Find the characteristic equation of $\displaystyle A$ by obtaining the determinant of $\displaystyle \lambda I - A$.

$\displaystyle \lambda I - A$

$\displaystyle = \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} -\begin{bmatrix} 1 & 2 & 3 \\ 1 & -1 & 4 \\ -2 & -4 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} -\begin{bmatrix} 1 & 2 & 3 \\ 1 & -1 & 4 \\ -2 & -4 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} \lambda -1 &- 2 & -3 \\ -1 & \lambda +1 &- 4 \\ 2 & 4 & \lambda -1 \end{bmatrix}$

The determinant of this $\displaystyle 3 \times 3$ matrix can be found by adding the upper-left to lower-right products:

$\displaystyle (\lambda - 1)(\lambda + 1) (\lambda - 1) + (-2)(-4)(2) + (-3 )(-1)(4)$

$\displaystyle =( \lambda^{3} - \lambda^{2} - \lambda + 1 )+ 16 +12$

$\displaystyle = \lambda^{3} - \lambda^{2} - \lambda + 29$

Adding the upper-right to lower-left products:

$\displaystyle (-3)(\lambda + 1)(2) + (-4)(4)(\lambda-1) + (\lambda - 1) (-2)(-1)$

$\displaystyle = (-6 \lambda -6) + (-16 \lambda+16) + (2\lambda - 2)$

$\displaystyle = -20 \lambda +8$

And subtracting the latter from the former:

$\displaystyle = \lambda^{3} - \lambda^{2} - \lambda + 29 - ( -20 \lambda +8)$

$\displaystyle = \lambda^{3} - \lambda^{2} +19 \lambda + 21$

The characteristic equation is 

$\displaystyle \lambda^{3} - \lambda^{2} +19 \lambda + 21 = 0$

Factor the polynomial completely

$\displaystyle (\lambda + 1) (\lambda ^{2} - 2 \lambda + 21) = 0$

The solution set of this equation is comprised of the zeroes of both polynomial factors: 

$\displaystyle \left \{ -1, 1 - i \sqrt{20}, 1 + i \sqrt{20} \right \}$

Only $\displaystyle -1$ is a choice.

Assume that $\displaystyle -2$ is an eigenvalue of $\displaystyle A$. Then, if $\displaystyle b = \begin{bmatrix} b_{1} \\ b_{2} \\b_{3} \end{bmatrix}$ is one of its eigenvectors, it follows that

$\displaystyle Ab = -2 b$, or, equivalently,

$\displaystyle (A + 2I) b =O$,

where $\displaystyle I,O$ are the $\displaystyle 3 \times 3$ identity and zero matrices, respectively.

$\displaystyle A= \begin{bmatrix} 2 & 3 & 6 \\ 0 & 4 & 4 \\ 2 & 0 & 4 \end{bmatrix}$, so

$\displaystyle A + 2 I = \begin{bmatrix} 2+2 & 3 & 6 \\ 0 & 4+2 & 4 \\ 2 & 0 & 4+2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4& 3 & 6 \\ 0 & 6 & 4 \\ 2 &6& 2\end{bmatrix}$

Changing to reduced row echelon form:

$\displaystyle R1 \leftrightarrow R3$

$\displaystyle \begin{bmatrix} 2 &6& 2 \\ 0 & 6 & 4 \\ 4& 3 & 6 \end{bmatrix}$

$\displaystyle \frac{1}{2} R1 \rightarrow R1$

$\displaystyle \begin{bmatrix} 1 &3&1 \\ 0 & 6 & 4 \\ 4& 3 & 6 \end{bmatrix}$

$\displaystyle -4R1+ R3 \rightarrow R3$

$\displaystyle \begin{bmatrix} 1 &3&1 \\ 0 & 6 & 4 \\ 0& -9 & 2 \end{bmatrix}$

$\displaystyle \frac{1}{6} R2 \rightarrow R2$

$\displaystyle \begin{bmatrix} 1 &3&1 \\ 0 & 1 & \frac{2}{3} \\ 0& -9 & 2 \end{bmatrix}$

$\displaystyle -3R2 + R1 \rightarrow R1$

$\displaystyle 9 R2 + R3 \rightarrow R3$

$\displaystyle \begin{bmatrix} 1 &0 &-1 \\ 0 & 1 & \frac{2}{3} \\ 0& 0& 8 \end{bmatrix}$

$\displaystyle \frac{1}{8} R3 \rightarrow R3$

$\displaystyle \begin{bmatrix} 1 &0 &-1 \\ 0 & 1 & \frac{2}{3} \\ 0& 0& 1 \end{bmatrix}$

We do not need to go further to see that this matrix will not have a row of zeroes. This means the rank of the matrix is 3, and the nullity is 0. If this happens, the tested value, in this case $\displaystyle -2$, is not an eigenvalue. 

