1.What is the rank of the matrix \(A\) = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \\ \end{bmatrix}\)
Ans-1: Performing row transformations to make the matrix into echelon form:
\(R_2 = R_2 + R_1,\) \(A\) = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \\ \end{bmatrix}\)
\(R_4 = R_4 - 5R_1,\) \(A\)= \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \\ \end{bmatrix}\)
\(R_2 = R_2/2,\) \(A\) = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \\ \end{bmatrix}\)
\(R_3 = R_3 - R_2,\) \(A\) = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & -6 & -17 & -23 \\ \end{bmatrix}\)
\(R_4 = R_4 + 6R_2,\) \(A\) = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & 0 & -5 & -2 \\ \end{bmatrix}\)
\(R_3 = R_3 / (-4),\) \(A\) = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & -5 & -2 \\ \end{bmatrix}\)
\(R_4 = R_4 + 5R_3,\) \(A\) = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 9/8 \\ \end{bmatrix}\)
The matrix above is in row echelon form and the number of non-zero row is 4.
Hence, the rank of the matrix A is 4.
library(matrixcalc)
A<-matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3),nrow=4)
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3matrix.rank(A)## [1] 4Ans-2: As the given matrix is non-zeron and the order is m x n , where, m > n , therefore, the maximum rank of the matrix could be n and the minimum rank of the matrix could be 1.
Ans-3: Performing row transformations to make the matrix into echelon form:
\(R_2 = R_2- 3R_1,\) \(B\) = \(\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 2 & 4 & 2\\ \end{bmatrix}\)
\(R_2= R_3 - 2R_1,\) \(B\) = \(\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{bmatrix}\)
The matrix above is in echelon form and has one non-zero row. Therefore, the rank of the matrix B is 1.
B<-matrix(c(1,3,2,2,6,4,1,3,2),nrow=3)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2matrix.rank(B)## [1] 1Ans-1: The given matrix is a 3 x 3 matrix. The eigenvalues will be the roots of the characteristic polynomial equation of the matrix A. The polynomial will be: \(f(\lambda)=det(A-\lambda I_3)\)
Eigenvalues:
Now, matrix \(A-\lambda I_3\) = \(\begin{bmatrix}1 & 2 & 3 \\ 0 & 4& 5 \\ 0 & 0 & 6\\ \end{bmatrix}\) \(-\) \(\lambda\) \(\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}\)
= \(\begin{bmatrix}1 & 2 & 3 \\ 0 & 4& 5 \\ 0 & 0 & 6\\ \end{bmatrix}\) \(-\) \(\begin{bmatrix} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda\\ \end{bmatrix}\)
= \(\begin{bmatrix} 1-\lambda & 2 & 3\\ 0 & 4-\lambda & 5\\ 0 & 0 & 6-\lambda\\ \end{bmatrix}\)
The determinant of matrix \(A-\lambda I_3\) \(=\) \((1-\lambda)(4-\lambda)(6-\lambda)\) \(-\) \(2(0-0)\) \(+\) \(3(0-0)\) \(=\) \((1-\lambda)(4-\lambda)(6-\lambda)\) \(=\) \((1-\lambda)(\lambda^2-10\lambda+24)\) \(=\) \(\lambda^2-10\lambda+24-\lambda^3+10\lambda^2-24\lambda\) \(=\) \(-\lambda^3+11\lambda^2-34\lambda+24\)
Therefore, the characteristic polynomial of the matrix A is \(-\lambda^3+11\lambda^2-34\lambda+24\)
Hence, the characteristic polynomial equation is \(-\lambda^3+11\lambda^2-34\lambda+24\) \(=\) \(0\). By solving this equation we get the eigenvalues.
