Question

How to do the inverse of a 3x3 matrix gaussian elimination.

Answer 1

To find the inverse of a 3x3 matrix using Gaussian elimination, write the matrix augmented with an identity matrix, transform the left side into the identity matrix, and the right side will be the inverse.

Step-by-step explanation:

To find the inverse of a 3x3 matrix using Gaussian elimination, you should start by writing the matrix augmented with an identity matrix of the same size. Apply elementary row operations to transform the left side of the augmented matrix into the identity matrix. The right side of the augmented matrix will then be the inverse of the original matrix. Finally, if the right side of the augmented matrix is indeed the inverse, you have found the inverse of the original matrix.

For example, let's say we have the matrix A:

A = [2 4 6] [1 3 5] [0 2 0]

We want to find the inverse of A. We first write the augmented matrix:

[2 4 6 | 1 0 0] [1 3 5 | 0 1 0] [0 2 0 | 0 0 1]

By applying elementary row operations, we can transform the left side into the identity matrix:

[1 0 0 | 2 -4 0] [0 1 0 | -1 3 0] [0 0 1 | -0.5 2 1]

The right side of the augmented matrix is the inverse of the original matrix A:

A-1 = [2 -4 0] [-1 3 0] [-0.5 2 1]

Answer 2

As an example, let's invert the matrix

`\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}`

We construct the augmented matrix,

`\left[ \begin{array}ccc -3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]`

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

`\left[ \begin{array}ccc -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]`

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

`\left[ \begin{array}ccc -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]`

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

`\left[ \begin{array}ccc -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]`

• Multiply row 3 by 1/3 :

`\left[ \begin{array}ccc -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]`

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

`\left[ \begin{array}ccc -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]`

• Multiply row 2 by 1/7 :

`\left[ \begin{array}ccc -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]`

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

`\left[ \begin{array}ccc -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]`

• Multiply row 1 by -1/3 :

`\left[ \begin{array}ccc 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]`

So, the inverse of our matrix is

`\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^(-1) = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}`