Apply matrix methods to determine rank, inverse, eigenvalues, eigenvectors, and solve systems of linear equations.
A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns.
Notation: \( A = [a_{ij}]_{m \times n} \) where \( i \) = row number, \( j \) = column number.
Order: \( m \times n \) (m rows, n columns)
| Type | Definition | Example |
|---|---|---|
| Row Matrix | 1 row only | \( [2 \;\; 3 \;\; 4] \) |
| Column Matrix | 1 column only | \( \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} \) |
| Square Matrix | rows = columns | \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) (2×2) |
| Diagonal Matrix | All non-diagonal elements are zero | \( \begin{bmatrix} 2 & 0 \\ 0 & 5 \end{bmatrix} \) |
| Identity Matrix (I) | Diagonal = 1, rest = 0 | \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) |
| Zero Matrix | All elements = 0 | \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \) |
| Symmetric Matrix | \( A = A^T \) | \( \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \) |
| Skew-Symmetric | \( A = -A^T \) | \( \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \) |
| Upper Triangular | All elements below diagonal = 0 | \( \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \) |
| Lower Triangular | All elements above diagonal = 0 | \( \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix} \) |
| Orthogonal | \( AA^T = A^TA = I \) | \( \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \) |
Only for same-order matrices: \( A + B = [a_{ij} + b_{ij}] \)
\( kA = [k \times a_{ij}] \)
\( A_{m \times n} \times B_{n \times p} = C_{m \times p} \)
\( C_{ij} = \sum_{k=1}^{n} a_{ik} \times b_{kj} \)
Problem: Find \( AB \) where \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), \( B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} \)
Solution:
\[ AB = \begin{bmatrix} 1(5) + 2(7) & 1(6) + 2(8) \\ 3(5) + 4(7) & 3(6) + 4(8) \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix} \]\( A^T \) is obtained by interchanging rows and columns.
| Property | Formula |
|---|---|
| Double Transpose | \( (A^T)^T = A \) |
| Addition | \( (A + B)^T = A^T + B^T \) |
| Multiplication | \( (AB)^T = B^T A^T \) |
| Scalar | \( (kA)^T = kA^T \) |
There are three elementary row operations:
| Operation | Notation | Description |
|---|---|---|
| Row Interchange | \( R_i \leftrightarrow R_j \) | Swap rows i and j |
| Row Scaling | \( R_i \to kR_i \) | Multiply row i by non-zero scalar k |
| Row Addition | \( R_i \to R_i + kR_j \) | Add k times row j to row i |
Similarly, three elementary column operations exist:
| Operation | Notation |
|---|---|
| Column Interchange | \( C_i \leftrightarrow C_j \) |
| Column Scaling | \( C_i \to kC_i \) |
| Column Addition | \( C_i \to C_i + kC_j \) |
A matrix is in Row Echelon Form if:
A matrix is in RREF if it satisfies all REF conditions and additionally:
Problem: Reduce to REF using elementary row operations:
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix} \]Solution:
Step 1: \( R_2 \to R_2 - 2R_1 \)
\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \]Step 2: \( R_3 \to R_3 - R_1 \)
\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -2 \end{bmatrix} \]Step 3: \( R_2 \leftrightarrow R_3 \)
\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{bmatrix} \]Step 4: \( R_2 \to -R_2 \)
\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \]This is in Row Echelon Form.
The rank of a matrix \( A \), denoted as \( \rho(A) \) or \( r(A) \), is the maximum number of linearly independent rows (or columns) in the matrix.
Equivalently, it is the order of the largest non-zero minor of the matrix.
