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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‘₯ βˆ’ 5𝑦 βˆ’ 5𝑧 = 35
βˆ’π‘₯ + 2𝑦 βˆ’ 3𝑧 = βˆ’12
3π‘₯ βˆ’ 2𝑦 + 7𝑧 = 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.
𝑀 + 4π‘₯ + 𝑧 = 2
π‘₯ + 2𝑦 βˆ’ 3𝑧 = 0
𝑀 βˆ’ 3𝑦 βˆ’ 8𝑧 = βˆ’1
3𝑀 + 2π‘₯ + 3𝑦 = 9
Determine whether each matrix is in row echelon form.
Solve the system using elementary row operations.
5π‘₯ βˆ’ 5𝑦 βˆ’ 5𝑧 = 35
βˆ’π‘₯ + 2𝑦 βˆ’ 3𝑧 = βˆ’12
3π‘₯ βˆ’ 2𝑦 + 7𝑧 = 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‘₯βˆ’π‘¦+𝑧 =0
βˆ’π‘₯ + 2𝑦 βˆ’ 3𝑧 = βˆ’5
2π‘₯ βˆ’ 3𝑦 + 5𝑧 = 8
Solve the system of equations.
βˆ’5π‘₯ βˆ’ 2𝑦 + 𝑧 = 2
4π‘₯ βˆ’ 𝑦 βˆ’ 6𝑧 = 2
βˆ’3π‘₯ βˆ’ 𝑦 + 𝑧 = 1
Solve the system of equations.
3π‘₯ + 5𝑦 βˆ’ 8𝑧 = βˆ’3
2π‘₯ + 5𝑦 βˆ’ 2𝑧 = βˆ’7
βˆ’π‘₯ βˆ’ 𝑦 + 4𝑧 = βˆ’1
Solve the system using row-echelon form.
1.
π‘₯ + 2𝑦 βˆ’ 3𝑧 = βˆ’28
3π‘₯ βˆ’ 𝑦 + 2𝑧 = 3
βˆ’π‘₯ + 𝑦 βˆ’ 𝑧 = βˆ’5
2.
3π‘₯ + 5𝑦 + 8𝑧 = βˆ’20
βˆ’π‘₯ + 2𝑦 βˆ’ 4𝑧 = 18
βˆ’6π‘₯ + 4𝑧 = 0
Solve using reduced row-echelon form.
3.
π‘₯ + 2𝑦 βˆ’ 3𝑧 = 7
βˆ’3π‘₯ βˆ’ 7𝑦 + 9𝑧 = βˆ’12
2π‘₯ + 𝑦 βˆ’ 5𝑧 = 8
4.
4π‘₯ + 9𝑦 + 16𝑧 = 2
βˆ’π‘₯ βˆ’ 2𝑦 βˆ’ 4𝑧 = βˆ’1
2π‘₯ + 4𝑦 + 9𝑧 = βˆ’5
6.
π‘₯ + 3𝑦 + 4𝑧 = 8
4π‘₯ βˆ’ 2𝑦 βˆ’ 𝑧 = 6
8π‘₯ βˆ’ 18𝑦 βˆ’ 19𝑧 = βˆ’2
Solve using any method.
5.
3π‘₯ βˆ’ 𝑦 βˆ’ 5𝑧 = 9
4π‘₯ + 2𝑦 βˆ’ 3𝑧 = 6
βˆ’7π‘₯ βˆ’ 11𝑦 βˆ’ 3𝑧 = 3