Question

For each matrix A, find a matrix C such that AC=0 or CA=0

$\mathit{A}\mathbf{=}\left(\begin{array}{cc}6& 9\\ 2& 3\end{array}\right)\mathbf{;}\mathit{A}\mathbf{=}\left(\begin{array}{cc}5& -10\\ -2& 4\end{array}\right)\mathbf{;}\mathit{A}\mathbf{=}\left(\begin{array}{cc}1/2& 1/4\\ 3& 3/2\end{array}\right)$

Short answer

The required corresponding matrices are

$\left(\begin{array}{cc}3& 6\\ -2& -4\end{array}\right),\left(\begin{array}{cc}2& 4\\ 1& -4\end{array}\right),and\left(\begin{array}{cc}1& 2\\ -2& -4\end{array}\right)$

Step-by-step solution

Product of general matrices

Let $A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]andC=\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]$

Then,localid="1653999940433" $AC=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right]=\left[\begin{array}{cc}{a}_{11}{c}_{11}+a_{12}{c}_{21}& a_{11}{c}_{12}+a_{12}{c}_{22}\\ a_{21}{c}_{11}+a_{22}{c}_{21}& a_{22}{c}_{12}+a_{22}{c}_{22}\end{array}\right]$

For . AC=0

localid="1654000012796" $$\begin{gathered} \left[\begin{array}{cc}{a}_{11}{c}_{11}+a_{12}{c}_{21}& a_{11}{c}_{12}+a_{12}{c}_{22}\\ a_{21}{c}_{11}+a_{22}{c}_{21}& a_{22}{c}_{12}+a_{22}{c}_{22}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ {a}_{11}{c}_{11}+a_{12}{c}_{21}=0 \\ {a}_{11}{c}_{12}+a_{12}{c}_{22}=0 \\ {a}_{21}{c}_{11}+a_{22}{c}_{21}=0 \\ {a}_{22}{c}_{12}+a_{22}{c}_{22}=0 \end{gathered}$$

Find C

The given matrices are

$A=\left(\begin{array}{cc}6& 9\\ 2& 3\end{array}\right);A=\left(\begin{array}{cc}5& -10\\ -2& 4\end{array}\right);A=\left(\begin{array}{cc}1/2& 1/4\\ 3& 3/2\end{array}\right)$

Now, find C by using explicit computations, which satisfies .AC=0

$$\begin{gathered} \left(\begin{array}{cc}6& 9\\ 2& 3\end{array}\right)\left(\begin{array}{cc}3& 6\\ -2& -4\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right) \\ \left(\begin{array}{cc}5& -10\\ -2& 4\end{array}\right)\left(\begin{array}{cc}2& 4\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right) \\ \left(\begin{array}{cc}\frac{1}{2}& \frac{1}{4}\\ 3& \frac{3}{2}\end{array}\right)\left(\begin{array}{cc}1& 2\\ -2& -4\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right) \end{gathered}$$

The required corresponding matrices are

localid="1654002042365" $\left(\begin{array}{cc}3& 6\\ -2& -4\end{array}\right),\left(\begin{array}{cc}2& 4\\ 1& 2\end{array}\right),and\left(\begin{array}{cc}1& 2\\ -2& -4\end{array}\right)$