| Property | Statement |
|---|---|
| Rank of zero matrix | \( 0 \) |
| Rank of identity matrix \( I_n \) | \( n \) |
| Rank bound | \( \rho(A) \leq \min(m, n) \) for \( A_{m \times n} \) |
| Rank of transpose | \( \rho(A) = \rho(A^T) \) |
| Rank of product | \( \rho(AB) \leq \min(\rho(A), \rho(B)) \) |
| Rank of sum | \( \rho(A + B) \leq \rho(A) + \rho(B) \) |
| Non-singular matrix | \( \rho(A) = n \) if \( |A| \neq 0 \) |
| Singular matrix | \( \rho(A) < n \) if \( |A| = 0 \) |
Problem: Find the rank of \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix} \)
Solution:
Step 1: \( R_2 \to R_2 - 2R_1 \)
\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 3 & 6 & 9 \end{bmatrix} \]Step 2: \( R_3 \to R_3 - 3R_1 \)
\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]Number of non-zero rows = 1
Therefore, \( \rho(A) = 1 \)
Problem: Find the rank of \( A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \)
Solution:
Step 1: Check 3×3 determinant:
\[ |A| = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) = -3 + 12 - 9 = 0 \]Step 2: Check 2×2 minors:
\[ \begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} = 5 - 8 = -3 \neq 0 \]Therefore, \( \rho(A) = 2 \)
Problem: Find the rank of \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 2 & k \end{bmatrix} \) for different values of \( k \)
Solution:
Step 1: \( R_2 \to R_2 - 2R_1 \): \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 2 & k \end{bmatrix} \)
Step 2: \( R_3 \to R_3 - R_1 \): \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & k-3 \end{bmatrix} \)
Case 1: If \( k \neq 3 \), then \( \rho(A) = 2 \)
Case 2: If \( k = 3 \), then \( \rho(A) = 1 \)
A set of vectors \( v_1, v_2, \ldots, v_n \) is linearly dependent if there exist scalars \( c_1, c_2, \ldots, c_n \) (not all zero) such that:
If the only solution is \( c_1 = c_2 = \cdots = c_n = 0 \), then the vectors are linearly independent.
Problem: Check if \( v_1 = [1, 2, 3] \), \( v_2 = [2, 4, 6] \), \( v_3 = [1, 1, 1] \) are linearly dependent or independent.
Solution:
Step 1: Form matrix \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix} \)
Step 2: \( R_2 \to R_2 - 2R_1 \): \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \)
Step 3: \( \rho(A) = 2 \) (two non-zero rows)
Number of vectors = 3
Since \( \rho(A) < 3 \) → Vectors are linearly dependent.
Problem: Check if \( v_1 = [1, 0, 0] \), \( v_2 = [0, 1, 0] \), \( v_3 = [0, 0, 1] \) are linearly independent.
Solution:
\[ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad |A| = 1 \neq 0 \]Since \( |A| \neq 0 \) → Vectors are linearly independent.
Problem: Show that \( v_1 = [1, 1, 1] \), \( v_2 = [1, 2, 3] \), \( v_3 = [1, 3, 5] \) are linearly dependent and find the relation.
Solution:
Solve \( c_1 v_1 + c_2 v_2 + c_3 v_3 = 0 \):
\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]Using row operations:
\[ R_2 \to R_2 - R_1, \quad R_3 \to R_3 - R_1 \] \[ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & 4 \end{bmatrix} \to \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \]From \( R_2 \): \( c_2 + 2c_3 = 0 \Rightarrow c_2 = -2c_3 \)
From \( R_1 \): \( c_1 + c_2 + c_3 = 0 \Rightarrow c_1 - 2c_3 + c_3 = 0 \Rightarrow c_1 = c_3 \)
Taking \( c_3 = 1 \): \( c_1 = 1, c_2 = -2, c_3 = 1 \)
Relation: \( v_1 - 2v_2 + v_3 = 0 \) → linearly dependent.
A system of \( m \) linear equations in \( n \) unknowns:
Matrix form: \( AX = B \) where:
| Type | Condition | Solution |
|---|---|---|
| Consistent | Has at least one solution | Unique or Infinite |
| Inconsistent | No solution | No solution exists |
where \( A_i \) is \( A \) with column \( i \) replaced by \( B \).
