What I tried to do was I made $2$ matrices $A$ and $B$ and set them not equal to the $2 \times 2$ identity matrix, I don't know what to do next and I am stuck on the proof so I need help on what I could do next.
If $A$ is a $2\times1$ matrix and $B$ is a $1\times {2}$, prove $C = AB$ is not invertible
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matrices
proof-writing
3 Answers
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We have$$C=AB=\left(\begin{array}{c} A_{1}\\ A_{2} \end{array}\right)\left(\begin{array}{cc} B_{1} & B_{2}\end{array}\right)=\left(\begin{array}{cc} A_{1}B_{1} & A_{1}B_{2}\\ A_{2}B_{1} & A_{2}B_{2} \end{array}\right)$$so $\det C=A_{1}B_{1}A_{2}B_{2}-A_{1}B_{2}A_{2}B_{1}=0$. Hence C is not invertible.
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For arbirtrary $v$, $Cv$ is a multiple of $A$.
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For $B$ there exists a non-zero vector $v$ such that $Bv=0$. If $B=(x,y)$ with $x\neq 0$ or $y\neq 0$ you can choose $v=\begin{pmatrix}y\\-x\end{pmatrix}$. If $B=0$ then any non-zero $v$ works.
Now $ABv=0$, so $AB$ can't be invertible.
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1Is there any way to prove it without vectors since I haven't learned about them yet – 2017-02-08
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0A vector is just a $2\times 1$ matrix. – 2017-02-08