Unit 1 Part B Solutions

Question 1

Find the eigenvalues and eigenvectors of the matrix: $$ A = \begin{bmatrix} 2 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 2 \end{bmatrix} $$

Solution

Step 1: Find the characteristic equation.
The characteristic equation is $ |A - \lambda I| = 0 $.

$$ \begin{vmatrix} 2-\lambda & 0 & 1 \\ 0 & 2-\lambda & 0 \\ 1 & 0 & 2-\lambda \end{vmatrix} = 0 $$

Expanding along the second row (R2):
$$ (2-\lambda) \left[ (2-\lambda)(2-\lambda) - (1)(1) \right] = 0 $$ $$ (2-\lambda) [ (2-\lambda)^2 - 1 ] = 0 $$ $$ (2-\lambda) [ \lambda^2 - 4\lambda + 4 - 1 ] = 0 $$ $$ (2-\lambda) (\lambda^2 - 4\lambda + 3) = 0 $$ $$ (2-\lambda) (\lambda - 1) (\lambda - 3) = 0 $$

The eigenvalues ($\lambda$) are 1, 2, 3.

Step 2: Find the eigenvectors for each eigenvalue.
The eigenvector $ X = [x_1, x_2, x_3]^T $ is found by solving $ (A - \lambda I)X = 0 $.

Case 1: $\lambda = 1$
$$ (A - 1I)X = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$ Equations:
1. $ x_1 + x_3 = 0 \Rightarrow x_1 = -x_3 $
2. $ x_2 = 0 $
Let $ x_3 = k $. Then $ x_1 = -k, x_2 = 0 $.
Eigenvector $ X_1 = k \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} $.

Case 2: $\lambda = 2$
$$ (A - 2I)X = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$ Equations:
1. $ x_3 = 0 $
2. $ x_1 = 0 $
$ x_2 $ is a free variable. Let $ x_2 = k $.
Eigenvector $ X_2 = k \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $.

Case 3: $\lambda = 3$
$$ (A - 3I)X = \begin{bmatrix} -1 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$ Equations:
1. $ -x_1 + x_3 = 0 \Rightarrow x_1 = x_3 $
2. $ -x_2 = 0 \Rightarrow x_2 = 0 $
Let $ x_3 = k $. Then $ x_1 = k $.
Eigenvector $ X_3 = k \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $.

Eigenvalues: $ \lambda = 1, 2, 3 $
Eigenvectors: $ X_1 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, X_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, X_3 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $

Question 2

Using Cayley-Hamilton theorem, find $ A^4 $ if: $$ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 4 \\ 3 & 1 & 1 \end{bmatrix} $$

Solution

Step 1: Find the characteristic equation of A.
The equation is $ \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 $.

$ S_1 = \text{Sum of diagonal elements} = 1 + (-1) + 1 = 1 $
$ S_2 = \text{Sum of minors of diagonal elements} $ $$ = \begin{vmatrix} -1 & 4 \\ 1 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} $$ $$ = (-1 - 4) + (1 - 9) + (-1 - 4) = -5 - 8 - 5 = -18 $$
$ S_3 = |A| $ $$ = 1(-1 - 4) - 2(2 - 12) + 3(2 + 3) $$ $$ = 1(-5) - 2(-10) + 3(5) = -5 + 20 + 15 = 30 $$

Characteristic Equation: $ \lambda^3 - \lambda^2 - 18\lambda - 30 = 0 $.

Step 2: Apply the Cayley-Hamilton Theorem.
Matrix A satisfies its own characteristic equation:
$ A^3 - A^2 - 18A - 30I = 0 $
Thus, $ A^3 = A^2 + 18A + 30I $.

