Question

1. let \\( \vec{a}_{1}=\left(\

$$\begin{array}{l}1 \\ 2\\end{array}$$

\
ight), \vec{a}_{2}=\left(\

$$\begin{array}{c}2 \\ -1\\end{array}$$

\
ight) \\), and \\( \vec{b}=\left(\

$$\begin{array}{l}4 \\ 3\\end{array}$$

\
ight) \\). show that \\( \vec{b} \in \operatorname{span}\left\\{\vec{a}_{1}, \vec{a}_{2}\
ight\\} \\).

Explanation:

Step1: Set up the linear combination equation

We want \(x\vec{a}_1 + y\vec{a}_2=\vec{b}\), which gives \(

$$\begin{pmatrix}x + 2y\\2x-y\end{pmatrix}$$

=

$$\begin{pmatrix}4\\3\end{pmatrix}$$

\), leading to the system \(

$$\begin{cases}x + 2y=4\\2x-y=3\end{cases}$$

\)

Step2: Use matrix - determinant (Cramer's rule)

Find the determinant of the coefficient matrix \(A=

$$\begin{pmatrix}1&2\\2&-1\end{pmatrix}$$

\), \(\text{det}(A)=1\times(-1)-2\times2=-5\)

For \(x\), the numerator determinant is \(

$$\begin{vmatrix}4&2\\3&-1\end{vmatrix}$$

=4\times(-1)-2\times3=-10\), so \(x = 2\)

For \(y\), the numerator determinant is \(

$$\begin{vmatrix}1&4\\2&3\end{vmatrix}$$

=1\times3 - 4\times2=-5\), so \(y = 1\)

Step3: Verify the solution

Substitute \(x = 2\) and \(y = 1\) into \(x\vec{a}_1+y\vec{a}_2\): \(2

$$\begin{pmatrix}1\\2\end{pmatrix}$$

+1

$$\begin{pmatrix}2\\-1\end{pmatrix}$$

=

$$\begin{pmatrix}2+2\\4 - 1\end{pmatrix}$$

=

$$\begin{pmatrix}4\\3\end{pmatrix}$$

=\vec{b}\)

Answer:

To show that \(\vec{b}\in\text{span}\{\vec{a}_1,\vec{a}_2\}\), we need to find scalars \(x\) and \(y\) such that \(x\vec{a}_1 + y\vec{a}_2=\vec{b}\).

This gives us the system of equations:

$$
LATEXBLOCK0
$$

We can write this system in matrix form \(A\vec{x}=\vec{b}\), where \(A=

$$\begin{pmatrix}1&2\\2&-1\end{pmatrix}$$

\), \(\vec{x}=

$$\begin{pmatrix}x\\y\end{pmatrix}$$

\) and \(\vec{b}=

$$\begin{pmatrix}4\\3\end{pmatrix}$$

\)

First, we find the determinant of \(A\), \(\text{det}(A)=(1\times(-1))-(2\times2)=-1 - 4=-5
eq0\)

We use Cramer's rule.

For \(x\):

$$ x=\frac{
LATEXBLOCK4
}{\text{det}(A)}=\frac{4\times(-1)-2\times3}{-5}=\frac{-4 - 6}{-5}=\frac{-10}{-5} = 2 $$

For \(y\):

$$ y=\frac{
LATEXBLOCK5
}{\text{det}(A)}=\frac{1\times3-4\times2}{-5}=\frac{3 - 8}{-5}=\frac{-5}{-5}=1 $$

Since \(x = 2\) and \(y = 1\) satisfy \(x\vec{a}_1+y\vec{a}_2=\vec{b}\) (because \(2

$$\begin{pmatrix}1\\2\end{pmatrix}$$

+1

$$\begin{pmatrix}2\\-1\end{pmatrix}$$

=

$$\begin{pmatrix}2 + 2\\4-1\end{pmatrix}$$

=

$$\begin{pmatrix}4\\3\end{pmatrix}$$

\)), we have shown that \(\vec{b}\in\text{span}\{\vec{a}_1,\vec{a}_2\}\)