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Gaussian elimination method for solving systems of linear equations.

Let's consider a system of m linear equations with n unknowns of the general form.

{a11x1+a12x2+...+a1nxn=b1a21x1+a22x2+...+a2nxn=b2...............................am1x1+am2x2+...+amnxn=bm(1)

Forward elimination of the Gauss method:

Using elementary row operations and column permutations, the augmented matrix of system (1) can be transformed into the form:

A=(a11′a12′...a1r′a1,r+1′...a1n′0a22′...a2r′a2,r+1′...a2n′.....................00...arr′ar,r+1′...arn′0000000.....................00...00...0);(2).

Matrix (2) is the augmented matrix of the system.

{a11′x1+a12′x2+...+a1r′xr+a1,r+1′xr+1+...+a1n′xn=b1′a22′x2+...+a2r′xr+a2,r+1′xr+1+...+a2n′xn=b2′..............................arr′xr+ar,r+1′xr+1+...+arn′xn=br′0=br+1′0=br+2′........0=bm′(3)

which is equivalent to the original system.

Backward substitution of the Gauss method:

If at least one of the numbers br+1′,...,b′m is nonzero, then systems (3) and, consequently, the original system (1) are inconsistent. However, if b′r+1=...=b′m=0, then the system is consistent, and from equations (3), we can express the basic unknowns in terms of the free variables xr+1,...,xn.

Examples:

Using the Gauss method, investigate the consistency and find the general solution of the following systems:

1.

{x1+2x2+3x3+4x4=07x1+14x2+20x3+27x4=05x1+10x2+16x3+19x4=−23x1+5x2+6x3+13x4=5

Solution.

Let's write down the augmented matrix:

(1234|07142027|05101619|−235613|5).

By performing elementary row operations on the augmented matrix, we get:

(1234|07142027|05101619|−235613|5)∼(I×3) (36912|07142027|05101619|−235613|5)∼(IV−I) (36912|07142027|05101619|−20−1−31|5)∼(I×53) (5101520|07142027|05101619|−20−1−31|5)∼(III−I;I:5) (1234|07142027|0001−1|−20−1−31|5)∼(I×7) (7142128|07142027|0001−1|−20−1−31|5)∼(II−I;I:7)

(1234|000−1−1|0001−1|−20−1−31|5)∼(III+II;II:−1) (1234|00011|0000−2|−20−1−31|5)∼ (1234|00−1−31|50011|0000−2|−2).

The system is consistent.

Back substitution step of the Gauss method:

−2x4=−2⇒x4=1;

x3+x4=0⇒x3=−x4=−1;

−x2−3x3+x4=5⇒x2=−3x3+x4−5=3+1−5=−1

x1+2x2+3x3+4x4=0⇒x1=−2x2−3x3−4x4=2+3−4=1.

Answer:x1=1;x2=−1;x3=−1;x4=1.

2.

{x1+x2=1x1+x2+x3=4x2+x3+x4=−3x3+x4+x5=2x4+x5=−1

Solution.

Let's write down the augmented matrix:

(11000|111100|401110|−300111|200011|−1).

By performing elementary row operations on the augmented matrix, we get:

(11000|111100|401110|−300111|200011|−1)∼(II−I)

(11000|100100|301110|−300111|200011|−1)∼(IV−II−V)

(11000|100100|301110|−300000|000011|−1)∼

(11000|101110|−300100|300011|−100000|0).

The system is consistent.

Backward substitution using the Gauss method:

x5=c;

x4+x5=−1⇒x4=−1−x5=−1−c;

x3=3

x2+x3+x4=−3⇒x2=−3−x3−x4=−3−3+1+c=−5+c.

x1+x2=1⇒x1=1−x2=1+5−c=6−c

Answer: x1=6−c;x2=−5+c;x3=3;x4=1−c;x5=c.

Jordan-Gauss method for solving systems of linear equations.

Using elementary row operations and column permutations, the augmented matrix of the system (1) can be transformed into the form:

A=(10...0a1r+1′...a1n′01...0a2,r+1′...a2n′.....................00...1ar,r+1′...arn′0000000.....................00...00...0);(4).