A $\displaystyle 4 \times 4$ matrix has four not necessarily distinct eigenvalues, $\displaystyle \lambda_{1} , \lambda_{2} , \lambda_{3} , \lambda_{4}$, which are the solutions of its characteristic polynomial equation 

$\displaystyle (\lambda - \lambda_{1}) (\lambda - \lambda_{2}) (\lambda - \lambda_{3} ) (\lambda - \lambda_{4})=0$

If there are only two distinct eigenvalues, 1 and $\displaystyle -1$, then one of the following three situations happens:

1 has multiplicity 3 and $\displaystyle -1$ has multiplicity 1, in which case $\displaystyle \lambda_{1} = \lambda_{2} = \lambda_{3}=1 , \lambda_{4} = -1$,

and the characteristic equation is 

$\displaystyle (\lambda - 1) (\lambda - 1) (\lambda - 1 ) (\lambda +1)=0$

or

$\displaystyle \lambda^{4}-2 \lambda^{3}+2 \lambda - 1 = 0$

1 has multiplicity 2 and $\displaystyle -1$ has multiplicity 2, in which case $\displaystyle \lambda_{1} = \lambda_{2} = 1 , \lambda_{3}= \lambda_{4} = -1$,

and the characteristic equation is 

$\displaystyle (\lambda - 1) (\lambda - 1) (\lambda + 1 ) (\lambda +1)=0$

or 

$\displaystyle \lambda ^{4} - 2 \lambda^{2} + 1 = 0$

1 has multiplicity 1 and $\displaystyle -1$ has multiplicity 3, in which case $\displaystyle \lambda_{1} = 1 , \lambda_{2} = \lambda_{3}= \lambda_{4} = -1$

and the characteristic equation is 

$\displaystyle (\lambda - 1) (\lambda + 1) (\lambda + 1 ) (\lambda +1)=0$

or

$\displaystyle \lambda^{4}+2 \lambda^{3}-2 \lambda - 1 = 0$

Of the four equations given as choices, only $\displaystyle \lambda ^{4}+ 2 \lambda^{2} + 1 = 0$ cannot be a characteristic equation of the matrix.

By the Cayley-Hamilton Theorem, a matrix is a solution of its own characteristic polynomial equation. None of the choices addresses the characteristic equation of $\displaystyle A$.

Assume that 2 is an eigenvalue of $\displaystyle A$. Then, if $\displaystyle b = \begin{bmatrix} b_{1} \\ b_{2} \\b_{3} \end{bmatrix}$ is one of its eigenvectors, it follows that

$\displaystyle Ab = 2 b$, or, equivalently,

$\displaystyle (A - 2I) b =O$,

where $\displaystyle I,O$ are the $\displaystyle 3 \times 3$ identity and zero matrices, respectively.

$\displaystyle A= \begin{bmatrix} 2 & 3 & 6 \\ 0 & 4 & 4 \\ 2 & 0 & 4 \end{bmatrix}$, so

$\displaystyle A - 2 I = \begin{bmatrix} 2-2 & 3 & 6 \\ 0 & 4-2 & 4 \\ 2 & 0 & 4-2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0& 3 & 6 \\ 0 & 2 & 4 \\ 2 & 0 & 2\end{bmatrix}$

Changing to reduced row echelon form:

$\displaystyle R1 \leftrightarrow R3$

$\displaystyle \begin{bmatrix}2 & 0 & 2 \\ 0 & 2 & 4 \\ 0& 3 & 6 \end{bmatrix}$

$\displaystyle \frac{1}{2}R1 \rightarrow R1$

$\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & 4 \\ 0& 3 & 6 \end{bmatrix}$

$\displaystyle \frac{1}{2}R2 \rightarrow R2$

$\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2\\ 0& 3 & 6 \end{bmatrix}$

$\displaystyle -3R2+ R3 \rightarrow R3$

$\displaystyle \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2\\ 0& 0 & 0 \end{bmatrix}$

The matrix is in reduced row echelon form. There is one column which does not contain leading 1s, so 2 is indeed an eigenvalue. The eigenspace is the set of all vectors $\displaystyle (x, y, z)$ such that $\displaystyle x= -z , y = -2z$ - that is, the set of vectors $\displaystyle t(-1,- 2, 1)$. The eigenspace has dimension 1.

