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Computing
Matrix Inverses

Master the row reduction method to find \( A^{-1} \) by transforming \( [A \mid I] \) into \( [I \mid A^{-1}] \) using elementary row operations.

Linear Algebra

What Is a Matrix Inverse?

For a square matrix \( A \), the inverse \( A^{-1} \) is the unique matrix satisfying:

$$ A \cdot A^{-1} = A^{-1} \cdot A = I $$

where \( I \) is the identity matrix. Think of it as the matrix equivalent of dividing: multiplying by \( A^{-1} \) "undoes" the transformation of \( A \).

The Identity Matrix \( I \)

The identity matrix has 1s on the main diagonal and 0s everywhere else. It acts like the number 1 in multiplication:

$$ I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
Left Identity
\( I \cdot A = A \) for any matrix \( A \)
Right Identity
\( A \cdot I = A \) for any matrix \( A \)

Why Do We Need Inverses?

Matrix inverses are essential for solving systems of linear equations. Given \( A\mathbf{x} = \mathbf{b} \), we can solve directly:

$$ \mathbf{x} = A^{-1}\mathbf{b} $$
Systems of Equations
Solve \( A\mathbf{x} = \mathbf{b} \) in one multiplication
Transformations
Reverse geometric transformations in graphics
Cryptography
Decode messages encrypted with matrix multiplication

The Augmented Matrix Method

To find \( A^{-1} \), form the augmented matrix \( [A \mid I] \) and row-reduce until the left side becomes \( I \). The right side then equals \( A^{-1} \):

$$ [A \mid I] \xrightarrow{\text{row ops}} [I \mid A^{-1}] $$

Elementary Row Operations

Three operations are allowed (they preserve equivalence):

1
Swap two rows: \( R_i \leftrightarrow R_j \)
2
Scale a row: \( R_i \to c \cdot R_i \) where \( c \neq 0 \)
3
Add a multiple of one row to another: \( R_i \to R_i + c \cdot R_j \)

Invertible vs Singular

How do you know if a matrix has an inverse? Check the determinant:

Invertible Singular
Determinant \( \det(A) \neq 0 \) \( \det(A) = 0 \)
Row Reduction Reaches \( I \) Row of zeros appears
Columns Linearly independent Linearly dependent
Inverse \( A^{-1} \) exists \( A^{-1} \) does not exist

How to Find \( A^{-1} \)

1
Write the augmented matrix \( [A \mid I] \), placing the identity to the right of a vertical bar.
2
Get leading 1s on the diagonal by scaling rows.
3
Eliminate below each pivot (forward elimination) to reach row echelon form.
4
Eliminate above each pivot (back substitution) to reach reduced row echelon form.
5
Read off \( A^{-1} \) from the right half. If a row of zeros appears on the left, \( A \) is singular.

Example 1: \( 2 \times 2 \) Matrix

Find the inverse of \( A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} \).

Step 1: Form the augmented matrix:

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

Example 1: Row Reduction

Step 2: Scale \( R_1 \) by \( \tfrac{1}{2} \) to get a leading 1:

$$ \left[\begin{array}{cc|cc} 1 & \tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 5 & 3 & 0 & 1 \end{array}\right] $$

Step 3: \( R_2 \to R_2 - 5R_1 \):

$$ \left[\begin{array}{cc|cc} 1 & \tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} & -\tfrac{5}{2} & 1 \end{array}\right] $$

Step 4: Scale \( R_2 \) by 2, then \( R_1 \to R_1 - \tfrac{1}{2}R_2 \):

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

Example 1: Result

Reading the right half gives us:

$$ A^{-1} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} $$

Example 2: \( 3 \times 3 \) Matrix

Find the inverse of \( A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 1 & 1 \end{bmatrix} \).

Step 1: Form the augmented matrix \( [A \mid I_3] \):

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

Example 2: Row Reduction

Step 2: \( R_3 \to R_3 - R_1 \):

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

Step 3: \( R_3 \to R_3 - R_2 \):

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

Step 4: Scale \( R_3 \to -\tfrac{1}{2} R_3 \):

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

Example 2: Back Substitution

Step 5: \( R_2 \to R_2 - 2R_3 \), then \( R_1 \to R_1 - R_3 \):

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

Result:

$$ A^{-1} = \begin{bmatrix} \tfrac{1}{2} & -\tfrac{1}{2} & \tfrac{1}{2} \\ -1 & 0 & 1 \\ \tfrac{1}{2} & \tfrac{1}{2} & -\tfrac{1}{2} \end{bmatrix} $$

Common Mistakes

✕ Forgetting to Check Determinant
Always verify \( \det(A) \neq 0 \) first. If it equals zero, stop: the matrix is singular and has no inverse.
✕ Applying Ops to One Side Only
Every row operation must be applied to the entire augmented row: both the left and right halves simultaneously.
✕ Stopping at Row Echelon Form
You need reduced row echelon form (RREF), not just REF. Eliminate above each pivot too, not just below.

Summary of Methods

\( 2 \times 2 \) Formula Row Reduction
Size \( 2 \times 2 \) only Any \( n \times n \)
Speed Very fast Slower for large \( n \)
Formula \( \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \) \( [A \mid I] \to [I \mid A^{-1}] \)
When to Use Quick calculations \( 3 \times 3 \) and larger
Detects Singular? \( ad - bc = 0 \) Row of zeros on left

Question 1 of 5

What is the first step in finding \( A^{-1} \) using row reduction?

Question 2 of 5

If \( A = \begin{bmatrix} 4 & 6 \\ 2 & 3 \end{bmatrix} \), what can you conclude about \( A^{-1} \)?

Question 3 of 5

Using the \( 2 \times 2 \) formula, what is the inverse of \( \begin{bmatrix} 1 & 2 \\ 3 & 7 \end{bmatrix} \)?

Question 4 of 5

During row reduction of \( [A \mid I] \), you obtain a row of all zeros on the left side. What does this mean?

Question 5 of 5

If \( A^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \), what is \( (A^{-1})^{-1} \)?