Answer :
Let's solve the problem step by step:
### Part (i): Determine [tex]\( \lambda \)[/tex] and [tex]\( Q^{-1} \)[/tex]
Given the matrix [tex]\( QQ^T = \lambda I \)[/tex], where [tex]\( I \)[/tex] is the identity matrix, we need to determine:
1. [tex]\( \lambda \)[/tex]
2. [tex]\( Q^{-1} \)[/tex]
#### Determining [tex]\( \lambda \)[/tex]
To find [tex]\( \lambda \)[/tex], we can see that [tex]\( QQ^T \)[/tex] is a product of the matrices [tex]\( Q \)[/tex] and [tex]\( Q^T \)[/tex]. This implies that [tex]\( QQ^T \)[/tex] is a symmetric matrix.
The eigenvalues of the matrix [tex]\( QQ^T \)[/tex] will help us find [tex]\( \lambda \)[/tex]. Since [tex]\( \lambda I \)[/tex] is a diagonal matrix with [tex]\( \lambda \)[/tex] as its repeated value, the eigenvalues of [tex]\( QQ^T \)[/tex] will give us [tex]\( \lambda \)[/tex].
Given the numerical solution, we have:
[tex]\[ \lambda = (8.999999999999993 + 15.588457268119885j) \][/tex]
#### Determining [tex]\( Q^{-1} \)[/tex]
[tex]\( Q \)[/tex] can be represented by the matrix of eigenvectors of [tex]\( QQ^T \)[/tex], and the inverse of [tex]\( Q \)[/tex] can be found by inverting the eigenvector matrix.
From the numerical solution, we have:
[tex]\[ Q^{-1} \approx \begin{pmatrix} -0.57735027 + 2.77555756e-16j & 0.28867513 + 0.5j & 0.28867513 - 0.5j \\ -0.57735027 - 3.05311332e-16j & 0.28867513 - 0.5j & 0.28867513 + 0.5j \\ 0.57735027 + 1.35456610e-17j & 0.57735027 - 1.85037171e-17j & 0.57735027 - 1.85037171e-17j \end{pmatrix} \][/tex]
### Part (ii): Solve the system of equations
Given the system of equations:
[tex]\[ \begin{array}{l} 12I_1 + 12I_2 - 6I_3 = -18 \\ -6I_1 + 12I_2 + 12I_3 = 0 \\ 12I_1 - 6I_2 + 12I_3 = 72 \end{array} \][/tex]
We can represent this system as the matrix equation [tex]\( A\mathbf{I} = \mathbf{B} \)[/tex], where:
[tex]\[ A = \begin{pmatrix} 12 & 12 & -6 \\ -6 & 12 & 12 \\ 12 & -6 & 12 \end{pmatrix} \][/tex]
and
[tex]\[ \mathbf{B} = \begin{pmatrix} -18 \\ 0 \\ 72 \end{pmatrix} \][/tex]
To solve for [tex]\( \mathbf{I} = \begin{pmatrix} I_1 \\ I_2 \\ I_3 \end{pmatrix} \)[/tex], we need to solve the equation [tex]\( \mathbf{I} = A^{-1}\mathbf{B} \)[/tex].
From the numerical solution, we have:
[tex]\[ \mathbf{I} \approx \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} \][/tex]
So the solutions for the system of equations are:
[tex]\[ I_1 = 2, \quad I_2 = -2, \quad I_3 = 3 \][/tex]
### Summary
1. [tex]\( \lambda \)[/tex] is approximately [tex]\( (8.999999999999993 + 15.588457268119885j) \)[/tex].
2. [tex]\( Q^{-1} \)[/tex] is:
[tex]\[ \begin{pmatrix} -0.57735027 + 2.77555756e-16j & 0.28867513 + 0.5j & 0.28867513 - 0.5j \\ -0.57735027 - 3.05311332e-16j & 0.28867513 - 0.5j & 0.28867513 + 0.5j \\ 0.57735027 + 1.35456610e-17j & 0.57735027 - 1.85037171e-17j & 0.57735027 - 1.85037171e-17j \end{pmatrix} \][/tex]
3. The solutions for [tex]\( I_1, I_2, \)[/tex] and [tex]\( I_3 \)[/tex] are:
[tex]\[ I_1 \approx 2, \quad I_2 \approx -2, \quad I_3 \approx 3 \][/tex]
This gives the complete solution to the problem.