| Condition | Result |
|---|---|
| \( \rho(A) = \rho([A|B]) = n \) | Unique solution |
| \( \rho(A) = \rho([A|B]) < n \) | Infinite solutions |
| \( \rho(A) < \rho([A|B]) \) | No solution (inconsistent) |
Problem: Solve \( x + y = 3 \), \( 2x + 3y = 7 \)
Solution:
\( A = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y \end{bmatrix} \), \( B = \begin{bmatrix} 3 \\ 7 \end{bmatrix} \)
\( |A| = 3 - 2 = 1 \neq 0 \) (invertible)
\[ A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -1 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ -2 & 1 \end{bmatrix} \] \[ X = A^{-1}B = \begin{bmatrix} 3 & -1 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 7 \end{bmatrix} = \begin{bmatrix} 9 - 7 \\ -6 + 7 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]Solution: \( x = 2, y = 1 \)
Problem: Solve \( x + 2y + z = 8 \), \( 2x + y + z = 7 \), \( x + y + 2z = 7 \)
Solution:
\[ |A| = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{vmatrix} = 1(2-1) - 2(4-1) + 1(2-1) = 1 - 6 + 1 = -4 \] \[ |A_1| = \begin{vmatrix} 8 & 2 & 1 \\ 7 & 1 & 1 \\ 7 & 1 & 2 \end{vmatrix} = 8(2-1) - 2(14-7) + 1(7-7) = 8 - 14 + 0 = -6 \] \[ |A_2| = \begin{vmatrix} 1 & 8 & 1 \\ 2 & 7 & 1 \\ 1 & 7 & 2 \end{vmatrix} = 1(14-7) - 8(4-1) + 1(14-7) = 7 - 24 + 7 = -10 \] \[ |A_3| = \begin{vmatrix} 1 & 2 & 8 \\ 2 & 1 & 7 \\ 1 & 1 & 7 \end{vmatrix} = 1(7-7) - 2(14-7) + 8(2-1) = 0 - 14 + 8 = -6 \] \[ x = \frac{|A_1|}{|A|} = \frac{-6}{-4} = \frac{3}{2}, \quad y = \frac{|A_2|}{|A|} = \frac{-10}{-4} = \frac{5}{2}, \quad z = \frac{|A_3|}{|A|} = \frac{-6}{-4} = \frac{3}{2} \]Solution: \( x = \frac{3}{2}, y = \frac{5}{2}, z = \frac{3}{2} \)
Problem: Solve \( x + y + z = 6 \), \( 2x + 2y + 2z = 12 \), \( 3x + 3y + 3z = 18 \)
Solution:
Augmented matrix:
\[ [A|B] = \begin{bmatrix} 1 & 1 & 1 & 6 \\ 2 & 2 & 2 & 12 \\ 3 & 3 & 3 & 18 \end{bmatrix} \]\( R_2 \to R_2 - 2R_1 \), \( R_3 \to R_3 - 3R_1 \):
\[ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]\( \rho(A) = 1 \), \( \rho([A|B]) = 1 \), \( n = 3 \)
Since \( \rho(A) = \rho([A|B]) < n \) → Infinite solutions
Let \( y = s, z = t \), then \( x = 6 - s - t \)
Problem: Check consistency of \( x + y = 2 \), \( 2x + 2y = 5 \)
Solution:
\[ [A|B] = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \]\( R_2 \to R_2 - 2R_1 \):
\[ \begin{bmatrix} 1 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} \]\( \rho(A) = 1 \), \( \rho([A|B]) = 2 \)
Since \( \rho(A) < \rho([A|B]) \) → No solution (inconsistent)
For a square matrix \( A \), if there exists a matrix \( A^{-1} \) such that:
then \( A^{-1} \) is the inverse of \( A \).
where \( \text{adj}(A) \) = transpose of cofactor matrix.
Steps:
| Property | Formula |
|---|---|
| Inverse of product | \( (AB)^{-1} = B^{-1}A^{-1} \) |
| Double inverse | \( (A^{-1})^{-1} = A \) |
| Transpose inverse | \( (A^T)^{-1} = (A^{-1})^T \) |
| Scalar inverse | \( (kA)^{-1} = \frac{1}{k}A^{-1} \) (k≠0) |
| Determinant | \( |A^{-1}| = \frac{1}{|A|} \) |
Problem: Find \( A^{-1} \) for \( A = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \)
Solution:
Step 1: \( |A| = (2)(3) - (1)(1) = 6 - 1 = 5 \neq 0 \)
Step 2: Cofactors:
\( C_{11} = 3, \quad C_{12} = -1, \quad C_{21} = -1, \quad C_{22} = 2 \)
Step 3: Cofactor matrix = \( \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix} \)
Step 4: \( \text{adj}(A) = \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}^T = \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix} \)
Step 5:
\[ A^{-1} = \frac{1}{5} \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3/5 & -1/5 \\ -1/5 & 2/5 \end{bmatrix} \]Problem: Find \( A^{-1} \) for \( A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix} \)
Solution:
Step 1: \( |A| = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1 \neq 0 \)
Step 2: Cofactors:
\[ C_{11} = \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} = -24, \quad C_{12} = -\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} = 20, \quad C_{13} = \begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix} = -5 \] \[ C_{21} = -\begin{vmatrix} 2 & 3 \\ 6 & 0 \end{vmatrix} = 18, \quad C_{22} = \begin{vmatrix} 1 & 3 \\ 5 & 0 \end{vmatrix} = -15, \quad C_{23} = -\begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix} = 4 \] \[ C_{31} = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 5, \quad C_{32} = -\begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -4, \quad C_{33} = \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \]Step 3: \( \text{adj}(A) = \begin{bmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{bmatrix}^T = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} \)
Step 4:
\[ A^{-1} = \frac{1}{1} \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} \]For a square matrix \( A \), if there exists a scalar \( \lambda \) (eigenvalue) and a non-zero vector \( X \) (eigenvector) such that:
then \( \lambda \) is called an eigenvalue of \( A \) and \( X \) is called the corresponding eigenvector.
