Find the inverse using Gaussian elimination of the 3x3 matrix: 1 2 -1; 3 7 -10; 7 16 -21.

Steps to Solve

1. Set Up the Augmented Matrix

Start by creating an augmented matrix from the original 3x3 matrix ( A ) and the 3x3 identity matrix ( I ).

If ( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix} ), then the augmented matrix will look like this:

$$ \left[ \begin{array}{ccc|ccc} a_{11} & a_{12} & a_{13} & 1 & 0 & 0 \ a_{21} & a_{22} & a_{23} & 0 & 1 & 0 \ a_{31} & a_{32} & a_{33} & 0 & 0 & 1 \end{array} \right] $$

2. Perform Row Operations to Achieve Row Echelon Form

Use row operations to create zeros below the main diagonal.

For example, if you want to eliminate the ( a_{21} ) entry, you can perform the operation:

$$ R_2 \leftarrow R_2 - \frac{a_{21}}{a_{11}}R_1 $$

Continue this for each column to create zeros where necessary.

3. Transform to Reduced Row Echelon Form

Continue performing row operations to get the rows in reduced row echelon form. This means making the leading coefficients of each row equal to 1 and ensuring that every pivot column has zeros in every other position.

For example, you might transform a row containing a leading 2 into:

$$ R_3 \leftarrow \frac{1}{a_{33}}R_3 $$

4. Extract the Inverse Matrix

Once you have the left side of the augmented matrix in the identity form, the right side will represent the inverse of the original matrix ( A ).

For example, if the augmented matrix looks like:

$$ \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & x_{11} & x_{12} & x_{13} \ 0 & 1 & 0 & x_{21} & x_{22} & x_{23} \ 0 & 0 & 1 & x_{31} & x_{32} & x_{33} \end{array} \right] $$

Then the inverse ( A^{-1} ) is given by:

$$ A^{-1} = \begin{bmatrix} x_{11} & x_{12} & x_{13} \ x_{21} & x_{22} & x_{23} \ x_{31} & x_{32} & x_{33} \end{bmatrix} $$