Revision Lecture 01 Revision Lecture 01 1 / 14 Review Gauss-Jordan Elimination Row equivalence Echelon form Rank Theorem Revision Lecture 01 2 / 14 System of Linear Equations Definition A linear equation in n variables x1 , x2 , . . . , xn is an equation that can be written in the form a1 x1 + a2 x2 + · · · + an xn = b where the coefficients a1 , a2 , . . . , an and the constant term b are constants. Definition A of a linear equation a1 x1 + a2 x2 + · · · + an xn = b is a vector solution s1 s2 · whose components satisfy the equation when we substitute x1 = s1 , · sn x2 = s2 , · · · , xn = sn . Revision Lecture 01 3 / 14 System of Linear Equations Definition A system of linear equations is a finite set of linear equations, each with the same variables. A solution of a system of linear equations is a vector that is simultaneously a solution of each equation in the system. Revision Lecture 01 4 / 14 System of Linear Equations Example 2.6 Solve the system x 3x 2x −y −3y −y −z +2z +z =2 = 16 =2 Solution The augmented matrix of the system is 1 −1 −1 2 1 −1 −1 2 R −3R1 R −2R1 3 −3 0 2 16 −−2−−−→ 0 5 10 −−3−−−→ 2 −1 1 9 2 −1 1 9 1 −1 −1 2 1 −1 −1 2 R2 ↔R3 0 0 5 10 −− 1 3 9 −−→ 0 0 1 3 9 0 0 5 10 Revision Lecture 01 5 / 14 System of Linear Equations Solution The matrix is in row echelom form. Using back substitution we have x −y y −z +3z 5z =2 =5 = 10 z = 2, y = −1, x = 3. 3 Therefore x = −1 . 2 Definition(Row equivalence) Matrices A and B are row equivalent if there is a sequence of elementary row operations that converts A into B. Revision Lecture 01 6 / 14 System of Linear Equations Elementary row operations (a) Ri ←→ Rj , i ̸= j. (b) kRi , k ̸= 0. (c) Ri + kRj , i ̸= j. Examples of row equivalent matrices 2 4 2 0 1 3 0 0 A= ,B= ,D= . ,C= 0 1 0 1 0 1 0 21 Revision Lecture 01 7 / 14 System of Linear Equations Theorem 2.1 Matrices A and B are row equivalent if and only if they can be reduced to the same row echelon form. Definition A matrix is in row echelon form if it satisfies the following properties (1) Any rows consisting entirely of zeros are at the bottom. (2) In each non-zero row, the first non-zero entry (leading entry) is a column to the left of any leading entries below it. Revision Lecture 01 8 / 14 System of Linear Equations Example 1 1 2 1 1 0 1 0 0 1 3 , 0 1 5 . 0 0 0 0 0 0 4 Example 2.11 Solve the system w 2w −w −x −2x +x −y −y −y Revision +2z +3z =1 =3 = −3 Lecture 01 9 / 14 System of Linear Equations Revision Lecture 01 10 / 14 System of Linear Equations w −x −y y Revision +2z −z =1 =1 Lecture 01 11 / 14 Elementary row operations Example 2 2 2 1 2 0 Let A = ,B= ,C= . 0 1 0 12 0 12 2 0 R1 +2R2 2 2 12 R2 2 2 A= −−−−−→ −−→ =B 0 1 0 1 0 21 2 0 12 R2 2 0 R1 +2R2 2 1 −−→ A= =C −−−−−→ 0 1 0 21 0 21 This shows that elementary row operations that do not commute. Row echelon form of a matrix is not unique. Example 1 0 2 0 , . 0 3 0 1 Revision Lecture 01 12 / 14 Rank of a matrix Definition The rank of a matrix A, denoted by rank(A), is the number of non-zero rows in its row echelon form. Example 1 1 A = 0 0 0 0 rank(A) = 3 2 1 1 0 1 1 3 , B = 0 1 5 . 0 1 0 1 5 and rank(B) = 2 Theorem 2.2(The Rank Theorem) Let A be the coefficient matrix of a system of linear equations with n variables. If the system is consistent, then number of free variables = n − rank(A). Revision Lecture 01 13 / 14 Rank of a matrix Example Solve the system x −y 3y 2y +2z −3z −2z =3 = −6 =1 Solution The augmented matrix of the system is 1 −1 2 3 1 −1 2 3 0 3 −3 −6 −−−−−→ 0 3 −3 −6 R3 − 32 R2 0 2 −2 0 0 0 1 5 The system has no solutions. Why? Revision Lecture 01 14 / 14