Let $A$ be a $3 \times 3$ real valued matrix. Find $t$ such that $A^3 = I$

Question

Let $$A = \begin{bmatrix} 1 & t & 0 \\ -2 & -2 & -1 \\ 0 & 1 & t \end{bmatrix}$$ Find $t$ such that $A^3 = I$.

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I've figured out $$\det(A) = 2t^2-2t+1$$ So I took determinant on both side of equation $A^3 = I$ then I got

$$ \det(A^3) = \det(A) \cdot \det(A) \cdot \det(A) = \det(I) = 1 $$

By $\det(A)^3 = 1$, I know that $\det(A)$ has to be $1$ which means

$$2t^2-2t+1 = 1 \Rightarrow 2t^2-2t=0 \Rightarrow t=1 \text{ or }t=0$$

But the answer is $t=1$ , I wonder why $t=0$ is incorrect ?

Answer 1

By direct calculation we have

$$A^3=\begin{bmatrix} 1 & 2 t-2 t^2 & t-t^2 \\ 4 t-4 & 5 t-4 & -t^2+4 t-3 \\ 2-2 t & t^2-4 t+3 & t^3-2 t+2 \\ \end{bmatrix}.$$

Your determinant calculation shows that $t = 1$ or $t= 0$ is a necessary condition for $A^3=I$. It does not show that $t=1$ or $t=0$ is a sufficient condition. If you plug in $t = 0$ in the above, clearly we don't get the identity matrix. Plugging in $t = 1$, on the other hand, gives us the identity matrix.

Of course, you don't have to calculate $A^3$ for arbitrary $t$, you can do it just for $t = 0$ and $t= 1$.

Answer 2

$p(x) =x^3-1$ is an anhilating polynomial.

Hence $m_A(x) \mid p(x) =(x-1) (x^2+x+1)$ implies $\textrm{spec}(A) \subseteq\{1,\omega,\omega^2\}$

Discard the possibility of $\textrm{spec}(A) =\{1\}$.

If $\textrm{spec}(A) =\{1\}$ then $\textrm{trace}(A)=-1+t=3$ implies $t=4$ and $\det(A) = 2\times 4 ^2-2\times 4 +1=25\neq 1$

Hence $\textrm{spec}(A) =\{1,\omega,\omega^2\}$

Then $\textrm{trace}(A)=-1+t=1+\omega+\omega^2=0$ implies $t=1$.

Answer 3

$A^3=\begin{bmatrix}1&t&0\\ \:-2&-2&-1\\ \:0&1&t\end{bmatrix}\begin{bmatrix}1&t&0\\ \:-2&-2&-1\\ \:0&1&t\end{bmatrix}\begin{bmatrix}1&t&0\\ \:-2&-2&-1\\ \:0&1&t\end{bmatrix}=$

$$\begin{pmatrix}1&-2t^2+2t&-t^2+t\\ \:4t-4&5t-4&-t^2+4t-3\\ \:-2t+2&t^2-4t+3&t^3-2t+2\end{pmatrix} \tag1$$

Your calculations are correct, the equation $2t^2-2t=0$ has $2$ roots, namely $t=1$ and/or $t=0$. Note that this does not mean that both solutions are valid for the original problem since $2t-2=0$ is a first degree equation and has 1 root only. The reason $t=0$ is not a valid solution for the main question is that when you use the value $t=0$ in the R.H.S (the product) in (1) you don't get the identity matrix, but when you use $t=1$ you get the desired identity matrix, sot $t=0$ is discarded and $t=1$ is accepted.