Let a1 = [1 -2 -5], a2 = [2 5 6], and b = [7 4 -3]. Determine whether b can be generated (or written) as a linear combination of a1 and a2. That is, determine whether weights x1 and x2 exist such that x1a1 + x2a2 = b (1) If vector equation (1) has a solution, find it.

Question

Let [latex] a_{1} = \left [ \begin{matrix} 1 \ -2 \ -5 \end{matrix} ight ] [/latex] , [latex] a_{2} = \left [ \begin{matrix} 2 \ 5 \ 6 \end{matrix} ight ] [/latex] , and [latex] b = \left [ \begin{matrix} 7 \ 4 \ -3 \end{matrix} ight ] [/latex] . Determine whether b can be generated (or written) as a linear combination of [latex] a_{1} [/latex] and [latex] a_{2} [/latex]. That is, determine whether weights [latex] x_{1} [/latex] and [latex] x_{2} [/latex] exist such that

[latex] x_{1} a_{1} + x_{2} a_{2} = b [/latex] (1)

If vector equation (1) has a solution, find it.

Accepted Answer

Use the definitions of scalar multiplication and vector addition to rewrite the vector equation

[latex] x_{1} \underset{\overset{\uparrow }{a_{1}} }{\left [ \begin{matrix} 1 \ -2 \ -5 \end{matrix} ight ]} + x_{2} \underset{\overset{\uparrow }{a_{2}} }{\left [ \begin{matrix} 2 \ 5 \ 6 \end{matrix} ight ]} = \underset{\overset{\uparrow }{b} } {\left [ \begin{matrix} 7 \ 4 \ -3 \end{matrix} ight ]} [/latex]

which is the same as

[latex] \left [ \begin{matrix} x_{1}\ -2x_{1} \ -5x_{1} \end{matrix} ight ] + \left [ \begin{matrix} 2x_{2} \ 5x_{2} \ 6x_{2} \end{matrix} ight ] = \left [ \begin{matrix} 7 \ 4 \ -3 \end{matrix} ight ] [/latex]

and

[latex] \left [ \begin{matrix} x_{1} + 2x_{2} \ -2x_{1} + 5x_{2} \ -5x_{1} + 6x_{2} \end{matrix} ight ] = \left [ \begin{matrix} 7 \ 4 \ -3 \end{matrix} ight ] [/latex] (2)

The vectors on the left and right sides of (2) are equal if and only if their corresponding entries are both equal. That is, [latex] x_{1} [/latex] and [latex] x_{2} [/latex] make the vector equation (1) true if and only if [latex] x_{1} [/latex] and [latex] x_{2} [/latex] satisfy the system

[latex] x_{1} + 2x_{2} = 7 [/latex]
[latex] -2x_{1} + 5x_{2} = 4 [/latex] (3)
[latex] -5x_{1} + 6x_{2} = -3 [/latex]

To solve this system, row reduce the augmented matrix of the system as follows:³

[latex] \left [ \begin{matrix} 1 & 2 & 7 \ -2 & 5 & 4 \ -5 & 6 & -3 \end{matrix} ight ] \sim \left [ \begin{matrix} 1 & 2 & 7 \ 0 & 9 & 18\ 0& 16& 32\end{matrix} ight ] \sim\left [ \begin{matrix} 1 & 2 & 7 \ 0& 1 & 2 \ 0 & 16& 32\end{matrix} ight ] \sim \left [ \begin{matrix} 1 & 0 & 3 \ 0 & 1 & 2 \ 0 & 0 & 0 \end{matrix} ight ] [/latex]

The solution of (3) is [latex] x_{1} = 3 [/latex] and [latex] x_{2} = 2 [/latex]. Hence b is a linear combination of [latex] a_{1} [/latex] and [latex] a_{2} [/latex], with weights [latex] x_{1} = 3 [/latex] and [latex] x_{2} = 2 [/latex]. That is,

[latex] 3 \left [ \begin{matrix} 1 \ -2 \ -5 \end{matrix} ight ] + 2 \left [ \begin{matrix} 2 \ 5 \ 6 \end{matrix} ight ] = \left [ \begin{matrix} 7 \ 4 \ -3 \end{matrix} ight ] [/latex]

³The symbol ∼ between matrices denotes row equivalence (Section 1.2).

Question 1.3.5: Let a1 = [1 -2 -5], a2 = [2 5 6], and b = [7 4 -3]. Determin...

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