Step 3: Express $ A^4 $ in terms of lower powers.
Multiply by A:
$ A^4 = A(A^3) = A(A^2 + 18A + 30I) $
$ A^4 = A^3 + 18A^2 + 30A $
Substitute $ A^3 $ back:
$ A^4 = (A^2 + 18A + 30I) + 18A^2 + 30A $
$ A^4 = 19A^2 + 48A + 30I $

Step 4: Calculate $ A^2 $.
$$ A^2 = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 4 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 4 \\ 3 & 1 & 1 \end{bmatrix} $$ $$ = \begin{bmatrix} 1+4+9 & 2-2+3 & 3+8+3 \\ 2-2+12 & 4+1+4 & 6-4+4 \\ 3+2+3 & 6-1+1 & 9+4+1 \end{bmatrix} = \begin{bmatrix} 14 & 3 & 14 \\ 12 & 9 & 6 \\ 8 & 6 & 14 \end{bmatrix} $$

Step 5: Calculate $ A^4 $.
$$ A^4 = 19 \begin{bmatrix} 14 & 3 & 14 \\ 12 & 9 & 6 \\ 8 & 6 & 14 \end{bmatrix} + 48 \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 4 \\ 3 & 1 & 1 \end{bmatrix} + 30 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

$$ = \begin{bmatrix} 266 & 57 & 266 \\ 228 & 171 & 114 \\ 152 & 114 & 266 \end{bmatrix} + \begin{bmatrix} 48 & 96 & 144 \\ 96 & -48 & 192 \\ 144 & 48 & 48 \end{bmatrix} + \begin{bmatrix} 30 & 0 & 0 \\ 0 & 30 & 0 \\ 0 & 0 & 30 \end{bmatrix} $$

$$ = \begin{bmatrix} 266+48+30 & 57+96+0 & 266+144+0 \\ 228+96+0 & 171-48+30 & 114+192+0 \\ 152+144+0 & 114+48+0 & 266+48+30 \end{bmatrix} $$

$$ A^4 = \begin{bmatrix} 344 & 153 & 410 \\ 324 & 153 & 306 \\ 296 & 162 & 344 \end{bmatrix} $$

Question 3

Reduce the quadratic form $ 3x_1^2 + 3x_2^2 + 3x_3^2 + 2x_1x_2 + 2x_1x_3 - 2x_2x_3 $ to canonical form through an orthogonal transformation. Also find its nature, index, signature.

Solution

Step 1: Write the matrix of the quadratic form.
$$ A = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & -1 & 3 \end{bmatrix} $$

Step 2: Find the eigenvalues of A.
$ \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 $

$ S_1 = 3 + 3 + 3 = 9 $
$ S_2 = (9 - 1) + (9 - 1) + (9 - 1) = 8 + 8 + 8 = 24 $
$ S_3 = |A| = 3(9-1) - 1(3 - (-1)) + 1(-1 - 3) = 3(8) - 4 - 4 = 24 - 8 = 16 $

Eq: $ \lambda^3 - 9\lambda^2 + 24\lambda - 16 = 0 $.
Inspection: For $ \lambda = 1 $: $ 1 - 9 + 24 - 16 = 0 $. Factor: $ (\lambda - 1) $.
For $ \lambda = 4 $: $ 64 - 144 + 96 - 16 = 0 $. Factor: $ (\lambda - 4) $.
Sum of eigenvalues = 9. So $ 1 + 4 + \lambda_3 = 9 \Rightarrow \lambda_3 = 4 $.
Eigenvalues: 1, 4, 4.

Step 3: Write the canonical form.
Formula: $ \lambda_1 y_1^2 + \lambda_2 y_2^2 + \lambda_3 y_3^2 $

Canonical Form: $ y_1^2 + 4y_2^2 + 4y_3^2 $

Step 4: Determine nature, index, and signature.

Nature: Positive Definite (all eigenvalues > 0).
Index (p): 3 (number of positive terms).
Signature (s): $ 3 - 0 = 3 $.

Question 4

Reduce the quadratic form $ 6x^2 + 3y^2 + 3z^2 - 4xy - 2yz + 4xz $ into a canonical form through an orthogonal reduction.

Solution

Step 1: Write the matrix of the quadratic form.
$$ A = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} $$

Step 2: Find the eigenvalues of A.
$ \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 $

$ S_1 = 6 + 3 + 3 = 12 $
$ S_2 = (9 - 1) + (18 - 4) + (18 - 4) = 8 + 14 + 14 = 36 $
$ S_3 = |A| = 6(9 - 1) - (-2)(-6 - (-2)) + 2(2 - 6) $
$ = 6(8) + 2(-4) + 2(-4) = 48 - 8 - 8 = 32 $

Eq: $ \lambda^3 - 12\lambda^2 + 36\lambda - 32 = 0 $.
By inspection:
If $ \lambda = 2 $: $ 8 - 48 + 72 - 32 = 0 $.
If $ \lambda = 8 $: $ 512 - 768 + 288 - 32 = 0 $.
Using sum: $ 2 + 8 + \lambda_3 = 12 \Rightarrow \lambda_3 = 2 $.
Eigenvalues: 2, 2, 8.