This polynomial equation gives all eigenvalues.
Problem: Find eigenvalues and eigenvectors of \( A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \)
Solution:
Step 1: Characteristic equation:
\[ |A - \lambda I| = \begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} = 0 \] \[ (2-\lambda)^2 - 1 = 0 \implies (2-\lambda)^2 = 1 \implies 2-\lambda = \pm 1 \]Step 2: \( \lambda_1 = 1, \lambda_2 = 3 \)
Step 3: For \( \lambda_1 = 1 \):
\[ (A - I)X = 0 \implies \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]\( x + y = 0 \Rightarrow x = -y \)
Eigenvector: \( X_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \)
Step 4: For \( \lambda_2 = 3 \):
\[ (A - 3I)X = 0 \implies \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]\( -x + y = 0 \Rightarrow x = y \)
Eigenvector: \( X_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)
Problem: Find eigenvalues of \( A = \begin{bmatrix} 4 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{bmatrix} \)
Solution:
Since \( A \) is upper triangular, eigenvalues are the diagonal elements:
Problem: Find eigenvalues and eigenvectors of \( A = \begin{bmatrix} 4 & 1 \\ 1 & 4 \end{bmatrix} \)
Solution:
\[ |A - \lambda I| = \begin{vmatrix} 4-\lambda & 1 \\ 1 & 4-\lambda \end{vmatrix} = (4-\lambda)^2 - 1 = 0 \] \[ 4 - \lambda = \pm 1 \implies \lambda_1 = 3, \lambda_2 = 5 \]For \( \lambda_1 = 3 \):
\[ \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies X_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \]For \( \lambda_2 = 5 \):
\[ \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies X_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]| Property | Statement |
|---|---|
| Sum of eigenvalues | \( \lambda_1 + \lambda_2 + \cdots + \lambda_n = \text{Trace}(A) = \sum a_{ii} \) |
| Product of eigenvalues | \( \lambda_1 \lambda_2 \cdots \lambda_n = |A| \) |
| Eigenvalues of \( A^{-1} \) | \( \frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \ldots, \frac{1}{\lambda_n} \) |
| Eigenvalues of \( A^T \) | Same as \( A \) |
| Eigenvalues of \( kA \) | \( k\lambda_1, k\lambda_2, \ldots, k\lambda_n \) |
| Eigenvalues of \( A^n \) | \( \lambda_1^n, \lambda_2^n, \ldots, \lambda_n^n \) |
| Eigenvalues of \( A - kI \) | \( \lambda_1 - k, \lambda_2 - k, \ldots, \lambda_n - k \) |
| Eigenvalues of \( A + kI \) | \( \lambda_1 + k, \lambda_2 + k, \ldots, \lambda_n + k \) |
| Matrix Type | Eigenvalues |
|---|---|
| Diagonal matrix | Diagonal elements |
| Upper/Lower triangular | Diagonal elements |
| Identity matrix \( I \) | All = 1 |
| Zero matrix | All = 0 |
| Idempotent (\( A^2 = A \)) | 0 or 1 |
| Nilpotent (\( A^n = 0 \)) | All = 0 |
| Involutory (\( A^2 = I \)) | 1 or -1 |
| Symmetric matrix | All eigenvalues are real |
| Skew-symmetric | Pure imaginary or zero |
| Orthogonal matrix | \( |\lambda| = 1 \) |
If \( B = P^{-1}AP \) (for some invertible \( P \)), then \( A \) and \( B \) have the same eigenvalues.
