9.7 Solving Systems with Inverses

Matrix Inverses and Systems of Linear Equations

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Calculation of matrix inverses

A matrix inverse, denoted $A^{-1}$, is the matrix that "undoes" what $A$ does. It satisfies the property $A \cdot A^{-1} = A^{-1} \cdot A = I$, where $I$ is the identity matrix. Not every matrix has an inverse, and knowing when one exists (and how to find it) is the foundation for this entire section.

Finding the inverse of a 2×2 matrix $A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$:

1. Calculate the determinant: $\det(A) = ad - bc$

2. Check that $\det(A) \neq 0$. If the determinant is zero, the matrix has no inverse and the system either has no solution or infinitely many.

3. Apply the formula: $A^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$

The pattern: swap $a$ and $d$ along the main diagonal, then flip the signs of $b$ and $c$.

Quick example: For $A = \begin{bmatrix}3 & 1 \\ 2 & 4\end{bmatrix}$, the determinant is $3(4) - 1(2) = 10$, so:

$A^{-1} = \frac{1}{10}\begin{bmatrix}4 & -1 \\ -2 & 3\end{bmatrix}$

Inverting larger matrices (3×3 and beyond):

1. Set up the augmented matrix $[A \mid I]$, placing the identity matrix of the same size next to $A$
2. Use row operations to transform the left side into the identity matrix
3. Once the left side is $I$, the right side has become $A^{-1}$

This is the same row reduction process you've used before, just applied to a wider augmented matrix. If at any point the left side can't be reduced to $I$ (you get a row of all zeros), the matrix is not invertible.

Application of inverse matrices

Any system of linear equations can be written in matrix form as $A\vec{x} = \vec{b}$, where:

• $A$ is the coefficient matrix (the numbers in front of the variables)
• $\vec{x}$ is the variable vector (what you're solving for)
• $\vec{b}$ is the constant vector (the numbers on the right side of each equation)

Solving the system using the inverse:

1. Find $A^{-1}$ (using the methods above)
2. Multiply both sides on the left by $A^{-1}$: $A^{-1}A\vec{x} = A^{-1}\vec{b}$
3. The left side simplifies because $A^{-1}A = I$, and $I\vec{x} = \vec{x}$
4. So the solution is: $\vec{x} = A^{-1}\vec{b}$

This method is especially efficient when you need to solve multiple systems that share the same coefficient matrix $A$ but have different constant vectors $\vec{b}$. You compute $A^{-1}$ once, then just multiply it by each new $\vec{b}$. This comes up in problems like varying supply-and-demand scenarios or mixture problems with changing quantities.

Interpretation of inverse matrix solutions

Real-world problems modeled by systems of equations require you to connect your numerical answers back to what the variables actually represent.

• Identify what each variable means. Before solving, be clear about what $x_1, x_2, x_3$ stand for in context (prices, quantities, rates, etc.).
• Check whether solutions are reasonable. A negative number of items produced or a flow rate of 10,000 gallons per second in a household pipe should raise a red flag. Always verify that your answers respect the problem's constraints.
• Communicate results with units and context. Don't just write $x = 5$. Write "the factory should produce 5 units of Product A per day" or whatever the problem calls for.

Matrix Properties and Alternative Methods

• Singular matrix: A square matrix with $\det(A) = 0$. It has no inverse, which means the corresponding system doesn't have a unique solution.
• Nonsingular (invertible) matrix: A square matrix with $\det(A) \neq 0$. It has an inverse, and the corresponding system has exactly one solution.
• Cramer's Rule: An alternative method for solving systems using determinants. For each variable, you replace one column of $A$ with $\vec{b}$ and take the ratio of that new determinant to $\det(A)$. It's useful for solving for a single variable without finding the full solution, but it gets tedious for large systems.