Step 3: Write the canonical form.

Canonical Form: $ 2y_1^2 + 2y_2^2 + 8y_3^2 $

Question 5

Find the eigenvalues and eigenvectors of a matrix: $$ A = \begin{bmatrix} 2 & 2 & 0 \\ 2 & 1 & 1 \\ -7 & 2 & -3 \end{bmatrix} $$

Solution

Step 1: Find the characteristic equation.

$ S_1 = 2 + 1 - 3 = 0 $
$ S_2 = (-3-2) + (-6-0) + (2-4) = -5 - 6 - 2 = -13 $
$ S_3 = |A| = 2(-3-2) - 2(-6+7) + 0 = -10 - 2 = -12 $

Eq: $ \lambda^3 - 0\lambda^2 - 13\lambda + 12 = 0 \Rightarrow \lambda^3 - 13\lambda + 12 = 0 $.
Factors of 12:
$ \lambda = 1 \Rightarrow 1 - 13 + 12 = 0 $.
$ \lambda = 3 \Rightarrow 27 - 39 + 12 = 0 $.
Sum = 0: $ 1 + 3 + \lambda_3 = 0 \Rightarrow \lambda_3 = -4 $.
Eigenvalues: 1, 3, -4.

Step 2: Find eigenvectors $ (A - \lambda I)X = 0 $.

Case 1: $\lambda = 1$
$$ \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ -7 & 2 & -4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 $$ $ x_1 + 2x_2 = 0 \Rightarrow x_1 = -2x_2 $
$ 2x_1 + x_3 = 0 \Rightarrow x_3 = -2x_1 = 4x_2 $
Let $ x_2 = k $. Eigenvector $ X_1 = k \begin{bmatrix} -2 \\ 1 \\ 4 \end{bmatrix} $.

Case 2: $\lambda = 3$
$$ \begin{bmatrix} -1 & 2 & 0 \\ 2 & -2 & 1 \\ -7 & 2 & -6 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 $$ $ -x_1 + 2x_2 = 0 \Rightarrow x_1 = 2x_2 $
$ 2x_1 - 2x_2 + x_3 = 0 \Rightarrow 4x_2 - 2x_2 + x_3 = 0 \Rightarrow x_3 = -2x_2 $
Let $ x_2 = k $. Eigenvector $ X_2 = k \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix} $.

Case 3: $\lambda = -4$
$$ \begin{bmatrix} 6 & 2 & 0 \\ 2 & 5 & 1 \\ -7 & 2 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0 $$ $ 6x_1 + 2x_2 = 0 \Rightarrow x_2 = -3x_1 $
$ 2x_1 + 5x_2 + x_3 = 0 \Rightarrow 2x_1 - 15x_1 + x_3 = 0 \Rightarrow x_3 = 13x_1 $
Let $ x_1 = k $. Eigenvector $ X_3 = k \begin{bmatrix} 1 \\ -3 \\ 13 \end{bmatrix} $.

Eigenvalues: 1, 3, -4
Eigenvectors: $ \begin{bmatrix} -2 \\ 1 \\ 4 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}, \begin{bmatrix} 1 \\ -3 \\ 13 \end{bmatrix} $

Question 6

Using Cayley-Hamilton theorem find the inverse of the matrix: $$ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} $$

Solution

Step 1: Characteristic Equation.

$ S_1 = 1 + 4 + 6 = 11 $
$ S_2 = (24-25) + (6-9) + (4-4) = -1 - 3 + 0 = -4 $
$ S_3 = 1(-1) - 2(-3) + 3(-2) = -1 + 6 - 6 = -1 $

Eq: $ \lambda^3 - 11\lambda^2 - 4\lambda + 1 = 0 $.

