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From an assignment:

Let $A = \left[ \begin{matrix} 3 & -6 \\ -2 & 4 \end{matrix}\right] $ Construct a $2 * 2$ matrix $B$ such that $AB$ is the zero matrix. Use two different nonzero columns for $B$.

The value of $AB$ would be:

$$ AB = \left[ \begin{matrix} 3b_{11} -6b_{12} & 3b_{21} -6b_{22} \\ -2b_{11} + 4b_{12} & -2b_{21} + 4b_{22} \end{matrix}\right] $$

I was thinking of using substitution, but the following equations just result in the variables equalling $0$:

$$\begin{align*} 3b_{11} -6b_{12} &= 0\\ -2b_{11} + 4b_{12} &= 0 \end{align*}$$

Any hints on how I can solve this?

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Recommendation: Don't write $+-6b_{22}$; write $-6b_{22}$. – Arturo Magidin Jan 24 '12 at 22:03
The two equations are the same, divide first by 3 and second by -2. – sdcvvc Jan 24 '12 at 23:34
Did you doublecheck if zero is the only solution to those equations? Or did you just notice zero was a solution and stop looking for other ones? The zero vector is always going to be a solution; the point of this problem is to make sure you know that other ones can exist and how to find them. – Chad Miller Jan 24 '12 at 23:43
up vote 1 down vote accepted

Your equations $$3b_{11}-6b_{12}=0,\qquad -2b_{11}+4b_{12}=0$$ are both telling you $b_{11}=2b_{12}$, which means you can let $b_{12}$ be your favorite nonzero number, and then you also know $b_{11}$. You also get two equations for $b_{21}$ and $b_{22}$; you can let $b_{22}$ be your second favorite nonzero number, and work out $b_{21}$. Then you're done.

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That helped. Thanks! – Tim Cooper Jan 25 '12 at 1:35

Consider the solutions to $A\mathbf{x}=\mathbf{0}$. These can be found easily by Gaussian elimination: $$\left(\begin{array}{rr} 3 & -6\\ -2 & 4 \end{array}\right)\rightarrow \left(\begin{array}{rr} 1 & -2\\ 1 & -2 \end{array}\right) \rightarrow \left(\begin{array}{rr} 1 & -2\\ 0 & 0 \end{array}\right).$$ So $\mathbf{x}=\left(\begin{array}{c}r\\s\end{array}\right)$ satisfies $A\mathbf{x}=\mathbf{0}$ if and only if $r-2s = 0$, if and only if $r=2s$.

Now notice that if $B=[\mathbf{b}_1|\mathbf{b}_2]$, where $\mathbf{b}_1$ is the first column and $\mathbf{b}_2$ is the second column, then $$AB = [A\mathbf{b}_1|A\mathbf{b}_2].$$

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I think for small matrices as the one above, it s good to just look for linear dependence of the columns in such a problem. (i.e. here you can quickly notice that in matrix $A$ column 2 is just column 1 multiplied by -2)

Looking at this kind of linear dependence between the columns gives you a quick way to understand why $B$ looks the way it does (i.e. take 1 of column 1 and -2 of column two and you arrive at the zero matrix, and this is exactly what multiplication by $B$ does )!

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