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Transcript
GSE Accelerated Pre-Calculus Keeper 5 A multivariable linear system is a system of linear equation in two or more variables. The substitution and elimination methods you have previously learned can be used to convert a multivariable linear system into an equivalent system in triangular or row-echelon form. The algorithm used to transform a system of linear equations into an equivalent system in row-echelon form is called Gaussian elimination. The operations used to produce equivalent systems are given below. Write the system of equations in triangular form using Gaussian elimination. Then solve the system. 5π₯ β 5π¦ β 5π§ = 35 βπ₯ + 2π¦ β 3π§ = β12 3π₯ β 2π¦ + 7π§ = 30 The augmented matrix of a system is derived from the coefficient and constant terms of the linear equations, each written in standard form with the constant terms to the right of the equal sign. If the column of constant terms is not included, the matrix reduces to that of the coefficient matrix of the system. Write the augmented matrix for the system of linear equations. π€ + 4π₯ + π§ = 2 π₯ + 2π¦ β 3π§ = 0 π€ β 3π¦ β 8π§ = β1 3π€ + 2π₯ + 3π¦ = 9 Determine whether each matrix is in row echelon form. Solve the system using elementary row operations. 5π₯ β 5π¦ β 5π§ = 35 βπ₯ + 2π¦ β 3π§ = β12 3π₯ β 2π¦ + 7π§ = 30 If you continue to apply elementary row operations to the row-echelon form of an augmented matrix, you can obtain a matrix in which the first nonzero element of each row is 1 and the rest of the elements in the same column are 0. This is called reduced row-echelon form of the matrix. Solving a system by transforming an augmented matrix so that it is in reduce row-echelon form is called Gauss-Jordan elimination. Solve the system of equations using Gauss-Jordan Elimination. π₯βπ¦+π§ =0 βπ₯ + 2π¦ β 3π§ = β5 2π₯ β 3π¦ + 5π§ = 8 Solve the system of equations. β5π₯ β 2π¦ + π§ = 2 4π₯ β π¦ β 6π§ = 2 β3π₯ β π¦ + π§ = 1 Solve the system of equations. 3π₯ + 5π¦ β 8π§ = β3 2π₯ + 5π¦ β 2π§ = β7 βπ₯ β π¦ + 4π§ = β1 Solve the system using row-echelon form. 1. π₯ + 2π¦ β 3π§ = β28 3π₯ β π¦ + 2π§ = 3 βπ₯ + π¦ β π§ = β5 2. 3π₯ + 5π¦ + 8π§ = β20 βπ₯ + 2π¦ β 4π§ = 18 β6π₯ + 4π§ = 0 Solve using reduced row-echelon form. 3. π₯ + 2π¦ β 3π§ = 7 β3π₯ β 7π¦ + 9π§ = β12 2π₯ + π¦ β 5π§ = 8 4. 4π₯ + 9π¦ + 16π§ = 2 βπ₯ β 2π¦ β 4π§ = β1 2π₯ + 4π¦ + 9π§ = β5 6. π₯ + 3π¦ + 4π§ = 8 4π₯ β 2π¦ β π§ = 6 8π₯ β 18π¦ β 19π§ = β2 Solve using any method. 5. 3π₯ β π¦ β 5π§ = 9 4π₯ + 2π¦ β 3π§ = 6 β7π₯ β 11π¦ β 3π§ = 3