If $A\in M_3\left(\mathbb C\right)$ is an invertible matrix such that $2A^2=4A+A^3$

Question

If $A\in M_3\left(\mathbb C\right)$ is an invertible matrix such that $$2A^2=4A+A^3$$

Then which of the following is(are)correct:

(A) $\det (A)=8;$ (B) $det\left(adj\left(\frac A2\right)\right)=1;$ (C) $tr\left(A-2I_3\right)^3=24;$ (D) $adj(A)=A^2$

My working:

From given information we get $A(A^2-2A+4I_3)=O_3$

$\implies A^2-2A+4I_3=O_3$ $\implies (A+2\omega I_3)(A+2\omega^2 I_3)=O_3$

$\implies$ Possible Eigen values of $A$ can be $-2\omega, -2\omega^2$

Now what to do

"we get $(A-2I_3)^3=-8I_3$" How? – Did Feb 10 '18 at 21:13 — Did, Feb 10 '18 at 21:13
$ A^2-2A+4=0 $. And you forgot binomial coefficients. – Arnaud Mortier Feb 10 '18 at 21:18 — Arnaud Mortier, Feb 10 '18 at 21:18
Sorry I made a foolish mistake I edited – Makar Feb 10 '18 at 21:43 — Makar, Feb 10 '18 at 21:43

Martin Argerami · Accepted Answer · 2018-02-11T16:09:29.633

Since $A^2-2A+4I=0$, we have $A^2-2A=-4I$. From the original equation together with this, $$ A^3=2A^2-4A=2(A^2-2A)=-8I. $$ Then $(\det A)^3=-8^3$, which tells us that $\det A$ is either $4-4i\sqrt3$ or $4+4i\sqrt3$. The real root is not an option, because no combination of the eigenvalues can have a real product. So $(a)$ is false.

From $A^2-2A+4I=0$, the eigenvalues are either $(1+i\sqrt3,1+i\sqrt3,1-i\sqrt3)$, or $(1-i\sqrt3,1-i\sqrt3,1+i\sqrt3)$.

For (c), $A-2I=A^2/2 $, so $(A-2I)^3=A^6/8=(A^3)^2/8=(-8I)^2/8=8I$. So $\text{Tr}\,(A-2I)^3=\text{Tr}\,(8I)=24$, and (c) is true.

Since $A/2-A^2/4=I $, we deduce that $A^{-1}=I/2-A/4$. Then $$\text {adj}\,A=(\det A)\,A^{-1}=(\det A)\, (I/2-A/4)=\frac18\,(\det A)\,(-A^2).$$ As we know from above that $\det A$ is not real, $(d)$ is false.

We can also use $A/2-A^2/4=I$, written as $(A/2)(I-A/2)=I$, to deduce that $(A/2)^{-1}=I-A/2$. And so $$ \text{adj}\,\frac{A}2=(\det\frac A2)\,(I-\frac A2)=\frac18\,(\det A)\,(I-\frac A2). $$ Then $$ \det(\text{adj}\,\frac A2)=\frac{(\det A)^3}{8^3}\,\det(I-\frac A2)=-\frac1{64}\,\det(I-\frac A2). $$ This cannot be $1$, as all the eigenvalues of $I-A/2$ have absolute value $1$.

There is mistake in the solution $\det(A^3)=(-8)^3\implies \det (A)= -8\omega,-8\omega^2,$ where $\omega$ is non-real cube root of unity. Excellent solution!! Thanks — Makar, Feb 11 '18 at 15:53

score 0 · Answer 2 · answered Feb 13 '18 at 17:08

From the given information, $A(A^2-2A+4I_3)=O_3$ $\implies A^2-2A+4I_3=O_3$ $\implies (A+2\omega I_3)(A+2\omega^2 I_3)=O_3$ $\implies$ eigen values of $A$ can be $-2\omega,-2\omega,-2\omega^2$ or $-2\omega,-2\omega^2,-2\omega^2.$ Where $\omega$ is a non real cube root of unity.

$\implies \det A=$ Product of eigen values of $A$ = $(-2\omega)\times(-2\omega)\times(-2\omega^2)$ or $(-2\omega)\times(-2\omega^2)\times(-2\omega^2)=-8\omega$ or $-8\omega^2.$

$\implies$ Option (A) is incorrect.

As in previous solution we have $A^3=-8I_3$ $\implies A^2=-8A^{-1}=-8\frac{1}{\det A}adj(A)=\omega^2 adj(A)$ or $\omega adj(A)\implies adj(A)=\omega A^2$ or $\omega^2 A^2$

$\implies$ option (D) is incorrect.

Now $adj\left(\frac A2\right)=\left(\frac 12\right)^2 adj(A)=\frac 14 \omega A^2$ or $\frac 14 \omega^2 A^2$

$\implies\det \left(adj\left(\frac A2\right)\right)=\left(\frac 14 \omega\right)^3 (\det A)^2$ or $\left(\frac 14 \omega^2\right)^3 (\det A)^2=\omega^2$ or $\omega$

$\implies$ Option (B) is incorrect.

So only correct option is (C) as demonstrated in previous solution.

If $A\in M_3\left(\mathbb C\right)$ is an invertible matrix such that $2A^2=4A+A^3$

2 Answers2