Prove that two vectors in R2 are linearly independent iff ad - bc is not zero

Let $(a,b)$ and $(c,d)$ be two vectors in $\mathbb{R}^2$ . Prove that they are linearly independent if and only if $ad-bc \neq 0$ .

Proof. Assume $ad - bc \neq 0$ . Then $x(a,b) + y(c,d) = (0,0)$ implies

$xa + yc = 0 \qquad \text{and} \qquad xb + yd = 0.$

Since $ad-bc \neq 0$ we know at least one of $a,d$ and at least one of $b,c$ are nonzero. Without loss of generality, assume $a, b \neq 0$ . Then,

$\begin{align*} xa + yc = 0 && \implies && x &= -\frac{c}{a} y \\ && \implies && \left( -\frac{c}{a} \right)by + yd &= 0 \\ && \implies && y \left( d + \frac{-bc}{a} \right) &= 0 \\ && \implies && y \left( \frac{ad-bc}{a} \right) &= 0 \\ && \implies && y &=0 &(\text{since } ad-bc \neq 0) \\ && \implies && x &=0. \end{align*}$

Hence, $(a,b)$ and $(c,d)$ are independent.

Conversely, assume $(a,b)$ and $(c,d)$ are independent. Then the system of equations

$\begin{align*} xa + yc &= 0 \\ xb + yd &= 0 \end{align*}$

has only the trivial solution. We know at least one of $a,b$ and $c,d$ are non-zero (otherwise one of the vectors is the zero vector, contrary to our hypothesis that the two vectors are independent). Without loss of generality, assume $a \neq 0$ . As before, we have

$y \left( \frac{ad-bc}{a} \right) = 0.$

But, if $ad-bc =0$ , then $y$ can take any value, contradicting that the system has only the trivial solution. Thus, $ad - bc \neq 0. \qquad \blacksquare$

Stumbling Robot

Prove that two vectors in R² are linearly independent iff ad – bc is not zero

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