Пример. Решить систему уравнений:

2x + y − z = 0

3y + 4z = −6

x +z =1

Решение

2

1

−1

x

0

0

3

4

−6

A =

X = y

B =

1

0

1

1

z

2

1

−1

x

0

0

3

4

−6

A =

X = y

B =

1

0

1

1

z

0

1

−1

x =

−6 3

4

= 0 3 1+1 1 4 +(−6) 0 (−1) −1 3 (−1) −(−6) 1 1−0 0 4 =13

1

0

1

2

0

−1

y =

0

−6 4

= 2 (−6) 1+0 +0 −(−1) (−6) 1−0 −2 4 1 = −26

1

1

1

2

1

0

z =

0

3

−6

= 2 3 1+1 1 (−6) +0 −0 −0 −0 = 0

1

0

1

x =

x

=

13

=1

13

y =

y

=

−26

= −2

13

z =

z

=

0

= 0

13

обратной матрицы

x − y + z = −2

2x + y + z =3

x + y +2z =1

Решение

1

−1 1

x

−2

2

1

1

3

A =

Х = y

B =

1

1

2

1

z

A = M

11

=

1

1

= 2 −1 =1

A = −M

12

= −

2

1

= −(4 −1) = −3

11

1

2

12

1

2

A = M

13

=

2

1

= 2 −1 =1

13

1

1

~

1

−3 1 T

1

3

−2

T

3

1

−2

−3 1

1

3. A = (Aij )

=

=

−2 1

3

1

−2 3

1

~

1

1

3

−2

1/ 5

3/ 5

−2 / 5

−1

4.

A

=

A =

−3 1

1

=

−3/ 5 1/ 5

1/ 5

A

5

1

−2 3

1/ 5

−2 / 5 3/ 5

1

−1 1

1/ 5

3/ 5

−2 / 5

1

0

0

A A−1

2

1

1

−3/ 5 1/ 5

1/ 5

0

1

0

=

=

1

1

2

1/ 5

−2 / 5 3/ 5

0

0

1