Finding an unknown value in a matrix
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?
$\endgroup$3 Answers
$\begingroup$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 $z$.
$\endgroup$ 1 $\begingroup$$$\det(A^2)=(\det(A))^2=(-9-5z)^2=0$$ Thus, $$9+5z=0\Rightarrow z=-\frac95$$
$\endgroup$ 4 $\begingroup$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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