\[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} \]
Answer:
Rank A = dim(colA) = # pivot columns in A
\[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} -> R2+R1 -> \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} -> R2-R3 -> \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 6 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} ->-5R1+R4-> \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 6 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix} \]
\[ ->6R2+R4 -> \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 6 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 19 & 13 \end{bmatrix} -> R2-R3 ->\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 6 \\ 0 & 0 & 8 & 5 \\ 0 & 0 & 19 & 13 \end{bmatrix} -> R4-2R3 -> \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 6 \\ 0 & 0 & 8 & 5 \\ 0 & 0 & 3 & 3 \end{bmatrix} \]
\[ -> (1/3)R4 -> \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 6 \\ 0 & 0 & 8 & 5 \\ 0 & 0 & 1 & 1 \end{bmatrix} -> (5/8)R3 ->\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 6 \\ 0 & 0 & 1 & \frac{5}{8} \\ 0 & 0 & 1 & 1 \end{bmatrix} ->R4-R3 ->\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 6 \\ 0 & 0 & 1 & \frac{5}{8} \\ 0 & 0 & 0 & \frac{3}{8} \end{bmatrix} \]
\[ -> (8/3)R4 -> \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 6 \\ 0 & 0 & 1 & \frac{5}{8} \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
No of Pivot column = 4
Dim column = 4
Rank A = 4
library(pracma)## Warning: package 'pracma' was built under R version 4.2.3A <- matrix(c(1, 2, 3, 4, -1, 0, 1, 3, 0, 1, -2, 1, 5, 4, -2, -3), nrow = 4, byrow = TRUE)
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3rank_A <- Rank(A)
cat("Rank of Matrix A is:", rank_A, "\n")## Rank of Matrix A is: 4If m > n, maximum rank is equal to n.
If m < n, maximum rank is equal to m.
Minimum rank is 1 for non-zero matrix. This is because a non-zero matrix always has at least one linearly independent row or column.
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix} \]
Answer:
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix} -> R2-3R1 -> \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 2 & 4 & 2 \end{bmatrix} ->R3-2R2-> \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\]
No of Pivot column = 1
Dim column = 1
Rank B = 1
B <- matrix(c(1, 2, 1, 3, 6, 3, 2, 4, 2), nrow = 3, byrow = TRUE)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2rank_B <- Rank(B)
cat("Rank of Matrix B is:", rank_B, "\n")## Rank of Matrix B is: 1Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your work. You’ll need to write out the characteristic polynomial and show your solution.
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \]
Answer:
\[ det(A- \lambda I) = det\begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{bmatrix} = -(\lambda-1)(\lambda-4)(\lambda-6) \]
\[ Characteristic\ Equation = -(\lambda-1)(\lambda-4)(\lambda-6) \] \[ Characteristic\ polynomial = -\lambda^3+11\lambda^2-34\lambda+24 \]
\[ Eigenvalues:\\ \lambda=1\\\lambda=4\\\lambda=6 \]
\[ Eigenvectors: \\(A-\lambda)\bar{x} = 0 \]
\[ \lambda=6 \]
\[ \begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{bmatrix} = \begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix} \]
The reduced row echelon form of the matrix is
\[ \begin{bmatrix} 1 & 0 & -\frac{8}{5} \\ 0 & 1 & -\frac{5}{2} \\ 0 & 0 & 0 \end{bmatrix} \]
Finding the null space will give us eigenvector
\[ \begin{bmatrix} 1 & 0 & -\frac{8}{5} \\ 0 & 1 & -\frac{5}{2} \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
\[ x1 = \frac{8}{5}x3,\ x2=\frac{5}{2}x3,\ x3\ is\ free\ variable \\ Thus, Null\ Space = x3\begin{bmatrix} \frac{8}{5} \\ \frac{5}{2} \\ 1 \end{bmatrix} \\ Eigenvector\ is\ \begin{bmatrix} \frac{8}{5} \\ \frac{5}{2} \\ 1 \end{bmatrix} \]
\[ \lambda=4 \]
\[ \begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{bmatrix} = \begin{bmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix} \]
The reduced row echelon form of the matrix is
\[ \begin{bmatrix} 1 & -\frac{2}{3} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \]
Finding the null space will give us eigenvector
\[ \begin{bmatrix} 1 & -\frac{2}{3} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
\[ x1 = \frac{2}{3}x2,\ x3=0,\ x2\ is\ free\ variable \\ Thus, Null\ Space = x2\begin{bmatrix} \frac{2}{3} \\ 1 \\ 0 \end{bmatrix} \\ Eigenvector\ is\ \begin{bmatrix} \frac{2}{3} \\ 1 \\ 0 \end{bmatrix} \]
\[ \lambda=1 \]
\[ \begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{bmatrix} \]
The reduced row echelon form of the matrix is
\[ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \]
Finding the null space will give us eigenvector
\[ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
\[ x2 = 0,\ x3=0,\ x1\ is\ free\ variable \\ Thus, Null\ Space = x1\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \\ Eigenvector\ is\ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \]
C <- matrix(c(1, 2, 3, 0, 4, 5, 0, 0, 6), nrow = 3, byrow = TRUE)
C## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6result <- eigen(C)
eigenvalues <- result$values
cat("Eigenvalues:\n")## Eigenvalues:print(eigenvalues)## [1] 6 4 1eigenvectors <- result$vectors
cat("Eigenvectors:\n")## Eigenvectors:print(eigenvectors)## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0