Trying to Prove three 2x2 Matrices are Linearly Independent

Hey,

Ive got a question asking;

The set of M_{2,2 }of all 2x2 matrices, with real entries, is a vector space. For what values of A (real) is the set

Z= {(
,

,

}

a linearly independent subset of M_{2,2}?

I have been trying to prove using a_{1}V_{1}+a_{2}v_{2}+a_{3}v_{3}=0 but I cant figure how to prove it or put it into a singular matrix form for row reduction... any help would be greatly appreciated.

Re: Trying to Prove three 2x2 Matrices are Linearly Independent

So you have $\displaystyle a_1\begin{bmatrix}1 & 2 \\ 1 & 0 \end{bmatrix}+ a_2\begin{bmatrix}3 & 7 \\ 0 & 0 \end{bmatrix}+ a_3\begin{bmatrix}2 & 6 \\ A & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix}a_1+ 3a_2+ 2a_3 & 2a_1+ 7a_2+ 6a_3 \\ a_1+ Aa_3 & 0\end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}$

so we must have $\displaystyle a_1+ 3a_2+ 2a_3= 0$, $\displaystyle 2a_1+ 7a_2+ 6a_3= 0$, and $\displaystyle a_1+ Aa_3= 0$ **only** if $\displaystyle a_1+ a_2+ a_3= 0$.

Okay, have you tried **solving** for $\displaystyle a_1$, $\displaystyle a_2$, and $\displaystyle a_3$? If you subtract twice the first equation from the second, you eliminate $\displaystyle a_1$- $\displaystyle a_2+ 2a_3= 0$. If you subtract the third equation from the first you again eliminate $\displaystyle a_1$- $\displaystyle 3a_2+ (2- A)a_3= 0$. Now eliminate $\displaystyle a_2$ by subtracting twice the first of those two equations from the other: $\displaystyle (2- A)a_3- 4a_3= (-2- A)a_3= 0$.

If $\displaystyle a_3= 0$ it follows that $\displaystyle a_2= 0$ and $\displaystyle a_1= 0$. What is the only value of A such that $\displaystyle a_3$ does not have to be 0?