Answer :
Let's go through the solutions to each question one by one.
### Question 7
We need to find the determinant of the matrix [tex]\( A = \frac{1}{2} \left( \begin{array}{ccc} \frac{k}{2} & \frac{9}{2} & 1 \\ 1 & k & 0 \\ 5 & -1 & 1 \end{array} \right) \)[/tex].
First, simplify the matrix:
[tex]\[ A = \frac{1}{2} \left( \begin{array}{ccc} \frac{k}{2} & \frac{9}{2} & 1 \\ 1 & k & 0 \\ 5 & -1 & 1 \end{array} \right) = \left( \begin{array}{ccc} \frac{k}{4} & \frac{9}{4} & \frac{1}{2} \\ \frac{1}{2} & \frac{k}{2} & 0 \\ \frac{5}{2} & -\frac{1}{2} & \frac{1}{2} \end{array} \right) \][/tex]
We can calculate the determinant using cofactor expansion along the first row.
So,
[tex]\[ |A| = \frac{k}{4} \left( \frac{k}{2} \cdot \frac{1}{2} - 0 \cdot -\frac{1}{2} \right) - \frac{9}{4} \left( \frac{1}{2} \cdot \frac{1}{2} - 0 \cdot \frac{5}{2} \right) + \frac{1}{2} \left( \frac{1}{2} \cdot -\frac{1}{2} - \frac{k}{2} \cdot \frac{5}{2} \right) \][/tex]
Simplify each of the terms for ease:
[tex]\[ \frac{k}{4} \left( \frac{k}{4} \right) - \frac{9}{4} \left( \frac{1}{4} \right) + \frac{1}{2} \left( -\frac{1}{4} - \frac{5k}{4} \right) = \frac{k^2}{16} - \frac{9}{16} + \frac{1}{2} \left( -\frac{1}{4} - \frac{5k}{4} \right) = \frac{k^2}{16} - \frac{9}{16} + \left( -\frac{1}{8} - \frac{5k}{8} \right) = \frac{k^2}{16} - \frac{9}{16} - \frac{1}{8} - \frac{5k}{8} = \frac{k^2 - 10k - 11}{16} \][/tex]
Thus, the correct answer is:
[tex]\[ (c) \ |A| = \frac{1}{16}(k^2 - 10k - 11) \][/tex]
### Question 8
Matrix [tex]\(B\)[/tex] needs to be singular (non-invertible), which means its determinant should be zero:
[tex]\[ B = \left( \begin{array}{ccc} 1 & k & 4 \\ 1 & 1 & 1 \\ -1 & -1 & 1 \end{array} \right) \][/tex]
Calculating the determinant of [tex]\( B \)[/tex]:
[tex]\[ |B|= 1\left(1 \cdot 1 - 1(-1) \right) - k\left(1 \cdot 1 - 1(-1) \right) + 4\left(1 \cdot (-1) - 1(-1)\right) = 1(2) - k(2) + 4(0) = 2 - 2k \][/tex]
Set the determinant to zero for the matrix to be singular:
[tex]\[ 2 - 2k = 0 \implies k = 1 \][/tex]
However, the options provided do not match. This implies there's a missing investigation or misinterpretion.
None of the given options matches.
### Question 9
Given [tex]\( A = \left( \begin{array}{ccc} 1 & -6 & 0 \\ -1 & 2 & -3 \\ 0 & 0 & -1 \end{array} \right) \)[/tex], we calculate the characteristic polynomial:
[tex]\[ |A - \lambda I| = \left| \begin{array}{ccc} 1 - \lambda & -6 & 0 \\ -1 & 2 - \lambda & -3 \\ 0 & 0 & -1 - \lambda \end{array} \right| \][/tex]
Expanding the determinant:
[tex]\[ (1-\lambda) \left| \begin{array}{cc} 2-\lambda & -3 \\ 0 & -1-\lambda \end{array} \right| - (-6) \left| \begin{array}{cc} -1 & -3 \\ 0 & -1-\lambda \end{array} \right| \][/tex]
[tex]\[ (1-\lambda) [(2-\lambda)(-1-\lambda)] - (-6) (-1) = (1-\lambda) (\lambda^2 - \lambda - 2) - 6 = \lambda^3 - \lambda^2 - 2\lambda - \lambda^2 + \lambda + 2 -6 =0 = \lambda^3 - 2 \lambda^2 -1 \lambda -4 = (\lambda +1, \lambda +1,\lambda -2) \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{(c) \ \lambda_1=-1, \text{ a double eigenvalue and } \lambda_2 = - 4} ] ### Question 10 The \((A-\lambda_1 I) u = 0\) Given \(\lambda_1 = -1\): \[A-(\lambda_{1}I) = \left( \begin{array}{ccc} 2 & -6 & 0 \\ -1 & 3 & -3 \\ 0 & 0 & 0 \end{array} \right)\][/tex]
Row reduce to get:
[tex]\[ u \to (\alpha (3, 1 , 0), alpha \in R)\][/tex]
The corresponding eigenvevctor is 3,1,0
\boxed{ (Final answer alpha(c) 3,1,0) none of the answer matches }]
### Question 11
Given [tex]\(A(A-\lambda_2)X=0\)[/tex] where [tex]\(\lambda_2=4\)[/tex]
\[
A-\lambda_2(X)=x0
Nullifying the third row : for \lambda2 there are
(2, -1,0)
So:
\[ \boxed{(a) -(2,1,1);ivial function [}
### Question 7
We need to find the determinant of the matrix [tex]\( A = \frac{1}{2} \left( \begin{array}{ccc} \frac{k}{2} & \frac{9}{2} & 1 \\ 1 & k & 0 \\ 5 & -1 & 1 \end{array} \right) \)[/tex].
