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Lesson 13: Solving Systems Using Matrix Inverses

1. The Inverse Matrix Method

If \(A\) is invertible \( (det(A) ≠ 0) \), the system \( Ax = b \) has unique solution:

\[ x = A^{-1} b \]

Steps:

  1. Check \( det(A) ≠ 0 \)
  2. Find \(A^{-1}\)
  3. Multiply \(A^{-1}\) by \(b\)

For \(2×2\) matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the inverse is:

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

Exercise 1

Find the inverse of \(A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)
\(det(A) = \)
\(A^{-1} = (1/det) \begin{pmatrix} x & y \\ z & w \end{pmatrix}\)
\(x = \), \(y = \)
\(z = \), \(w = \)

2. Solving \(Ax = b\) with Inverse

Example system:

\[ \begin{cases} 3x + y = 7 \\ 5x + 2y = 13 \end{cases} \quad A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}, \quad b = \begin{pmatrix} 7 \\ 13 \end{pmatrix} \]

\(A⁻¹ = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\)

\(x = A⁻¹b = \begin{pmatrix} 2·7 + (-1)·13 \\ -5·7 + 3·13 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\)

Exercise 2

Solve using inverse:
\(A = \begin{pmatrix} 4 & -1 \\ 2 & 3 \end{pmatrix}\), b = \(\begin{pmatrix} 5 \\ 11 \end{pmatrix}\)
\(det(A) = \)
\(x = \), \(y = \)

Summary – Lesson 13

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