The matrix (4) is the augmented matrix of the system

{x1+a1,r+1′xr+1+...+a1n′xn=b1′x2+a2,r+1′xr+1+...+a2n′xn=b2′..........................xr+ar,r+1′xr+1+...+arn′xn=br′0=br+1′0=br+2′........0=bm′(5) which is equivalent to the original system.

If at least one of the numbers br+1′,...,b′m is nonzero, then systems (5) and, consequently, the original system (1) are inconsistent. However, if b′r+1=...=b′m=0, then the system is consistent, and formulas (5) provide an explicit expression for the basic variables in terms of the free variables xr+1,...,xn.

Examples:

Using the Jordan-Gauss method, investigate the consistency and find the general solution of the following systems:

3.

{x1+2x2+3x3+4x4=07x1+14x2+20x3+27x4=05x1+10x2+16x3+19x4=−23x1+5x2+6x3+13x4=5

Solution.

Let's write down the augmented matrix:

(1234|07142027|05101619|−235613|5).

Performing elementary row operations on the augmented matrix, we get:

(1234|07142027|05101619|−235613|5)∼(I×3) (36912|07142027|05101619|−235613|5)∼(IV−I) (36912|07142027|05101619|−20−1−31|5)∼(I×53) (5101520|07142027|05101619|−20−1−31|5)∼(III−I;I:5) (1234|07142027|0001−1|−20−1−31|5)∼(I×7) (7142128|07142027|0001−1|−20−1−31|5)∼(II−I;I:7) (1234|000−1−1|0001−1|−20−1−31|5)∼(III+II;II:−1) (1234|00011|0000−2|−20−1−31|5)∼ (1234|00−1−31|50011|0000−2|−2)∼(I+II×2) (10−36|100−1−31|50011|0000−2|−2)∼(I+III×3;II+III×3;IV:−2) (1009|100−104|50011|00001|1)∼(I−IV×9;II−IV×4;III−IV) (1000|10−100|10010|−10001|1)∼(II:−1) (1000|10100|−10010|−10001|1).

From here, we immediately get the answer.

Answer:x1=1;x2=−1;x3=−1;x4=1.

4.

{x1+x2=1x1+x2+x3=4x2+x3+x4=−3x3+x4+x5=2x4+x5=−1

Solution.

Let's write down the augmented matrix:

(11000|111100|401110|−300111|200011|−1).

Performing elementary row operations on the augmented matrix, we obtain:

(11000|111100|401110|−300111|200011|−1)∼(II−I) (11000|100100|301110|−300111|200011|−1)∼(IV−II−V) (11000|100100|301110|−300000|000011|−1)∼(11000|101110|−300100|300011|−100000|0)∼(I−II+III;II−III−IV) (100−10|70100−1|−500100|300011|−100000|0)∼(I+IV) (10001|60100−1|−500100|300011|−100000|0). Considering x1,x2,x3,x4− ​

as basic variables and x5− as a free variable, we get

x5=c;

x4+x5=−1⇒x4=−1−x5=−1−c;

x3=3

x2−x5=−5⇒x2=−5+x5=−5+c.

x1+x5=6⇒x1=6−x5=6−c.

Answer: x1=6−c;x2=−5+c;x3=3;x4=1−c;x5=c.

Homework.

1. Using the Gauss method and the Jordan-Gauss method, investigate the compatibility and find the general solution of the following system: {x1−2x2+3x3=62x1+3x2−4x3=183x1−2x2+5x3=6

2. Using the Gauss method, investigate the compatibility and find the general solution of the following system

{105x1−175x2−315x3+245x4=8490x1−150x2−270x3+210x4=7275x1−125x2−225x3+175x4=59

Answer: The system is inconsistent.

3. Investigate the compatibility using the Jordan-Gauss method and find the general solution of the following system.

{8x1+12x2=2014x1+21x2=359x3+11x4=016x3+20x4=010x3+12x6=2215x5+18x6=33

Answer: x1=52−32c1;x2=c1;x3=0;x4=0;x5=115−65c2;x6=c2.

Tags: Gaussian method, Jordan-Gauss method, linear algebra, linear equations, matrix

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