The general form of a skew symmetric matrix is given by,
\(\begin{bmatrix}
0 & w1 & -w2\\
-w1 & 0 & w3 \\
w2&-w3 & 0
\end{bmatrix} \) Therefore, from given matrix,
y = 5,
\(x+y = 4 \rightarrow x+5=4 \rightarrow x = -1\)

The sum of entries on main diagonal is called trace of matrix A.
Therefore, trace = 1+5+4 = 9.

For given Matrix,
A=\(\begin{bmatrix}-1&2&8\\-2&3&0\\-4&5&1\end{bmatrix}\)

The characteristic polynomial is given by-
α³-(Sum of diagonal elements) α²+(Sum of minor of diagonal element)α-|A|=0

The canonical form is also known as sum of squares form since after reducing to canonical form, we get terms as sum of squares.

The reversal rule of matrix multiplication states that, ? inverse of product of two matrices is equal to product of ir individual inverses, taken in reverse order?.

The eigen values of given matrix can be computed as,
?A ? ?I ? = 0
\(\begin{vmatrix}
1-? & 0 & 1\\
3 & 3-? & 0\\
0 & 0 & 1-?
\end{vmatrix}\)= 0
(1-?) ((3-?) (1-?)) = 0
(1-?)² (3-?) = 0
? = 1, 1, 3 are eigen values of matrix. So, algebraic multiplicity of ? = 1 is two.
For ? = 3,
(A – ?I) \(\vec{X} = \vec{0} \)
\(\begin{bmatrix}
-2 & 0 & 1 \\
3 & 0 & 0 \\
0 & 0 & -2
\end{bmatrix}\vec{X} = \vec{0}\quad ^{\underrightarrow{Reducing \,furr, \,we \,get,}}
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 1\\
0 & 0 & 0
\end{bmatrix}
\vec{X} = \vec{0} \)
x1 = 0, x3 = 0
x2 is free variable, refore let x2 = s,
Hence, \(\vec{X_1}= s\begin{bmatrix}0 \\ 1\\ 0\end{bmatrix} ? \begin{bmatrix}0 \\ 1 \\0 \end{bmatrix} \)
For ? = 1,
(A – ?I) \(\vec{X} = \vec{0} \)
\(\begin{bmatrix}
0 & 0 & 1 \\
3 & 2 & 0 \\
0 & 0 & 0
\end{bmatrix} \vec{X} = \vec{0} \quad ^{\underrightarrow{Reducing \,furr, \,we \,get,}}
\begin{bmatrix}
1 & 2/3 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{bmatrix} \vec{X} = \vec{0} \)
x1 + \(\frac{2}{3}\) x2 = 0, x3 = 0
Let x1 = -2 and x3 = 3,
\(\vec{X_2} = \begin{bmatrix}-2 \\ 3\\ 0 \end{bmatrix}\)
Thus, re corresponds only one eigen vector for repeated eigen value ?=1. Thus, geometric multiplicity of ? = 2 is one.

We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
Given that, λ=-2
[A-λI]=\(\begin{bmatrix}3&5\\3&1\end{bmatrix}-(-2)\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}3+2&5\\3&1+2\end{bmatrix}\)
[A-λI]=\(\begin{bmatrix}5&5\\3&3\end{bmatrix}\)
Since, [A-λI]X=0
\(\begin{bmatrix}5&5\\3&3\end{bmatrix}
\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}\)
Thus,
5x+5y=0 and 3x+3y=0
Let x=t,
Then, y=-t
X = \(\begin{bmatrix}t\\-t\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}\)

Quadratic forms can be classified based on nature of eigen values of matrix into 5 types:
i. Positive definite
ii. Negative definite
iii. Positive Semi-definite
iv. Negative Semi-definite
v. Indefinite

The product of eigen values of a matrix gives determinant of matrix,
Therefore, ? = 63.

To find nature of matrix, we find eigenvalues, |A- ?I|=0
\(\begin{vmatrix}2-? & 0 & 0 \\1 & 2-? & 1 \\ 0 & 0 & 1-? \end{vmatrix} = 0 \)

(2- ?) ((2- ?)(1- ?)) = 0
? = 1, 2, 2
Since all eigenvalues are positive, condition for positive definite quadratic form is satisfied.