Revision Lecture on Linear Algebra PDF

Revision Lecture on Linear Algebra Dr. Moataz Mostafa Elkhateeb Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 1 / 23 Introduction to Matrices A matrix is a rectangular array of numbers arranged in rows and columns. Matrices...

Revision Lecture on Linear Algebra Dr. Moataz Mostafa Elkhateeb Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 1 / 23 Introduction to Matrices A matrix is a rectangular array of numbers arranged in rows and columns. Matrices can represent linear transformations and systems of linear equations. Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 2 / 23 Square Matrix Definition: A matrix is called square if it has the same number of rows and columns, i.e., n × n. Example: " # 2 3 1 4 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 3 / 23 Lower Triangular Matrix Definition: A square matrix where all entries above the diagonal are zero. Example:   4 0 0 2 3 0   1 −1 5 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 4 / 23 Upper Triangular Matrix Definition: A square matrix where all entries below the diagonal are zero. Example:   3 5 2 0 4 −1   0 0 6 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 5 / 23 Diagonal Matrix Definition: A matrix with all non-diagonal elements equal to zero. Example:   5 0 0 0 8 0   0 0 3 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 6 / 23 Identity Matrix Definition: A diagonal matrix where all diagonal entries are 1. It acts as a multiplicative identity in matrix operations. Example:   1 0 0 0 1 0   0 0 1 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 7 / 23 Transpose of a Matrix Definition: The transpose of a matrix A, denoted AT , is obtained by swapping rows and columns. Example: " # " # 1 2 T1 3 A= , A = 3 4 2 4 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 8 / 23 Orthogonal Matrix Definition: A matrix Q is orthogonal if QQ T = I. Example: " # 0 −1 Q= 1 0 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 9 / 23 Symmetric and Skew-Symmetric Matrices Symmetric Matrix: A = AT. Skew-Symmetric Matrix: A = −AT. Example: " # " # 2 1 0 2 Symmetric: , Skew-Symmetric: 1 3 −2 0 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 10 / 23 Product of Matrices Definition: The product of matrices A and B is defined if the number of columns of A equals the number of rows of B. Example: " # " # " # 1 2 2 0 4 2 A= , B= , AB = 3 4 1 1 10 4 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 11 / 23 Properties of the Inverse If A and B are invertible matrices, then: (AB)−1 = B −1 A−1 (cA−1 ) = c1 A−1 (A−1 )−1 = A (AT )−1 = (A−1 )T Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 12 / 23 Finding the Inverse of a 2 × 2 Matrix " # a b For a matrix A = : c d 1 Compute the determinant: det(A) = ad − bc " # d −b 2 If det(A) ̸= 0, then A−1 = 1 det(A) −c a ! 4 7 Given the matrix A = , 2 6 det(A) = (4)(6) − (7)(2) = 24 − 14 = 10 ̸= 0 ! ! −1 1 6 −7 0.6 −0.7 A = = 10 −2 4 −0.2 0.4 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 13 / 23 Inverse by Adjoint Method To find the inverse of a matrix A using the adjoint method, follow these steps: 1 Find the cofactor matrix of A: For each element aij in matrix A, calculate the cofactor Cij. The cofactor Cij is given by (−1)i+j · Mij , where Mij is the determinant of the submatrix formed by removing the i-th row and j-th column from A. Repeat this for each element in A to form the cofactor matrix cof(A). Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 14 / 23 2 Transpose the cofactor matrix to get the adjoint: Take the transpose of the cofactor matrix cof(A), which means swapping rows and columns. The resulting matrix is called the adjoint of A, denoted as adj(A). 3 Calculate the inverse: Compute the determinant of A, det(A). If det(A) ̸= 0, the inverse of A exists and is given by: 1 A−1 = adj(A) det(A) Multiply each entry of the adjoint adj(A) by 1 det(A) to obtain A−1. Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 15 / 23 Example: Finding the Inverse Using the Adjoint Method   2 −1 0 Consider the matrix A = 1 3 −1.   4 1 2 1 Find the cofactor matrix: ! ! 3 −1 1 −1 C11 = det = 7, C12 = − det = −6 1 2 4 2 ! ! 1 3 −1 0 C13 = det = −11 C21 = − det = 2, 4 1 1 2 ! ! 2 0 2 −1 C22 = det =4 C23 = − det = −6 4 2 4 1 ! ! −1 0 2 0 C31 = det = 1, C32 = − det =2 3 −1 1 −1 ! 2 −1 C33 = det =7 1 3 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 16 / 23  The cofactor matrix is:   7 −6 −11 cof(A) = 2 4 −6    1 2 7 Transpose the cofactor matrix:   7 2 1 adj(A) =  −6 4 2   −11 −6 7 Calculate the inverse: det(A) = 2·(3·2−(−1)·1)−(−1)·(1·2−(−1)·4)+0·(1·1−3·4) = 28     7 2 1 7 2 1 1    286 28 28 A−1 =  −6 4 2 = − 28 4 28 2  28  28 11 6 7 −11 −6 7 − 28 − 28 28 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 17 / 23 Conditions for a Matrix to be in Reduced Row Echelon Form (RREF) 3 2 Pivots must be 1: The elements on the leading diagonal (pivot elements) must be equal to 1. All other elements in the pivot’s column must be zero: All elements in the same column as the pivot (except the pivot itself) must be zero. If all elements below a 0 pivot are zeros: If a pivot element is 0 and has no non-zero elements below it, then the pivot must shift to the next column to the right. Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 18 / 23 Matrices in RREF Examples ! 1 0 A= 0 1   1 0 0 B = 0 1 3   0 0 0   1 1 0 C = 0 0 1   0 0 0 ! 1 0 4 D= 0 1 −3 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 19 / 23 Steps of Finding the Inverse using Reduced Row Echelon Form (RREF) 1 Set up the augmented matrix: Write A next to the identity matrix, forming [A|I]. 2 Identify the first pivot: Start with the leftmost column. The first non-zero entry in this column is the pivot. 3 Make the pivot 1: If the pivot is not 1, divide the entire row by the pivot’s value. If the pivot is 0, exchange with a lower row that has a non-zero entry in this column (if possible). 4 Clear the pivot column: Use row operations to make all other entries in the pivot’s column 0. For each row above and below the pivot row, subtract a suitable multiple of the pivot row. Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 20 / 23 5 Move to the next pivot: Shift to the next column and find the pivot entry in the next row. If the pivot entry is 0, exchange with a lower row that has a non-zero entry. If all entries below are 0, shift the pivot to the next column. 6 Repeat until the left block is the identity matrix: Continue the above steps for each row until A is transformed into I on the left side. 7 The right block is A−1 : When [A|I] becomes [I|A−1 ], the right side is the inverse of A. Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 21 / 23 Example: Finding the Inverse of a Matrix Using RREF ! 1 −1 1 0 3 −2 0 1 ! 1 −1 1 0 −3 · R1 + R2 : 0 1 −3 1 ! 1 0 −2 1 R2 + R1 : 0 1 −3 1 ! −1 −2 1 A = −3 1 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 22 / 23 Solving Systems of Linear Equations using Inverse To solve AX = B: 1 Find A−1. 2 Multiply both sides from left by A−1 : X = A−1 B. 3 Substitute A−1 and B to find X. ( x −y =5 3x − 2y = 7 ! ! ! −2 1 5 −3 X= = −3 1 7 −8 Dr. Moataz Mostafa Elkhateeb Revision Lecture on Linear Algebra 23 / 23