Problem: If \( A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \) has eigenvalues \( \lambda_1 = 1, \lambda_2 = 3 \), find eigenvalues of:
(a) \( A^{-1} \) (b) \( A^3 \) (c) \( 2A \) (d) \( A + 2I \)
Solution:
Every square matrix \( A \) satisfies its own characteristic equation.
If the characteristic equation is:
Then:
From Cayley-Hamilton:
\[ A^n + c_1A^{n-1} + \cdots + c_nI = 0 \]Multiply by \( A^{-1} \):
\[ A^{n-1} + c_1A^{n-2} + \cdots + c_nA^{-1} = 0 \]Use the characteristic equation to reduce higher powers to lower powers.
Problem: Using Cayley-Hamilton theorem, find \( A^{-1} \) for \( A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \)
Solution:
Step 1: Characteristic equation:
\[ |A - \lambda I| = \begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0 \]Step 2: By Cayley-Hamilton:
\[ A^2 - 4A + 3I = 0 \]Step 3: Multiply by \( A^{-1} \):
\[ A - 4I + 3A^{-1} = 0 \implies 3A^{-1} = 4I - A \] \[ A^{-1} = \frac{1}{3}(4I - A) = \frac{1}{3}\left(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\right) \] \[ A^{-1} = \frac{1}{3}\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2/3 & -1/3 \\ -1/3 & 2/3 \end{bmatrix} \]Verify: \( AA^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 2/3 & -1/3 \\ -1/3 & 2/3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) ✓
Problem: For \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), use Cayley-Hamilton to find \( A^4 \).
Solution:
Step 1: Characteristic equation:
\[ |A - \lambda I| = (1-\lambda)(4-\lambda) - 6 = \lambda^2 - 5\lambda - 2 = 0 \]Step 2: By Cayley-Hamilton: \( A^2 - 5A - 2I = 0 \implies A^2 = 5A + 2I \)
Step 3: \( A^3 = A \cdot A^2 = A(5A + 2I) = 5A^2 + 2A = 5(5A + 2I) + 2A = 27A + 10I \)
Step 4: \( A^4 = A \cdot A^3 = A(27A + 10I) = 27A^2 + 10A = 27(5A + 2I) + 10A = 145A + 54I \)
\[ A^4 = 145\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + 54\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 145 & 290 \\ 435 & 580 \end{bmatrix} + \begin{bmatrix} 54 & 0 \\ 0 & 54 \end{bmatrix} = \begin{bmatrix} 199 & 290 \\ 435 & 634 \end{bmatrix} \]Problem: Verify Cayley-Hamilton for \( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \)
Solution:
Characteristic equation: \( (1-\lambda)(2-\lambda)(3-\lambda) = 0 \)
\[ \lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0 \]By Cayley-Hamilton: \( A^3 - 6A^2 + 11A - 6I = 0 \)
Compute:
\[ A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 9 \end{bmatrix}, \quad A^3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 27 \end{bmatrix} \] \[ A^3 - 6A^2 + 11A - 6I = \begin{bmatrix} 1-6+11-6 & 0 & 0 \\ 0 & 8-24+22-6 & 0 \\ 0 & 0 & 27-54+33-6 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]Verified ✓
| Topic | Key Formula / Concept |
|---|---|
| Matrix Multiplication | \( C_{ij} = \sum a_{ik}b_{kj} \) |
| Transpose Properties | \( (AB)^T = B^TA^T \), \( (A^T)^T = A \) |
| Rank (Echelon Form) | Number of non-zero rows |
| Rank (Minors) | Order of largest non-zero minor |
| Linear Independence | \( \rho(A) = \) number of vectors |
| Linear Dependence | \( \rho(A) < \) number of vectors |
| System Consistency | \( \rho(A) = \rho([A|B]) \) |
| Unique Solution | \( \rho(A) = \rho([A|B]) = n \) |
| Infinite Solutions | \( \rho(A) = \rho([A|B]) < n \) |
| No Solution | \( \rho(A) < \rho([A|B]) \) |
| Cramer's Rule | \( x_i = \dfrac{|A_i|}{|A|} \) |
| Inverse (Adjoint) | \( A^{-1} = \dfrac{\text{adj}(A)}{|A|} \) |
| Inverse Properties | \( (AB)^{-1} = B^{-1}A^{-1} \) |
