Inverse Matrices Practice and Review

Inverse Matrix Formulas & Methods

  • 2×22 \times 2 Matrix Inverse Formula:

    • For a matrix A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is det⁡(A)=ad−bc\det(A) = ad - bc.

    • If det⁡(A)≠0\det(A) \neq 0, the inverse matrix is A−1=1ad−bc(d−b−ca)A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.

    • If det⁡(A)=0\det(A) = 0, the matrix is singular and the inverse does not exist.

  • Solving Systems using Inverse Matrices:

    • Convert a system of linear equations into matrix form AX=BA X = B.

    • Solve for the variable matrix XX using X = A^{-1} B$.\n\n# Inverse Matrix Practice Problems\n\n- **Problem 1**: Find the inverse of \begin{pmatrix} 2 & 0 \ 3 & 5 \end{pmatrix}.\n - \det(A) = (2)(5) - (0)(3) = 10\n - A^{-1} = \frac{1}{10} \begin{pmatrix} 5 & 0 \ -3 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \ -\frac{3}{10} & \frac{1}{5} \end{pmatrix}\n\n- **Problem 2**: Find the inverse of \begin{pmatrix} 4 & 6 \ 6 & 9 \end{pmatrix}.\n - \det(B) = (4)(9) - (6)(6) = 36 - 36 = 0\n - Inverse does not exist because the determinant is 0\n\n- **Problem 3**: Solve the system using inverse matrices:\n - System: 4x + 2y = 10andandx - y = 13\n - Matrix equation: \begin{pmatrix} 4 & 2 \ 1 & -1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 10 \ 13 \end{pmatrix}\n - Determinant: \det(A) = (4)(-1) - (2)(1) = -6\n - Inverse matrix: A^{-1} = \begin{pmatrix} \frac{1}{6} & \frac{1}{3} \ \frac{1}{6} & -\frac{2}{3} \end{pmatrix}\n - Solution: \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} \frac{1}{6} & \frac{1}{3} \ \frac{1}{6} & -\frac{2}{3} \end{pmatrix} \begin{pmatrix} 10 \ 13 \end{pmatrix} = \begin{pmatrix} 6 \ -7 \end{pmatrix}((x = 6,,y = -7)\n\n# Word Problems with Inverse Matrices\n\n- **Problem 4**: Kelly and Mike's Fundraiser\n - **System of Equations**:\n - Kelly: 7p + 9s = 225 (7 rolls of plain wrapping paper, 9 rolls of shiny wrapping paper for $225)\n - Mike: 1p + 9s = 171 (1 roll of plain wrapping paper, 9 rolls of shiny wrapping paper for $171)\n - **Matrix Equation**:\n - \begin{pmatrix} 7 & 9 \ 1 & 9 \end{pmatrix} \begin{pmatrix} p \ s \end{pmatrix} = \begin{pmatrix} 225 \ 171 \end{pmatrix}\n - **Solution using Inverse Matrix**:\n - \det(A) = (7)(9) - (9)(1) = 63 - 9 = 54\n - A^{-1} = \frac{1}{54} \begin{pmatrix} 9 & -9 \ -1 & 7 \end{pmatrix}\n - \begin{pmatrix} p \ s \end{pmatrix} = \frac{1}{54} \begin{pmatrix} 9 & -9 \ -1 & 7 \end{pmatrix} \begin{pmatrix} 225 \ 171 \end{pmatrix} = \begin{pmatrix} 9 \ 18 \end{pmatrix}\n - Cost of one plain wrapping paper roll p = \$9\n - Cost of one shiny wrapping paper roll s = \$18\n\n# Matrix Operations Review\n\n- **Problem 5**: Scalar Multiplication and Matrix Addition\n - Expression: -3 \begin{pmatrix} 0 & -5 & -6 \ -4 & -3 & 2 \end{pmatrix} + 4 \begin{pmatrix} 1 & 0 & 3 \ 0 & -2 & 5 \end{pmatrix}\n - Scalar multiples: \begin{pmatrix} 0 & 15 & 18 \ 12 & 9 & -6 \end{pmatrix} + \begin{pmatrix} 4 & 0 & 12 \ 0 & -8 & 20 \end{pmatrix}\n - Result: \begin{pmatrix} 4 & 15 & 30 \ 12 & 1 & 14 \end{pmatrix}\n\n- **Problem 6**: Matrix Multiplication\n - Expression: \begin{pmatrix} 0 & -1 \ -6 & 0 \ -3 & 3 \end{pmatrix} \begin{pmatrix} 3 & -4 \ -6 & 2 \end{pmatrix}\n - Row calculations:\n - Row 1: (0)(3) + (-1)(-6) = 6,,(0)(-4) + (-1)(2) = -2\n - Row 2: (-6)(3) + (0)(-6) = -18,,(-6)(-4) + (0)(2) = 24\n - Row 3: (-3)(3) + (3)(-6) = -27,,(-3)(-4) + (3)(2) = 18\n - Result: \begin{pmatrix} 6 & -2 \ -18 & 24 \ -27 & 18 \end{pmatrix}$$