The matrix which remains unchanged under transposition is known as symmetric matrix.
For example, if we consider a symmetric matrix A = \(\begin{bmatrix} 2 & 3 & 4 \\
3 & 5 & 6\\
4 & 6 & 7 \end{bmatrix} \) and take transpose of it, we get,
A^T = \(\begin{bmatrix}
2 & 3 & 4 \\
3 & 5 & 6 \\
4 & 6 & 7 \end{bmatrix} \)

The diagonalization orem in terms of multiplicities of eigen values is defined as follows,
The algebraic multiplicity of ? is its multiplicity as a root of characteristic polynomial of A.
The geometric multiplicity of ? is dimension of ?-eigenspace.

The product of eigen values of a matrix gives determinant of matrix,
Therefore, ? = 24.

An orthogonal matrix can be defined as ?a matrix having its entries as orthogonal unit vectors? or it can also be defined as ?a matrix whose transpose is equal to its inverse?. Since this property is satisfied by an identity matrix, every identity matrix is an orthogonal matrix.

Given quadratic form is, \(3x_1^2+ 2x_2^2+8x_{12}+8x_{23}+8x_{31}=0\)
General form of matrix can be written as, \(\begin{bmatrix}x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33}\end{bmatrix} \)
Hence, matrix form can be obtained by,
Placing square term coefficients in diagonal of matrix such that, \(x_{ii}= x_i^2,\)
\(\begin{bmatrix}3 & x_{12} & x_{13} \\ x_{21} & 2 & x_{23} \\ x_{31} & x_{32} & 0\end{bmatrix} \)
Dividing coefficients of terms x_ij between x_ij and x_ji positions, for example, coefficient of x₁₂ is 8, hence term x₁₂ =x₂₁= 8/2 = 4.
Therefore, matrix form of given quadratic equation is,
\(\begin{bmatrix}3 & 4 & 4 \\ 4 & 2 & 4 \\ 4 & 4 & 0\end{bmatrix} \)

We know that for any given matrix
[A-λI]X=0 and |A-λI|=0
[A-λI]=\(\begin{bmatrix}4&1&3\\1&3&1\\2&0&5\end{bmatrix}-\lambda \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
The characteristic polynomial is given by-
λ³-(Sum of diagonal elements) λ²+(Sum of minor of diagonal element)λ-|A|=0
λ³-12λ²+40λ-39=0
λ=7.1 or λ=3.

From ory of diagonalization, we know that,
A = PDP^-1
Aⁿ = PDⁿP^-1
Given, P= \(\begin{bmatrix}
2 & -1\\
5 & 1,\end{bmatrix} hence P^{-1} = \frac{1}{7}
\begin{bmatrix}
1 & 1\\
-5 & 2\end{bmatrix}\)
Therefore, \(A^3 = \frac{1}{7} \begin{bmatrix}
2 & -1\\
5 & 1
\end{bmatrix}
\begin{bmatrix}
6 & 0\\
0 & -1 \end{bmatrix}^3
\begin{bmatrix}
1 & 1\\
-5 & 2 \end{bmatrix}\)??????????. since n=3
\(A^3 = \frac{1}{7}
\begin{bmatrix}2 & -1 \\ 5 & 1 \end{bmatrix}
\begin{bmatrix}216 & 0 \\ 0 & -1 \end{bmatrix}
\begin{bmatrix}1 & 1 \\ -5 & 2 \end{bmatrix}\)
\(A^3 = \frac{1}{7}
\begin{bmatrix} 2 & -1 \\ 5 & 1 \end{bmatrix}
\begin{bmatrix} 216 & 216 \\ 5 & -2 \end{bmatrix}\)
\(A^3 = \frac{1}{7}
\begin{bmatrix}427 & 434\\ 1085 & 1078 \end{bmatrix}\)
\(A^3 = \begin{bmatrix} 61 & 62\\ 155 & 154 \end{bmatrix}\)

Out of given options, \(\begin{bmatrix}0.33 & 0.67 & -0.67\\ -0.67 & 0.67 & 0.33\\ 0.67 & 0.33 & 0.67\end{bmatrix} \) satisfies condition for orthogonality, i.e. AA^T = I
\(\begin{bmatrix}0.33 & 0.67 & -0.67\\ -0.67 & 0.67 & 0.33\\ 0.67 & 0.33 & 0.67\end{bmatrix}
\begin{bmatrix}0.33 & -0.67 & 0.67 \\ 0.67 & 0.67 & 0.33\\-0.67 & 0.33 & 0.67\end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \)
Since \(\begin{bmatrix}0.33 & -0.67 & 0.67\\ 0.67 & 0.67 & 0.33\\ -0.67 & 0.33 & 0.67\end{bmatrix} \)is transpose of A, it is also orthogonal.
Coming to \(\begin{bmatrix}cosx & sinx\\ -sin x &cosx\end{bmatrix}, \)
\(\begin{bmatrix}cosx & sinx\\-sinx & cosx\end{bmatrix} \begin{bmatrix}cosx & -sinx \\ sinx & cosx\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} \)
The remaining option \(\begin{bmatrix}cosx & sinx\\ -sinx & -cosx\end{bmatrix}, \)which does not satisfy condition for orthogonality.

Signature of a quadratic form is defined as ? difference between number of positive and negative square terms in canonical form.?

Let A = \(\begin{bmatrix}3 & 0 & 2\\ 6 & 1 & 1\\ 2 & 8 & 91\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix} ? \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix} \)
Hence, given matrix equation can be written in form,
AX = B
Multiplying both sides by A^-1, we get
X = A^-1 B
But since B = 0, X = 0 and hence solution is,
x₁ = 0, x₂ = 0, x₃ = 0.