Step 2: Apply Cayley-Hamilton.
$ A^3 - 11A^2 - 4A + I = 0 $

Step 3: Find $ A^{-1} $.
Multiply by $ A^{-1} $:
$ A^2 - 11A - 4I + A^{-1} = 0 $
$ A^{-1} = -A^2 + 11A + 4I $

Step 4: Calculate $ A^2 $.
$$ A^2 = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} $$ $$ = \begin{bmatrix} 1+4+9 & 2+8+15 & 3+10+18 \\ 2+8+15 & 4+16+25 & 6+20+30 \\ 3+10+18 & 6+20+30 & 9+25+36 \end{bmatrix} = \begin{bmatrix} 14 & 25 & 31 \\ 25 & 45 & 56 \\ 31 & 56 & 70 \end{bmatrix} $$

Step 5: Calculate $ A^{-1} $.
$$ A^{-1} = -\begin{bmatrix} 14 & 25 & 31 \\ 25 & 45 & 56 \\ 31 & 56 & 70 \end{bmatrix} + 11\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix} + 4\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ $$ = \begin{bmatrix} -14+11+4 & -25+22+0 & -31+33+0 \\ -25+22+0 & -45+44+4 & -56+55+0 \\ -31+33+0 & -56+55+0 & -70+66+4 \end{bmatrix} $$

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

Question 7

Diagonalise the matrix by Orthogonality transformation: $$ A = \begin{bmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{bmatrix} $$

Solution

Step 1: Find Eigenvalues.
$ S_1 = 18 $
$ S_2 = (21-16) + (24-4) + (56-36) = 5 + 20 + 20 = 45 $
$ S_3 = 8(5) + 6(-10) + 2(10) = 40 - 60 + 20 = 0 $
Eq: $ \lambda^3 - 18\lambda^2 + 45\lambda = 0 \Rightarrow \lambda(\lambda^2 - 18\lambda + 45) = 0 $.
Factors: $ \lambda(\lambda - 3)(\lambda - 15) = 0 $.
Eigenvalues: 0, 3, 15.

Step 2: Find Eigenvectors.
For $\lambda = 0$:
Using cross-multiplication on first two rows of A:
$ \frac{x_1}{24-14} = \frac{x_2}{-12+32} = \frac{x_3}{56-36} \Rightarrow \frac{x_1}{10} = \frac{x_2}{20} = \frac{x_3}{20} $
$ X_1 = [1, 2, 2]^T $.

For $\lambda = 3$:
Rows imply $ 2x_1 - 4x_2 = 0 \Rightarrow x_1 = 2x_2 $ and $ 5x_1 - 6x_2 + 2x_3 = 0 $.
Sub $ x_1 $: $ 10x_2 - 6x_2 + 2x_3 = 0 \Rightarrow 4x_2 = -2x_3 \Rightarrow x_3 = -2x_2 $.
Let $ x_2 = 1 $. $ X_2 = [2, 1, -2]^T $.

For $\lambda = 15$:
Cross multiplication on $ -7x_1 - 6x_2 + 2x_3 = 0 $ and $ x_1 - 2x_2 - 6x_3 = 0 $ (simplified R3):
$ \frac{x_1}{36+4} = \frac{x_2}{2-42} = \frac{x_3}{14+6} \Rightarrow \frac{x_1}{40} = \frac{x_2}{-40} = \frac{x_3}{20} $
$ X_3 = [2, -2, 1]^T $.

Step 3: Normalize Eigenvectors.
$ |X_1| = \sqrt{1+4+4} = 3 $
$ |X_2| = \sqrt{4+1+4} = 3 $
$ |X_3| = \sqrt{4+4+1} = 3 $
Normalized Matrix N: $$ N = \frac{1}{3} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix} $$

$$ D = N^T A N = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 15 \end{bmatrix} $$

Question 8

Reduce the quadratic form $ x_1^2 + 2x_2^2 + x_3^2 - 2x_1x_2 + 2x_2x_3 $ to canonical form and find its nature.

Solution

Step 1: Matrix A.
$$ A = \begin{bmatrix} 1 & -1 & 0 \\ -1 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} $$

Step 2: Eigenvalues.
$ S_1 = 4 $, $ S_2 = (2-1) + (1-0) + (2-1) = 3 $, $ S_3 = 1(1) - (-1)(-1) = 1 - 1 = 0 $.
Eq: $ \lambda^3 - 4\lambda^2 + 3\lambda = 0 \Rightarrow \lambda(\lambda-1)(\lambda-3) = 0 $.
Eigenvalues: 0, 1, 3.