| Characteristic Equation | \( |A - \lambda I| = 0 \) |
| Eigenvalue Sum | \( \sum \lambda_i = \text{Trace}(A) \) |
| Eigenvalue Product | \( \prod \lambda_i = |A| \) |
| Eigenvalues of \( A^{-1} \) | \( 1/\lambda_i \) |
| Eigenvalues of \( A^n \) | \( \lambda_i^n \) |
| Cayley-Hamilton | \( p(A) = 0 \) (zero matrix) |
| \( A^{-1} \) (Cayley-Hamilton) | \( A^{-1} = -\dfrac{1}{c_n}(A^{n-1} + c_1A^{n-2} + \cdots + c_{n-1}I) \) |
| Matrix | Property | Eigenvalues |
|---|---|---|
| Diagonal | Non-diagonal = 0 | Diagonal elements |
| Triangular | Above or below diagonal = 0 | Diagonal elements |
| Symmetric | \( A = A^T \) | All real |
| Skew-Symmetric | \( A = -A^T \) | Pure imaginary or zero |
| Orthogonal | \( AA^T = I \) | \( \|\lambda\| = 1 \) |
| Idempotent | \( A^2 = A \) | 0 or 1 |
| Nilpotent | \( A^n = 0 \) | All 0 |
| Involutory | \( A^2 = I \) | 1 or -1 |
Find the rank of \( A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \end{bmatrix} \)
Solve using Gauss-Jordan method: \( x + y + z = 6 \), \( 2x + 3y + z = 10 \), \( x + 2y + 3z = 12 \)
Find eigenvalues and eigenvectors of \( A = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix} \)
Using Cayley-Hamilton theorem, find \( A^{-1} \) for \( A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} \)
Check if vectors \( v_1 = [1, 2, 3] \), \( v_2 = [2, 4, 6] \), \( v_3 = [1, 3, 5] \) are linearly dependent or independent.
Find \( A^{-1} \) for \( A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix} \) using the adjoint method.
If \( A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \) has eigenvalues \( \lambda_1, \lambda_2 \), find \( \lambda_1 + \lambda_2 \) and \( \lambda_1\lambda_2 \) without finding eigenvalues explicitly.
Determine the consistency of the system: \( x + 2y - z = 3 \), \( 2x - y + z = 2 \), \( 3x + y = 5 \).
\( A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \end{bmatrix} \)
\( R_2 \to R_2 - 2R_1 \), \( R_3 \to R_3 - 3R_1 \):
\[ \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]Number of non-zero rows = 1
\( \rho(A) = 1 \)
Augmented matrix:
\[ [A|B] = \begin{bmatrix} 1 & 1 & 1 & 6 \\ 2 & 3 & 1 & 10 \\ 1 & 2 & 3 & 12 \end{bmatrix} \]\( R_2 \to R_2 - 2R_1 \), \( R_3 \to R_3 - R_1 \):
\[ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & 1 & -1 & -2 \\ 0 & 1 & 2 & 6 \end{bmatrix} \]\( R_3 \to R_3 - R_2 \):
\[ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 3 & 8 \end{bmatrix} \]Back-substitute:
\( 3z = 8 \Rightarrow z = \frac{8}{3} \)
\( y - z = -2 \Rightarrow y = -2 + \frac{8}{3} = \frac{2}{3} \)
\( x + y + z = 6 \Rightarrow x = 6 - \frac{2}{3} - \frac{8}{3} = 6 - \frac{10}{3} = \frac{8}{3} \)
Solution: \( x = \frac{8}{3}, y = \frac{2}{3}, z = \frac{8}{3} \)
\( A = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix} \)
\[ |A - \lambda I| = \begin{vmatrix} 3-\lambda & 1 \\ 1 & 3-\lambda \end{vmatrix} = (3-\lambda)^2 - 1 = 0 \] \[ 3-\lambda = \pm 1 \implies \lambda_1 = 2, \lambda_2 = 4 \]For \( \lambda_1 = 2 \): \( \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = 0 \Rightarrow X_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \)
For \( \lambda_2 = 4 \): \( \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = 0 \Rightarrow X_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)
\( A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} \)
Characteristic equation: \( (2-\lambda)^3 = 0 \Rightarrow \lambda^3 - 6\lambda^2 + 12\lambda - 8 = 0 \)
