Elimination and Inverse Matrices

The Essentials

Matrix multiplication performs a function that changes the matrix. Each of the row operations can be performed by left multiplying the matrix by another matrix called an elimination or permutation matrix. For example these three matrices each perform the following functions: swap the first two rows, scale the second row by 2, and add the first row to the second row:

\[ \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 & 0\\ 1 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \]

Note that if these permutation matrices are on the right instead of the left, then the columns will be manipulated, not the rows. Permutation matrices can be combined to form a matrix that performs many operations. For example, if you had a matrix A and matrix B swapped two rows and matrix C added two of the new first row to the third row and matrix D scaled the second row by 5, you could write the operations like this:

\[ DCBA \]

Where all the permutation matrices are on the left and the ones that correspond with the operations that you want to perform first are closest to A. At this point, D, C, and B can be multiplied into one matrix that performs the given operations to any matrix of the same size with one matrix multiplication.

Permutation matrices can be used to prove inverse matrices. A matrix multiplied (right or left multiplication) by its inverse is equal to the identity matrix. A matrix is only invertible iff it is linearly independent. Therefore, a matrix is invertible iff it has a non zero determinant. If the operations performed by permutations matrices B, C, and D reduce A to the Identity matrix such that:

\[ DCBA = I \]

Then, the result of the matrix multiplication, DCB is equal to the inverse of A:

\[ A^{-1}A = I \]

Therefore, to find the inverse of A, all that is needed is to perform the row operations that reduce A to I on the matrix I to get the inverse of A. This can be done by augmenting the matrix A with the identity matrix and reducing it until the part that was A is I and the part that was I will be the inverse of I.

\[ \begin{bmatrix} A|I \end{bmatrix} \xrightarrow{} \begin{bmatrix} I|A^{-1} \end{bmatrix} \]

A shortcut for finding the inverse matrix of a two by two matrix is given below:

\[ \begin{bmatrix} a & b\\ c & d \end{bmatrix}^{-1} = \frac{1}{\det} \begin{bmatrix} d & - b \\ - c & a \end{bmatrix} = \frac{1}{ad-bc} \begin{bmatrix} d & - b \\ - c & a \end{bmatrix} \]

Example

Find the inverse matrix:

\[ \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 2\\ 2 & 1 & 1 \end{bmatrix} \]

First, we augment the matrix with the identity matrix:

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

Now we perform matrix operations: \( R_3=R_3-2R_1 \)

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

\( R_3=R_3+3R_2 \)

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

\( R_2=R_2-2R_3 \)

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

\( R_1=R_1-2R_2 \)

\( R_1=R_1-3R_3 \)

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

Now that we reduced the left side of the matrix to I, the right side is the inverse of A:

\[ \begin{bmatrix} -1 & 1 & 3 \\ 4 & -5 & -2 \\ -2 & 3 & 1 \end{bmatrix} \]

Practice

  1. Find the three by three matrix that performs all of these operations in this order: Scale the first row by a half, scale the second row by 3, swap the second and third rows, subtract 2 of the first row from the second row, add three of the second row to the third row.
  2. Find the inverse of this 2 by 2 matrix using the method for 2 by 2 matrices:
    \[ \begin{bmatrix} 2 & 3\\ 5 & 12 \end{bmatrix} \]
  3. Find the inverse matrix:
    \[ \begin{bmatrix} 7 & 8 & 9\\ 3 & 4 & 6\\ 7 & 6 & 1 \end{bmatrix} \]

Solutions:

  1. \[ \begin{bmatrix} 1/2 & 0 & 0\\ -1 & 0 & 1\\ -3 & 3 & 3 \end{bmatrix} \]
  2. \[ \begin{bmatrix} 4/3 & -1/3\\ -5/9 & 2/9 \end{bmatrix} \]
  3. \[ \begin{bmatrix} 16 & -23 & -6 \\ -39/2 & 28 & 15/2 \\ 5 & -7 & -2 \end{bmatrix} \]