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Linear Combinations (Selected HW Problems)

Problem 8

Find the value of $a$ for which $\vv = \bm{ 4 \\ a \\ -16 \\ -6}$ is in the span of $A = \left\{ \bm{2 \\4 \\-1 \\-1}, \bm{0 \\ -2 \\ 3\\5}, \bm{0 \\ 0 \\ -4 \\ 3} \right\}.$

By Proposition 2 in Notes 14, $\vv \in \spp{A}$ if and only if $A\vx = \vv$ is consistent.

The latter condition can be decided by row reduction on, $B=[A|\vv]$,the augmented matrix of the system

var('a'); v = column_matrix([4,a,-16,-6]); A = [(2,4,-1,-1),(0,-2,3,5),(0,0,-4,3)]

MA = column_matrix(A); B = block_matrix([MA,v], ncols=2); B

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} 2 & 0 & 0 & 4 \\ 4 & -2 & 0 & a \\ -1 & 3 & -4 & -16 \\ -1 & 5 & 3 & -6 \end{array}\right]$$

Applying row reduction, we get

B.add_multiple_of_row(1,0,-2); B.add_multiple_of_row(2,0,1/2); B.add_multiple_of_row(3,0,1/2); B

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} 2 & 0 & 0 & 4 \\ 0 & -2 & 0 & a - 8 \\ 0 & 3 & -4 & -14 \\ 0 & 5 & 3 & -4 \end{array}\right]$$

B.swap_rows(1,3); B

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} 2 & 0 & 0 & 4 \\ 0 & 5 & 3 & -4 \\ 0 & 3 & -4 & -14 \\ 0 & -2 & 0 & a - 8 \end{array}\right]$$

B.add_multiple_of_row(2,1,-3/5); B.add_multiple_of_row(3,1,2/5); B

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} 2 & 0 & 0 & 4 \\ 0 & 5 & 3 & -4 \\ 0 & 0 & -\frac{29}{5} & -\frac{58}{5} \\ 0 & 0 & \frac{6}{5} & a - \frac{48}{5} \end{array}\right]$$

B.add_multiple_of_row(3,2,6/29); B

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} 2 & 0 & 0 & 4 \\ 0 & 5 & 3 & -4 \\ 0 & 0 & -\frac{29}{5} & -\frac{58}{5} \\ 0 & 0 & 0 & a - 12 \end{array}\right]$$

Thus the system is consistent if and only if $a = 12$. In other words, $\vv \in \spp{A}$ if and only if $a = 12$.

Problem 4

Let $A = \left[\begin{array}{ccc} 5 & -10 & -15 \\ 4 & -11 & -15 \\ -1 & 5 & 5 \end{array}\right]$ and $\vb = \left[\begin{array}{c} 5 \\ 4 \\ 2 \end{array}\right]$

Is $\vb$ a linear combination of $\va_1, \va_2$ and $a_3$, the columns of $A$?

If $\vb$ is a linear combination of the columns of $A$, determine a non-trivial linear relation between $\va_1,\va_2,\va_3$ and $\vb$.

By Proposition 2 of Notes 14, the solutions of $A\vx = \vb$ are in 1-to-1 correspondence with the linear combinations $x_1\va_1 + x_2\va_2 + x_3\va_3 = \vb$ of the columns of $A$ that sum to $\vb$.

So to answer the question, we solve the system. As usual, form the augmented matrix $B = [A|\vb]$ of the system then find its rref.

A=matrix([(5,-10,-15),(4,-11,-15),(-1,5,5)]); b = column_matrix((5,4,2)); 
B = block_matrix([A,b], ncols=2); B,B.rref()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left[\begin{array}{ccc|c} 5 & -10 & -15 & 5 \\ 4 & -11 & -15 & 4 \\ -1 & 5 & 5 & 2 \end{array}\right], \left[\begin{array}{ccc|c} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -3 \end{array}\right]\right)$$

From the rref of $B$, we know that the system $A\vx = \vb$ has a unqiue solution, namely $\bm{-2 \\3 \\-3}$. Therefore, $\vb$ is a linear combination of $\va_1, \va_2, \va_3$ and

$$\vb = (-2)\va_1 + 3\va_2 + (-3)\va_3.$$

Check: $$ \bm{5 \\4 \\2} = (-2)\bm{5 \\4 \\-1} + 3\bm{ -10 \\-11 \\5} + (-3)\bm{-15 \\ -15 \\5}$$