# Find inverse matrix using determinants and adjoints

• ducmod

#### ducmod

Hello!

Please, help me to see my mistake - for quite a while I can't solve a very easy matrix.

I have to
find the inverse of the given matrix using their determinants and adjoints.
4 6 -3
3 4 -3
1 2 6

to find adjoint matrix I need to find cofactors 11, 12, etc till 33.
Cofactor11 = (-1)^(1+1) x det([4 -3] => C11 = 24 + 6 = 30
2 6
Cofactor12 = (-1)^(1+2) x det([3 -3] => C12 = -(18 + 3) = -21
1 6

Cofactor13 = (-1)^(1+3) x det([3 4] => C13 = 6 - 4 = 2
1 2

Cofactor21 = (-1)^(2+1) x det([6 -3] => C21 = -(36 + 6) = -42
2 6

Cofactor22 = (-1)^(2+2) x det([4 -3] => C22 = 24 + 3 = 27
1 6
Cofactor23 = (-1)^(2+3) x det([4 6] => C23 = -(8 - 6) = -2
1 2

Cofactor31 = (-1)^(3+1) x det([6 -3] => C31 = -6
4 -3

Cofactor32 = (-1)^(3+2) x det([4 -3] => C32 = -(-12 + 9) = 3
3 -3
Cofactor33 = (-1)^(3+3) x det([4 6] => C33 = 16 - 18 = -2
3 4

30 -42 -6
-21 27 3
2 -2 -2

det(of initial matrix taken by the first row) = 4 x (-1)^(1+1) x det([A11)] + 6 x (-1)^(1+2) x det([A12)] + (-3) x (-1)^(1+3) x det([A13)] = 4 x 30 + 6 x (-21) + (-3) x (-2) = 0

and if I try to find the det of intial matrix by expanding over the third row I get 1 x (-6) + 2 x (-1) x (-3) + 6 x (-2) = -2

and if I try to find the det of intial matrix by expanding over the second row I get (-3) x 42 + 4 x 27 + 3 x 2 = -12

I have tried multiple times, with different rows for initial matrix, and each time I get a different result.
Am I doing something wrong with cofactors?

Thank you!

Last edited:
• ducmod
Hello!

Please, help me to see my mistake - for quite a while I can't solve a very easy matrix.

I have to
find the inverse of the given matrix using their determinants and adjoints.
4 6 -3
3 4 -3
1 2 6

to find adjoint matrix I need to find cofactors 11, 12, etc till 33.
Cofactor11 = (-1)^(1+1) x det([4 -3] => C11 = 24 + 6 = 30
2 6
Cofactor12 = (-1)^(1+2) x det([3 -3] => C12 = -(18 + 3) = -21
1 6

Cofactor13 = (-1)^(1+3) x det([3 4] => C13 = 6 - 4 = 2
1 2

Cofactor21 = (-1)^(2+1) x det([6 -3] => C21 = -(36 + 6) = -42
2 6

Cofactor22 = (-1)^(2+2) x det([4 -3] => C22 = 24 + 3 = 27
1 6
Cofactor23 = (-1)^(2+3) x det([4 6] => C23 = -(8 - 6) = -2
1 2

Cofactor31 = (-1)^(3+1) x det([6 -3] => C31 = -6
4 -3

Cofactor32 = (-1)^(3+2) x det([4 -3] => C32 = -(-12 + 9) = 3
3 -3
Cofactor33 = (-1)^(3+3) x det([4 6] => C33 = 16 - 18 = -2
3 4

30 -42 -6
-21 27 3
2 -2 -2

det(of initial matrix taken by the first row) = 4 x (-1)^(1+1) x det([A11)] + 6 x (-1)^(1+2) x det([A12)] + (-3) x (-1)^(1+3) x det([A13)] = 4 x 30 + 6 x (-21) + (-3) x (-2) = 0

I have tried multiple times, with different rows for initial matrix, and each time I get a different result.
Am I doing something wrong with cofactors?

Thank you!
Your calculation of the det(A) = 0 is incorrect. If A has a zero determinant, can it have an inverse?

• ducmod
The last sign you typed in the determinant calculation is wrong.

• ducmod
Consider that you are in fact in front of three systems with each three linear combinations depending on three unknown scalars; for example, the system corresponding to the first column is:

4 . x11 + 6 . x12 - 3 . x13 = 1

3 . x11 + 4 . x12 - 3 . x13 = 0

1 . x11 + 2 . x12 + 6 . x13 = 0

First verify that the discriminant of the system doesn’t vanish; that is let calculate:

+ 4 . [4 . 6 – 2 . (-3)] – (+6) . [3 . 6 – 1 . (-3)] + (-3) . [3 . 2 – 1 . 4]

=

4 . (24 + 6) – 6 . (18 + 3) + 3 . (6 – 4)

=

4 . 30 – 6 . 21 + 3 . 2

=

120 – 126 + 6

=

0

Unfortunately, yes that discriminant vanishes. The matrix cannot be inverted. The most important is to know how a discriminant of a system /determinant of a matrix must be calculated.

Consider that you are in fact in front of three systems with each three linear combinations depending on three unknown scalars; for example, the system corresponding to the first column is:

4 . x11 + 6 . x12 - 3 . x13 = 1

3 . x11 + 4 . x12 - 3 . x13 = 0

1 . x11 + 2 . x12 + 6 . x13 = 0

First verify that the discriminant of the system doesn’t vanish; that is let calculate:

+ 4 . [4 . 6 – 2 . (-3)] – (+6) . [3 . 6 – 1 . (-3)] + (-3) . [3 . 2 – 1 . 4]

=

4 . (24 + 6) – 6 . (18 + 3) + 3 . (6 – 4)
You flipped a sign here. The matrix can be inverted.

• ducmod
Thank you everyone! I see - my mistake is in the sign, as you have pointed out. Expanding by the first and second row both produce -12.

Once again - thank you very much for your help! I have found the inverse matrix by correcting the determinant of the initial matrix.