### Part (i): Determine [tex]\( \lambda \)[/tex] and [tex]\( Q^{-1} \)[/tex]
Given the matrix [tex]\( QQ^T = \lambda I \)[/tex], where [tex]\( I \)[/tex] is the identity matrix, we need to determine:
1. [tex]\( \lambda \)[/tex]
2. [tex]\( Q^{-1} \)[/tex]
#### Determining [tex]\( \lambda \)[/tex]
To find [tex]\( \lambda \)[/tex], we can see that [tex]\( QQ^T \)[/tex] is a product of the matrices [tex]\( Q \)[/tex] and [tex]\( Q^T \)[/tex]. This implies that [tex]\( QQ^T \)[/tex] is a symmetric matrix.
The eigenvalues of the matrix [tex]\( QQ^T \)[/tex] will help us find [tex]\( \lambda \)[/tex]. Since [tex]\( \lambda I \)[/tex] is a diagonal matrix with [tex]\( \lambda \)[/tex] as its repeated value, the eigenvalues of [tex]\( QQ^T \)[/tex] will give us [tex]\( \lambda \)[/tex].
Given the numerical solution, we have:
[tex]\[ \lambda = (8.999999999999993 + 15.588457268119885j) \][/tex]
#### Determining [tex]\( Q^{-1} \)[/tex]
[tex]\( Q \)[/tex] can be represented by the matrix of eigenvectors of [tex]\( QQ^T \)[/tex], and the inverse of [tex]\( Q \)[/tex] can be found by inverting the eigenvector matrix.
From the numerical solution, we have:
[tex]\[ Q^{-1} \approx \begin{pmatrix} -0.57735027 + 2.77555756e-16j & 0.28867513 + 0.5j & 0.28867513 - 0.5j \\ -0.57735027 - 3.05311332e-16j & 0.28867513 - 0.5j & 0.28867513 + 0.5j \\ 0.57735027 + 1.35456610e-17j & 0.57735027 - 1.85037171e-17j & 0.57735027 - 1.85037171e-17j \end{pmatrix} \][/tex]
### Part (ii): Solve the system of equations
Given the system of equations:
[tex]\[ \begin{array}{l} 12I_1 + 12I_2 - 6I_3 = -18 \\ -6I_1 + 12I_2 + 12I_3 = 0 \\ 12I_1 - 6I_2 + 12I_3 = 72 \end{array} \][/tex]
We can represent this system as the matrix equation [tex]\( A\mathbf{I} = \mathbf{B} \)[/tex], where:
[tex]\[ A = \begin{pmatrix} 12 & 12 & -6 \\ -6 & 12 & 12 \\ 12 & -6 & 12 \end{pmatrix} \][/tex]
and
[tex]\[ \mathbf{B} = \begin{pmatrix} -18 \\ 0 \\ 72 \end{pmatrix} \][/tex]
To solve for [tex]\( \mathbf{I} = \begin{pmatrix} I_1 \\ I_2 \\ I_3 \end{pmatrix} \)[/tex], we need to solve the equation [tex]\( \mathbf{I} = A^{-1}\mathbf{B} \)[/tex].
From the numerical solution, we have:
[tex]\[ \mathbf{I} \approx \begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} \][/tex]
So the solutions for the system of equations are:
[tex]\[ I_1 = 2, \quad I_2 = -2, \quad I_3 = 3 \][/tex]
### Summary
1. [tex]\( \lambda \)[/tex] is approximately [tex]\( (8.999999999999993 + 15.588457268119885j) \)[/tex].
2. [tex]\( Q^{-1} \)[/tex] is:
[tex]\[ \begin{pmatrix} -0.57735027 + 2.77555756e-16j & 0.28867513 + 0.5j & 0.28867513 - 0.5j \\ -0.57735027 - 3.05311332e-16j & 0.28867513 - 0.5j & 0.28867513 + 0.5j \\ 0.57735027 + 1.35456610e-17j & 0.57735027 - 1.85037171e-17j & 0.57735027 - 1.85037171e-17j \end{pmatrix} \][/tex]
3. The solutions for [tex]\( I_1, I_2, \)[/tex] and [tex]\( I_3 \)[/tex] are:
[tex]\[ I_1 \approx 2, \quad I_2 \approx -2, \quad I_3 \approx 3 \][/tex]
This gives the complete solution to the problem.
