x^2-1=0

i am not able to do it here but !

EXAMPLE 3 Determine if the following sets of vectors are linearly independent.

a. $v_1=\left[\begin{array}{c}3\\ 1\end{array}\right],v_2=\left[\begin{array}{c}6\\ 2\end{array}\right]$

b. $v_1=\left[\begin{array}{c}3\\ 2\end{array}\right],v_2=\left[\begin{array}{c}6\\ 2\end{array}\right]$

$v_1+v_2=$

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SOLUTION

a. Notice that $v_2$ is a multiple of $v_1$, namely $v_2=2v_1$. Hence $-2v_1+v_2=0$, which shows that $\{v_1,v_2\}$ is linearly dependent.

b. The vectors $v_1$ and $v_2$ are certainly not multiples of one another. Could they be linearly dependent? Suppose $c$ and $d$ satisfy

$c{v}_1+d{v}_2=0$

If $c\ne 0$, then we can solve for $v_1$ in terms of $v_2$, namely $v_1=(-d/c)v_2$. This result is impossible because $v_1$ is not a multiple of $v_2$. So $c$ must be zero. Similarly, $d$ must also be zero. Thus $\{v_1,v_2\}$ is a linearly independent set.

The arguments in Example 3 show that you can always decide by inspection when a set of two vectors is linearly dependent. Row operations are unnecessary. Simply check whether at least one of the vectors is a scalar times the other. (The test applies only to sets of two vectors.)

A set of two vectors $\{v_1,v_2\}$ is linearly dependent if at least one of the vectors is a multiple of the other. The set is linearly independent if and only if neither of the vectors is a multiple of the other.

$a=0$

$b=1$

$a+b=1$

${\sin x}^{\prime }=\cos x$