**Problem 1: Analysis of a Real Symmetric Matrix**Consider the real symmetric matrix:\[A = \begin{pmatrix} 3 & -2 & 4 \\ -2 & 6 & 2 \\ 4 & 2 & 3 \end{pmatrix}.\]Tasks:1. **Determine Eigenvalues and Eigenvectors:** - Calculate the eigenvalues and corresponding eigenvectors of matrix \( A \).2. **Orthogonal Matrix Construction:** - Use the Gram-Schmidt process to construct an orthogonal matrix \( Q \) from the eigenvectors. This orthogonal matrix will facilitate the diagonalization of matrix \( A \).3. **Classification of the Quadratic Form:** - Analyze the quadratic form given by: \[ 3x_1^2 + 6x_2^2 + 3x_3^2 - 4x_1x_2 + 8x_1x_3 + 4x_2x_3 = 98 \] - Identify whether the quadratic form represents: - An ellipsoid - A hyperboloid (one sheet or two sheets) - A paraboloid**Note:** The classification requires examining the coefficients and determining the nature of the conic section represented by the quadratic form.

Given real symmetric matrix is A=3-24-262423 let λ be the Eigenvalue of the given matrix. So A-λI=3-λ-24-26-λ2423-λ=0⇒3-λ-24-26-λ2423-λ=0⇒-λ3+12λ2-21-98=0⇒-λ-7λ2-5λ-14=0⇒(λ-7)(λ-7)(λ+2)=0⇒λ=7, 7, -2 Now Eigenvector corresponding to different Eigenvalues. for λ=7 [A-7I]X=-4-24-2-1242-4x1x2x3=000 Solving we get x1+12x2-x3=0 So general solution is X=-12x2+x3x2x3 for x2=1, x3=0 v1=-1210 for x2=0, x3=1 v2=101 Similarly for λ=-2 v3=-1-121 Now we have to find orthogonal matrix by Gram-Schimidt process. Given v1, v2, v3=-1210,101,-1-120 let u1=v1=-1210 u2=v2-u1T v2u1u1Tu1⇒u2=101--1210101-1210-1210-1210⇒u2=101+-15250⇒u2=45251 u3=v3-u1T v3u1u1Tu1-u2T v3u2u2Tu2⇒u3=-1-120--1210-1-120-1210-1210-1210-45251-1-120452514525145251⇒u3=-1-120+5945251⇒u3=59-5181 Hence orthogonal matrix is -124559125-518011