Step 3: Canonical Form.

$ 0y_1^2 + 1y_2^2 + 3y_3^2 = y_2^2 + 3y_3^2 $
Nature: Positive Semi-definite.

Question 9

Obtain an orthogonal transformation which will transform the quadratic form $ Q = 2xy + 2yz + 2xz $ to canonical form.

Solution

Step 1: Matrix A.
$$ A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} $$

Step 2: Eigenvalues.
$ S_1 = 0 $, $ S_2 = -3 $, $ S_3 = 2 $.
Eq: $ \lambda^3 - 3\lambda - 2 = 0 $.
Roots: $ (\lambda-2)(\lambda+1)^2 = 0 $.
Eigenvalues: 2, -1, -1.

Step 3: Eigenvectors.
Case 1: $\lambda = 2$
$ -2x + y + z = 0 $ and $ x - 2y + z = 0 $. Cross multiply: $ x/3 = y/3 = z/3 $.
$ X_1 = [1, 1, 1]^T $.

Case 2: $\lambda = -1$ (Repeated)
The equation reduces to single form $ x + y + z = 0 $.
Let $ X_2 = [1, -1, 0]^T $ (satisfies eq).
To find $ X_3 $, it must be orthogonal to $ X_1 $ and $ X_2 $.
$ X_3 = [l, m, n]^T $.
1. $ l+m+n = 0 $
2. $ l-m = 0 \Rightarrow l=m $
Sub into 1: $ 2l + n = 0 \Rightarrow n = -2l $. Let $ l=1 $.
$ X_3 = [1, 1, -2]^T $.

Step 4: Normalize.
$ |X_1| = \sqrt{3} $, $ |X_2| = \sqrt{2} $, $ |X_3| = \sqrt{6} $.

Transformation $ X = NY $: $$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1/\sqrt{3} & 1/\sqrt{2} & 1/\sqrt{6} \\ 1/\sqrt{3} & -1/\sqrt{2} & 1/\sqrt{6} \\ 1/\sqrt{3} & 0 & -2/\sqrt{6} \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} $$ Canonical Form: $ 2y_1^2 - y_2^2 - y_3^2 $

Question 10

Diagonalise the symmetric matrix by Orthogonality transformation: $$ A = \begin{bmatrix} 2 & 0 & 4 \\ 0 & 6 & 0 \\ 4 & 0 & 2 \end{bmatrix} $$

Solution

Step 1: Eigenvalues.
$ |A-\lambda I| = (6-\lambda)[(2-\lambda)^2 - 16] = 0 $
$ (6-\lambda)(\lambda^2 - 4\lambda - 12) = 0 $
$ (6-\lambda)(\lambda-6)(\lambda+2) = 0 $
Eigenvalues: 6, 6, -2.

Step 2: Eigenvectors.
Case 1: $\lambda = -2$
$ 4x_1 + 4x_3 = 0 \Rightarrow x_1 = -x_3 $, $ 8x_2 = 0 \Rightarrow x_2 = 0 $.
$ X_1 = [-1, 0, 1]^T $.

Case 2: $\lambda = 6$ (Repeated)
$ -4x_1 + 4x_3 = 0 \Rightarrow x_1 = x_3 $. $ x_2 $ is free.
We need two orthogonal vectors satisfying $ x_1 = x_3 $.
Let $ X_2 = [0, 1, 0]^T $ (x2=1, x1=x3=0).
Let $ X_3 = [1, 0, 1]^T $ (x2=0, x1=x3=1).
Check orthogonality: $ X_2 \cdot X_3 = 0 $. Correct.

Step 3: Normalize and Form N.
$ |X_1| = \sqrt{2} $, $ |X_2| = 1 $, $ |X_3| = \sqrt{2} $.
$$ N = \begin{bmatrix} -1/\sqrt{2} & 0 & 1/\sqrt{2} \\ 0 & 1 & 0 \\ 1/\sqrt{2} & 0 & 1/\sqrt{2} \end{bmatrix} $$

$$ D = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} $$

Unit 1: Part B Solutions (Matrices and Calculus)

Question 1

Solution

Question 2

Solution

Question 3

Solution

Question 4

Solution

Question 5

Solution

Question 6

Solution

Question 7

Solution

Question 8

Solution

Question 9

Solution

Question 10

Solution