First, simplify the matrix:
[tex]\[ A = \frac{1}{2} \left( \begin{array}{ccc} \frac{k}{2} & \frac{9}{2} & 1 \\ 1 & k & 0 \\ 5 & -1 & 1 \end{array} \right) = \left( \begin{array}{ccc} \frac{k}{4} & \frac{9}{4} & \frac{1}{2} \\ \frac{1}{2} & \frac{k}{2} & 0 \\ \frac{5}{2} & -\frac{1}{2} & \frac{1}{2} \end{array} \right) \][/tex]
We can calculate the determinant using cofactor expansion along the first row.
So,
[tex]\[ |A| = \frac{k}{4} \left( \frac{k}{2} \cdot \frac{1}{2} - 0 \cdot -\frac{1}{2} \right) - \frac{9}{4} \left( \frac{1}{2} \cdot \frac{1}{2} - 0 \cdot \frac{5}{2} \right) + \frac{1}{2} \left( \frac{1}{2} \cdot -\frac{1}{2} - \frac{k}{2} \cdot \frac{5}{2} \right) \][/tex]
Simplify each of the terms for ease:
[tex]\[ \frac{k}{4} \left( \frac{k}{4} \right) - \frac{9}{4} \left( \frac{1}{4} \right) + \frac{1}{2} \left( -\frac{1}{4} - \frac{5k}{4} \right) = \frac{k^2}{16} - \frac{9}{16} + \frac{1}{2} \left( -\frac{1}{4} - \frac{5k}{4} \right) = \frac{k^2}{16} - \frac{9}{16} + \left( -\frac{1}{8} - \frac{5k}{8} \right) = \frac{k^2}{16} - \frac{9}{16} - \frac{1}{8} - \frac{5k}{8} = \frac{k^2 - 10k - 11}{16} \][/tex]
Thus, the correct answer is:
[tex]\[ (c) \ |A| = \frac{1}{16}(k^2 - 10k - 11) \][/tex]
### Question 8
Matrix [tex]\(B\)[/tex] needs to be singular (non-invertible), which means its determinant should be zero:
[tex]\[ B = \left( \begin{array}{ccc} 1 & k & 4 \\ 1 & 1 & 1 \\ -1 & -1 & 1 \end{array} \right) \][/tex]
Calculating the determinant of [tex]\( B \)[/tex]:
[tex]\[ |B|= 1\left(1 \cdot 1 - 1(-1) \right) - k\left(1 \cdot 1 - 1(-1) \right) + 4\left(1 \cdot (-1) - 1(-1)\right) = 1(2) - k(2) + 4(0) = 2 - 2k \][/tex]
Set the determinant to zero for the matrix to be singular:
[tex]\[ 2 - 2k = 0 \implies k = 1 \][/tex]
However, the options provided do not match. This implies there's a missing investigation or misinterpretion.
None of the given options matches.
### Question 9
Given [tex]\( A = \left( \begin{array}{ccc} 1 & -6 & 0 \\ -1 & 2 & -3 \\ 0 & 0 & -1 \end{array} \right) \)[/tex], we calculate the characteristic polynomial:
[tex]\[ |A - \lambda I| = \left| \begin{array}{ccc} 1 - \lambda & -6 & 0 \\ -1 & 2 - \lambda & -3 \\ 0 & 0 & -1 - \lambda \end{array} \right| \][/tex]
Expanding the determinant:
[tex]\[ (1-\lambda) \left| \begin{array}{cc} 2-\lambda & -3 \\ 0 & -1-\lambda \end{array} \right| - (-6) \left| \begin{array}{cc} -1 & -3 \\ 0 & -1-\lambda \end{array} \right| \][/tex]
[tex]\[ (1-\lambda) [(2-\lambda)(-1-\lambda)] - (-6) (-1) = (1-\lambda) (\lambda^2 - \lambda - 2) - 6 = \lambda^3 - \lambda^2 - 2\lambda - \lambda^2 + \lambda + 2 -6 =0 = \lambda^3 - 2 \lambda^2 -1 \lambda -4 = (\lambda +1, \lambda +1,\lambda -2) \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{(c) \ \lambda_1=-1, \text{ a double eigenvalue and } \lambda_2 = - 4} ] ### Question 10 The \((A-\lambda_1 I) u = 0\) Given \(\lambda_1 = -1\): \[A-(\lambda_{1}I) = \left( \begin{array}{ccc} 2 & -6 & 0 \\ -1 & 3 & -3 \\ 0 & 0 & 0 \end{array} \right)\][/tex]
Row reduce to get:
[tex]\[ u \to (\alpha (3, 1 , 0), alpha \in R)\][/tex]
The corresponding eigenvevctor is 3,1,0
\boxed{ (Final answer alpha(c) 3,1,0) none of the answer matches }]
### Question 11
Given [tex]\(A(A-\lambda_2)X=0\)[/tex] where [tex]\(\lambda_2=4\)[/tex]
\[
A-\lambda_2(X)=x0
Nullifying the third row : for \lambda2 there are
(2, -1,0)
So:
\[ \boxed{(a) -(2,1,1);ivial function [}
