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3.5 Matrix Inverse

scalar: \( a \cdot a^{-1} = a^{-1} \cdot a = 1 \) for \( a \neq 0 \), \( a^{-1} = \frac{1}{a} \)

matrix: \( AA^{-1} = A^{-1}A = I \)

\( A^{-1} \) is the inverse of A

\( I \) is the identity matrix

I: Square matrix w/ 1 on its main diagonal and 0 everywhere else

\( 2 \times 2 \) identity: \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

\( 4 \times 4 \) identity: \( I = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \)

any square matrix A

\( AI = IA = A \)

for example,

\( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

\( AI = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = A \)

\( IA = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = A \)

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notice \( AA^{-1} = A^{-1}A = I \) \( \rightarrow \) only square A can have \( A^{-1} \)

but not all square A have \( A^{-1} \)
(just like not all scalar a have \( a^{-1} \))

if \( A^{-1} \) exists, then we say A is invertible

if not, A is not invertible or A is singular

for \( 2 \times 2 \) A, \( A^{-1} \) has a very simple formula:

\( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)

\( A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)

ad-bc is the determinant of A

(\( ad-bc = 0 \rightarrow \) A is not invertible)

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\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad A^{-1} = \frac{1}{1 \cdot 4 - 3 \cdot 2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix} \]

\[ AA^{-1} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

\[ A^{-1}A = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Inverse can be used to solve systems

\[ \begin{matrix} x_1 + 2x_2 = 5 \\ 3x_1 + 4x_2 = 6 \end{matrix} \longrightarrow \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix} \]

\[ A\vec{x} = \vec{b} \]

Solution: Gaussian elimination → always possible

or \( A\vec{x} = \vec{b} \) so \( \vec{x} = A^{-1}\vec{b} \) → only if \( A^{-1} \) exists

order is important

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here, \( A^{-1} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix} \) (previous example)

\[ \vec{x} = A^{-1}\vec{b} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix} \begin{bmatrix} 5 \\ 6 \end{bmatrix} = \begin{bmatrix} -4 \\ 9/2 \end{bmatrix} \quad \begin{matrix} x_1 = -4 \\ x_2 = 9/2 \end{matrix} \]

for \( n \times n \) \( A \) where \( n > 2 \), no nice formulas exist for \( A^{-1} \) but there is a general algorithm that works for all \( n \) (including \( n = 2 \))

given \( n \times n \) \( A \), make an augmented matrix with left half being \( A \) and right half \( I \) of the same size

\[ \left[ \begin{array}{c:c} A & I \\ \text{\tiny } n \times n & \text{\tiny } n \times n \end{array} \right] \quad \text{this matrix is } n \times 2n \]

→ row operations →

\[ \left[ \begin{array}{c:c} I & A^{-1} \end{array} \right] \]

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Matrix Inversion Example: 2x2 Case

Try it on matrix \( A \):

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \]

Set up the augmented matrix \( [A | I] \):

\[ \left[ \begin{array}{cc:cc} 1 & 2 & 1 & 0 \\ 3 & 4 & 0 & 1 \end{array} \right] \]

Perform row operations to reach reduced row echelon form:

\[ \xrightarrow{(-3)R_1 + R_2} \left[ \begin{array}{cc:cc} 1 & 2 & 1 & 0 \\ 0 & -2 & -3 & 1 \end{array} \right] \xrightarrow{R_2 + R_1} \left[ \begin{array}{cc:cc} 1 & 0 & -2 & 1 \\ 0 & -2 & -3 & 1 \end{array} \right] \]

\[ \xrightarrow{(-\frac{1}{2})R_2} \left[ \begin{array}{cc:cc} 1 & 0 & -2 & 1 \\ 0 & 1 & 3/2 & -1/2 \end{array} \right] \]

The right side of the augmented matrix is now \( A^{-1} \):

\[ A^{-1} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix} \]

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Matrix Inversion Example: 3x3 Case

Try matrix \( A \):

\[ A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix} \]

Set up the augmented matrix \( [A | I] \):

\[ \left[ \begin{array}{ccc:ccc} 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right] \]

Apply row operations:

\[ \xrightarrow[(-1)R_1 + R_3]{(-1)R_1 + R_2} \left[ \begin{array}{ccc:ccc} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1 & 1 & 0 \\ 0 & 1 & 1 & -1 & 0 & 1 \end{array} \right] \]

\[ \xrightarrow{(-1)R_2 + R_3} \left[ \begin{array}{ccc:ccc} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1 & 1 & 0 \\ 0 & 0 & 1 & 0 & -1 & 1 \end{array} \right] \]

The resulting augmented matrix is in the form \( [I | A^{-1}] \), where:

\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \]

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Non-Existence of Matrix Inverse

If \( A^{-1} \) does not exist, then we won't be able to make the left half into an \( I \).

Example Matrix

\[ A = \begin{bmatrix} 1 & -2 & -1 \\ -1 & 5 & 6 \\ 5 & -4 & 5 \end{bmatrix} \]

Augmented Matrix Row Reduction

We start with the augmented matrix \( [A \mid I] \):

\[ \begin{bmatrix} 1 & -2 & -1 & 1 & 0 & 0 \\ -1 & 5 & 6 & 0 & 1 & 0 \\ 5 & -4 & 5 & 0 & 0 & 1 \end{bmatrix} \]

Performing row operations:

\[ \rightarrow \begin{bmatrix} 1 & -2 & -1 & 1 & 0 & 0 \\ 0 & 3 & 5 & 1 & 1 & 0 \\ 0 & 6 & 10 & -5 & 0 & 1 \end{bmatrix} \]

\[ \rightarrow \begin{bmatrix} 1 & -2 & -1 & 1 & 0 & 0 \\ 0 & 3 & 5 & 1 & 1 & 0 \\ 0 & 0 & 0 & x & x & x \end{bmatrix} \]

A is not invertible

no pivot in row 3

not possible to become \( I \)

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Why does \( [A \mid I] \rightarrow \dots \rightarrow [I \mid A^{-1}] \) work?

Revisiting Systems of Equations

Consider the system:

\[ a_1 x_1 + b_1 x_2 = c_1 \]\[ a_2 x_1 + b_2 x_2 = c_2 \]

To find \( x_1, x_2 \), we use the augmented matrix:

\[ \begin{bmatrix} a_1 & b_1 & \vdots & c_1 \\ a_2 & b_2 & \vdots & c_2 \end{bmatrix} \rightarrow \dots \rightarrow \begin{bmatrix} 1 & 0 & \vdots & x_1 \\ 0 & 1 & \vdots & x_2 \end{bmatrix} \]

Application to Matrix Inverses

For the inverse of \( A \), the same principle is used.

\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \quad A^{-1} = \begin{bmatrix} e & f \\ g & h \end{bmatrix} \]

Since \( AA^{-1} = I \):

\[ AA^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \rightarrow \begin{cases} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} e \\ g \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} f \\ h \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{cases} \]

Note:

These represent two columns solved together.

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Matrix Inversion via Row Operations

\[ \left[ \begin{array}{cc:cc} a & b & 1 & 0 \\ c & d & 0 & 1 \end{array} \right] \]

Solving first system twice, one for each column but together (columns do not change using row ops)

\[ \rightarrow \dots \rightarrow \left[ \begin{array}{c:c} I & A^{-1} \end{array} \right] \]