrix} + \mu \begin{pmatrix}-3\\0\\1\end{pmatrix}. \] Equate the corresponding components to obtain the system of equations: 1. \(x\)-component: \[ 4 = \alpha - \beta - 3\mu. \] 2. \(y\)-component: \[ -1 = \alpha + 3\beta. \] 3. \(z\)-component: \[ 1 = \alpha + 2\beta + \mu. \] ### Step 1. Express \(\alpha\) in terms of \(\beta\) from the \(y\)-component From \[ -1 = \alpha + 3\beta, \] we have: \[ \alpha = -1 - 3\beta. \] ### Step 2. Substitute \(\alpha\) into the \(x\)-component equation Substitute \(\alpha = -1 - 3\beta\) into \[ 4 = \alpha - \beta - 3\mu: \] \[ 4 = (-1 - 3\beta) - \beta - 3\mu. \] Combine like terms: \[ 4 = -1 - 4\beta - 3\mu. \] Add \(1 + 4\beta\) to both sides: \[ 4 + 1 + 4\beta = -3\mu, \] \[ 5 + 4\beta = -3\mu. \] Thus, \[ \mu = -\frac{5+4\beta}{3}. \] ### Step 3. Substitute \(\alpha\) and \(\mu\) into the \(z\)-component equation Now substitute \(\alpha = -1 - 3\beta\) and \(\mu = -\frac{5+4\beta}{3}\) into \[ 1 = \alpha + 2\beta + \mu: \] \[ 1 = (-1 - 3\beta) + 2\beta -\frac{5+4\beta}{3}. \] Simplify the expression: \[ 1 = -1 - \beta - \frac{5+4\beta}{3}. \] Multiply both sides by \(3\) to eliminate the fraction: \[ 3 = -3 - 3\beta - (5+4\beta). \] Combine like terms on the right-hand side: \[ 3 = -3 - 3\beta - 5 - 4\beta, \] \[ 3 = -8 - 7\beta. \] Add \(8\) to both sides: \[ 11 = -7\beta. \] Thus, \[ \beta = -\frac{11}{7}. \] ### Step 4. Solve for \(\alpha\) and \(\mu\) Using \(\alpha = -1 - 3\beta\): \[ \alpha = -1 - 3\left(-\frac{11}{7}\right) = -1 + \frac{33}{7}. \] Express \(-1\) as \(-\frac{7}{7}\): \[ \alpha = \frac{-7+33}{7} = \frac{26}{7}. \] Now using \(\mu = -\frac{5+4\beta}{3}\): \[ \mu = -\frac{5+4\left(-\frac{11}{7}\right)}{3} = -\frac{5 - \frac{44}{7}}{3}. \] Express \(5\) as \(\frac{35}{7}\): \[ \mu = -\frac{\frac{35-44}{7}}{3} = -\frac{-\frac{9}{7}}{3} = \frac{9}{21} = \frac{3}{7}. \] ### Final Answer The scalars are: \[ \alpha = \frac{26}{7}, \quad \beta = -\frac{11}{7}, \quad \mu = \frac{3}{7}. \]