The sum of the eigenvalues of a square matrix is equal to its trace, the sum of its diagonal elements. Examine these elements, which are in red below:

$\displaystyle A = \begin{bmatrix} {\color{Red}\textbf{a} } & 4 & -2 \\ 1 & {\color{Red}\textbf{3}} & 0 \\ - 6 & 4 & {\color{Red}\textbf{a}} \end{bmatrix}$

Set the trace equal to 10 and solve for $\displaystyle a$:

$\displaystyle \textup{Tr} (A ) = 10$

$\displaystyle a+ a + 3 = 10$

$\displaystyle 2 a+ 3 = 10$

$\displaystyle 2a = 7$

$\displaystyle a = 3.5$

Examine the columns of $\displaystyle A$:

$\displaystyle A= \begin{bmatrix} {\color{Red} 0.1} & {\color{DarkGreen} 0.4} & {\color{Blue} 0. 7} \\ {\color{Red} 0. 5} & {\color{DarkGreen} 0. 2} &{\color{Blue} 0.3} \\ {\color{Red} 0.4 }&{\color{DarkGreen} 0.4} & {\color{Blue} 0} \end{bmatrix}$

The entries of each column add up to 1:

Column 1: $\displaystyle 0.1 + 0.5 + 0.4 = 1$

Column 2: $\displaystyle 0.4+ 0.2 + 0.4 = 1$

Column 3: $\displaystyle 0.7+0.3 + 0 = 1$

It follows that $\displaystyle A$ could be a stochastic matrix for a state system; one of the properties of such a matrix is that one of its eigenvalues must be 1.

A $\displaystyle 4 \times 4$ matrix will have four eigenvalues, which are the zeroes of its characteristic polynomial equation. Since all of the entries of the matrix are real, all of the coefficients will be real as well. It follows that any imaginary zeroes must occur in conjugate pairs, so, since $\displaystyle 2-3i$ is a zero, so it $\displaystyle 2+3i$. 

We now know three of the zeroes, and since only one eigenvalue is unknown, it must be a real value. However, there is no restriction on this zero. Therefore, we have no way of determining the fourth zero - and, consequently, no way of figuring out the characteristic equation with any certainty. 

The sum of the eigenvalues of a matrix is equal to the trace of a matrix - the sum of its diagonal elements; the product of a matrix is equal to its determinant. 

Therefore, to find $\displaystyle a$, it is necessary to first find $\displaystyle b$. The trace of $\displaystyle A$ is equal to diagonal sum $\displaystyle b+7$, so set that equal to the sum of the given eigenvalues and solve:

$\displaystyle b+7 = (4+ \sqrt{3}) + (4- \sqrt{3})$

$\displaystyle b+7 = 8$

$\displaystyle b =1$

The matrix is therefore

$\displaystyle A = \begin{bmatrix} 7 & a \\ -3 & 1 \end{bmatrix}$

$\displaystyle a$ can be found by setting the determinant - the upper-left to lower-right product minus the upper-right to lower-left product - equal to the product of the eigenvalues:

$\displaystyle 7(1)-a(-3) = (4+ \sqrt{3}) (4- \sqrt{3})$

$\displaystyle 3a+ 7 = 13$

$\displaystyle 3a = 6$

$\displaystyle a= 2$,

the correct choice.

$\displaystyle A$ is a lower triangular matrix - all elements above its main diagonal are equal to 0. Such a matrix has as its eigenvalues its diagonal elements. All three diagonal elements in $\displaystyle A$ are 1, so 1 is the only eigenvalue of $\displaystyle A$.

To find the eigenvector(s) of $\displaystyle A$ corresponding to $\displaystyle \lambda = 1$, first, find the matrix

$\displaystyle A - \lambda I = A - I$. 

$\displaystyle A-I = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}- \begin{bmatrix} 1 & 0 & 0 \\ 0& 1 & 0 \\ 0&0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 & 0 \\ 2 & 0 & 0 \\ 3 & 2 & 0 \end{bmatrix}$

Perform the Gauss-Jordan elimination process to get this matrix in reduced row-echelon form:

$\displaystyle R1 \leftrightarrow R2$

$\displaystyle R2 \leftrightarrow R3$

$\displaystyle \begin{bmatrix} 2 & 0 & 0 \\ 3 & 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\displaystyle \frac{1}{2} R1 \rightarrow R1$

$\displaystyle \begin{bmatrix} 1 & 0 & 0 \\ 3 & 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\displaystyle -3R1 + R2 \rightarrow R2$

$\displaystyle \begin{bmatrix} 1 & 0 & 0 \\ 0& 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\displaystyle \frac{1}{2} R2 \rightarrow R2$

$\displaystyle \begin{bmatrix} 1 & 0 & 0 \\ 0&1 & 0\\ 0 & 0 & 0 \end{bmatrix}$

This matrix is interpreted as

$\displaystyle x = 0$, $\displaystyle y = 0$, $\displaystyle z$ arbitrary.

The eigenvectors must all take the form

$\displaystyle \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix}$

for some scalar $\displaystyle z$ - equivalently, all eigenvectors of $\displaystyle A$ are scalar multiples of 

$\displaystyle \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$.