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I've been doing lots of matrices practice today and I've come across this one which I'm finding quite tricky. $$ 𝐴 = \begin{pmatrix} 3 & 5 \\ z & -3 \end{pmatrix} $$ All I'm told is that $A^2$ is a matrix whose entries are all $0$, and I need to find z. $$ A^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} $$ So I suppose I need to write out the second matrix and then I need to find a way of getting to k. Would I first be right to find the determinant? So $$ (A_{11} \times A_{22}) - (A_{12} \times A_{21}) = -9 - 5z $$

I was wondering how I might bring the second mentioned matrix into my working out, because obviously it is important?

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You know that $A^2 = 0$. If $A$ were invertible, you could multiply both sides by the inverse of $A$: $$ A^{-1} A^2 = A = A^{-1} 0 = 0 $$ but you know that $A \ne 0$, so $A$ has no inverse, thus its determinant must vanish. Which helps you to determine $z$.

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  • $\begingroup$ Thank you very much for explaining, I appreciate the clear help! $\endgroup$ – New Zealand's finest Mar 16 '16 at 19:16
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$$\det(A^2)=(\det(A))^2=(-9-5z)^2=0$$ Thus, $$9+5z=0\Rightarrow z=-\frac95$$

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  • $\begingroup$ Thank you for this. So just to clear up, A and A^2 are two different vectors. How do they fit into each other in terms of the determinant of A being squared as above? Are they the same? I might just be being incredibly stupid but I can't see the link anywhere. $\endgroup$ – New Zealand's finest Mar 16 '16 at 19:00
  • $\begingroup$ Thank you very much for your clear help! $\endgroup$ – New Zealand's finest Mar 16 '16 at 19:15
  • $\begingroup$ GoodDeeds May I ask two very simple questions? Where is the equation above derived from given that A^2 is not A squared, just another matrix? Also, I can see why it would equal 0 if the equations were the same, but they're not so where does the 0 come from? Would it be possible to have a breakdown of the steps above? Many thanks. $\endgroup$ – New Zealand's finest Mar 16 '16 at 19:39
  • $\begingroup$ Why is the determinant of A^2 equal to the determinant of A, Squared? $\endgroup$ – New Zealand's finest Mar 16 '16 at 19:41
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Actually, since you are asking why $\det(A^2)=\det(A)^2$, it might be an easier solution to simply compute $A^2$, which is $$ \begin{pmatrix} 9+5z & 0 \cr 0 & 9+5z\end{pmatrix}. $$ Then $0=A^2=(9+5z)I_2$, hence $z=-9/5$. If you would like to use determinants, then we will need that $\det$ is multiplicative, i.e, $$ \det(AB)=\det(A)\det(B) $$ for all $A,B$.

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  • $\begingroup$ Hello, thanks for this. My question states that all of the values for A^2 are 0, so surely it would be 0, 0, 0, 0? $\endgroup$ – New Zealand's finest Mar 17 '16 at 13:34
  • $\begingroup$ Yes, it says that all four entries are zero. This is the most direct solution. $\endgroup$ – Dietrich Burde Mar 22 '16 at 22:41

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