By Cayley-Hamilton: \( A^3 - 6A^2 + 12A - 8I = 0 \)
Multiply by \( A^{-1} \): \( A^2 - 6A + 12I - 8A^{-1} = 0 \)
\[ A^{-1} = \frac{1}{8}(A^2 - 6A + 12I) \]\( A^2 = \begin{bmatrix} 4 & 4 & 1 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end{bmatrix} \)
\[ A^{-1} = \frac{1}{8}\left(\begin{bmatrix} 4 & 4 & 1 \\ 0 & 4 & 4 \\ 0 & 0 & 4 \end{bmatrix} - \begin{bmatrix} 12 & 6 & 0 \\ 0 & 12 & 6 \\ 0 & 0 & 12 \end{bmatrix} + \begin{bmatrix} 12 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 12 \end{bmatrix}\right) \] \[ A^{-1} = \frac{1}{8}\begin{bmatrix} 4 & -2 & 1 \\ 0 & 4 & -2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 1/2 & -1/4 & 1/8 \\ 0 & 1/2 & -1/4 \\ 0 & 0 & 1/2 \end{bmatrix} \]\( v_1 = [1, 2, 3], v_2 = [2, 4, 6], v_3 = [1, 3, 5] \)
Form matrix \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 3 & 5 \end{bmatrix} \)
\( R_2 \to R_2 - 2R_1 \): \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 1 & 3 & 5 \end{bmatrix} \)
\( R_3 \to R_3 - R_1 \): \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 1 & 2 \end{bmatrix} \)
Rank = 2 (two non-zero rows)
Number of vectors = 3
Since \( \rho(A) < 3 \) → Linearly dependent
\( A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix} \)
\( |A| = 1(0-24) - 2(0-20) + 3(0-5) = -24 + 40 - 15 = 1 \neq 0 \)
Cofactors:
\[ C_{11} = -24, \quad C_{12} = 20, \quad C_{13} = -5 \] \[ C_{21} = 18, \quad C_{22} = -15, \quad C_{23} = 4 \] \[ C_{31} = 5, \quad C_{32} = -4, \quad C_{33} = 1 \]\( \text{adj}(A) = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} \)
\[ A^{-1} = \frac{1}{1}\begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} = \begin{bmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{bmatrix} \]\( A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \)
Sum of eigenvalues = Trace(A) = \( 4 + 3 = 7 \)
Product of eigenvalues = \( |A| = 4(3) - 2(1) = 12 - 2 = 10 \)
System: \( x + 2y - z = 3 \), \( 2x - y + z = 2 \), \( 3x + y = 5 \)
Augmented matrix:
\[ [A|B] = \begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & -1 & 1 & 2 \\ 3 & 1 & 0 & 5 \end{bmatrix} \]\( R_2 \to R_2 - 2R_1 \), \( R_3 \to R_3 - 3R_1 \):
\[ \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & -5 & 3 & -4 \\ 0 & -5 & 3 & -4 \end{bmatrix} \]\( R_3 \to R_3 - R_2 \):
\[ \begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & -5 & 3 & -4 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]\( \rho(A) = 2 \), \( \rho([A|B]) = 2 \), \( n = 3 \)
Since \( \rho(A) = \rho([A|B]) < n \) → Infinite solutions (consistent)
| Reference | Book |
|---|---|
| T-1 | Advanced Engineering Mathematics – R.K. Jain, S.R.K. Iyengar (Narosa Publishing House) |
| R-1 | Higher Engineering Mathematics – B.S. Grewal (Khanna Publishers) |
| R-2 | Mathematics Textbook for Class XII Part I – NCERT |
| R-3 | Mathematics Textbook for Class XII Part II – NCERT |
| OR-1 | https://ncert.nic.in/textbook.php?lemh1=5-6 |
| OR-2 | https://ncert.nic.in/textbook.php?lemh2=1-7 |
After studying this unit, students should be able to:
| CO1 Components | Covered in Section |
|---|---|
| Determine rank of a matrix | III |
| Find inverse of a matrix | VI |
| Find eigenvalues and eigenvectors | VII, VIII |
| Solve systems of linear equations | V |
| Apply Cayley-Hamilton theorem | IX |
| Check linear dependence/independence | IV |
⭐ End of Unit I Notes ⭐
Matrix Methods and Linear Systems
MTH165 – Mathematics for Engineers
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