Now, \(-\lambda^3+11\lambda^2-34\lambda+24\) \(=\) \(0\)
or \(\lambda^2-\lambda^3-10\lambda+10\lambda^2+24-24\lambda\) \(=\) \(0\)
or \(\lambda^2(1-\lambda)-10\lambda(1-\lambda)+24(1-\lambda)\) \(=\) \(0\)
or \((1-\lambda)(\lambda^2-10\lambda+24)\) \(=\) \(0\)
or \((1-\lambda)(4-\lambda)(6-\lambda)\) \(=\) \(0\)
So, \(\lambda\) \(=\) \(1\), \(\lambda\) \(=\) \(4\), and \(\lambda\) \(=\) \(6\)
Therefore, the eigenvalues are 1,4 and 6
Eigenvectors:
Case-1: \(\lambda = 1\),
\(\begin{bmatrix}A-\lambda I \end{bmatrix}\) \(\begin{bmatrix} X \end{bmatrix}\) \(=\) \(0\)
or \(\begin{bmatrix} 1-\lambda & 2 & 3\\ 0 & 4-\lambda & 5\\ 0 & 0 & 6-\lambda\\ \end{bmatrix}\) \(\begin{bmatrix} X \end{bmatrix}\) \(=\) \(0\)
Putting \(\lambda = 1\) we get,
\(\begin{bmatrix} 0 & 2 & 3\\ 0 & 3 & 5\\ 0 & 0 & 5\\ \end{bmatrix}\) \(\begin{bmatrix} X \end{bmatrix}\) \(=\) \(0\)
\(X_1\) is free as first pivot is 0. Let, \(X_1\) \(=\) \(1\)
\(so\), \(3X_2\) \(+\) \(5X_3\) \(=\) \(0\) and \(5X_3\) \(=\) \(0\)
Solving the equations we get, \(X_2\) \(=\) \(0\) and \(X_3\) \(=\) \(0\)
Hence, the eigenvector for \(\lambda = 1\) is \(\begin{bmatrix}X \end{bmatrix}\) \(=\) \(\begin{bmatrix} 1\\0\\0 \end{bmatrix}\)
Case-2:
\(\lambda = 4\),
\(\begin{bmatrix}A-\lambda I \end{bmatrix}\) \(\begin{bmatrix} X \end{bmatrix}\) \(=\) \(0\)
or \(\begin{bmatrix} 1-\lambda & 2 & 3\\ 0 & 4-\lambda & 5\\ 0 & 0 & 6-\lambda\\ \end{bmatrix}\) \(\begin{bmatrix} X \end{bmatrix}\) \(=\) \(0\)
Putting \(\lambda = 4\) we get,
\(\begin{bmatrix} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 2\\ \end{bmatrix}\) \(\begin{bmatrix} X \end{bmatrix}\) \(=\) \(0\)
\(X_2\) is free as second pivot is 0. Let, \(X_2\) \(=\) \(1\)
\(so\), \(-3X_1\) \(+\) \(2X_2\) \(+\) \(3X_3\) \(=\) \(0\) and \(2X_3\) \(=\) \(0\)
Solving the equations we get, \(X_1\) \(=\) \(2/3\) and \(X_3\) \(=\) \(0\)
Hence, the eigenvector for \(\lambda = 4\) is \(\begin{bmatrix}X \end{bmatrix}\) \(=\) \(\begin{bmatrix} 2/3\\1\\0 \end{bmatrix}\)
Case-3:
\(\lambda = 6\),
\(\begin{bmatrix}A-\lambda I \end{bmatrix}\) \(\begin{bmatrix} X \end{bmatrix}\) \(=\) \(0\)
or \(\begin{bmatrix} 1-\lambda & 2 & 3\\ 0 & 4-\lambda & 5\\ 0 & 0 & 6-\lambda\\ \end{bmatrix}\) \(\begin{bmatrix} X \end{bmatrix}\) \(=\) \(0\)
Putting \(\lambda = 6\) we get,
\(\begin{bmatrix} -5 & 2 & 3\\ 0 & -2 & 5\\ 0 & 0 & 0\\ \end{bmatrix}\) \(\begin{bmatrix} X \end{bmatrix}\) \(=\) \(0\)
\(X_3\) is free as third pivot is 0. Let, \(X_3\) \(=\) \(1\)
\(so\), \(-5X_1\) \(+\) \(2X_2\) \(+\) \(3X_3\) \(=\) \(0\) and \(-2X_2\) \(+\) \(5X_3\) \(=\) \(0\)
Solving the equations we get, \(X_1\) \(=\) \(8/5\) and \(X_2\) \(=\) \(5/2\)
Hence, the eigenvector for \(\lambda = 6\) is \(\begin{bmatrix}X \end{bmatrix}\) \(=\) \(\begin{bmatrix} 8/5\\5/2\\1 \end{bmatrix}\)
A<-matrix(c(1,0,0,2,4,0,3,5,6),nrow=3)
A## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6E<-eigen(A)
E$values## [1] 6 4 1E$vectors## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0