Solving Systems with Inverses: Fresh Take

  • Find the inverse of a matrix when possible
  • Solve a system of linear equations using an inverse matrix

Finding the Inverse of a Matrix

The Main Idea

  • Matrix Inverses:
    • For a square matrix [latex]A[/latex], its inverse [latex]A^{-1}[/latex] satisfies: [latex]AA^{-1} = A^{-1}A = I[/latex]
    • [latex]I[/latex] is the identity matrix, with [latex]1[/latex]s on the main diagonal and [latex]0[/latex]s elsewhere
    • Not all square matrices have inverses; those that do are called invertible
  • Properties of Matrix Inverses:
    • Only square matrices can have inverses
    • If [latex]A[/latex] is invertible, its inverse [latex]A^{-1}[/latex] is unique
    • The identity matrix acts like [latex]1[/latex] in matrix algebra: [latex]AI = IA = A[/latex]
  • Identity Matrix:
    • 2×2 Identity Matrix: [latex]I_2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}[/latex]
    • 3×3 Identity Matrix: [latex]I_3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}[/latex]
  • Solving Systems of Equations:
    • For a system [latex]AX = B[/latex], the solution is [latex]X = A^{-1}B[/latex]
    • This method is efficient for larger systems of equations
  • Verifying Inverses:
    • To show that [latex]B[/latex] is the inverse of [latex]A[/latex], prove that [latex]AB = BA = I[/latex]
Show that the following two matrices are inverses of each other.

[latex]A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right],B=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right][/latex]

You can view the transcript for “The Identity Matrix” here (opens in new window).

You can view the transcript for “Determining Inverse Matrices Using Augmented Matrices” here (opens in new window).

 

Finding the Multiplicative Inverse

The Main Idea

  • Definition of Matrix Inverse:
    • For a square matrix [latex]A[/latex], its inverse [latex]A^{-1}[/latex] satisfies: [latex]AA^{-1} = A^{-1}A = I[/latex]
    • [latex]I[/latex] is the identity matrix
  • Methods to Find Matrix Inverses:
    • Matrix Multiplication Method
    • Augmenting with Identity Method
    • Inverse Formula (for 2×2 matrices)
  • Matrix Multiplication Method:
    • Set up equation: [latex]A \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}[/latex]
    • Solve system of equations to find [latex]a, b, c, d[/latex]
  • Augmenting with Identity Method:
    • Create augmented matrix [latex][A|I][/latex]
    • Perform row operations to transform [latex]A[/latex] into [latex]I[/latex]
    • The transformed I becomes [latex]A^{-1}[/latex]
  • Inverse Formula for 2×2 Matrices:
    • For [latex]A = \begin{bmatrix} a & b \ c & d \end{bmatrix}[/latex]
    • [latex]A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}[/latex]
    • Only valid if [latex]ad-bc \neq 0[/latex]
Use the formula to find the inverse of matrix [latex]A[/latex]. Verify your answer by augmenting with the identity matrix.

[latex]A=\left[\begin{array}{cc} \hfill 1& \hfill -1\\ \hfill 2& \hfill 3\end{array}\right][/latex]

Find the inverse, if it exists, of the given matrix.

[latex]A=\left[\begin{array}{cc}3& 6\\ 1& 2\end{array}\right][/latex]

Multiplicative Inverse of 3×3 Matrices

The Main Idea

  • Definition of Matrix Inverse:
    • For a square matrix [latex]A[/latex], its inverse [latex]A^{-1}[/latex] satisfies: [latex]AA^{-1} = A^{-1}A = I[/latex]
    • [latex]I[/latex] is the 3×3 identity matrix: [latex]I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}[/latex]
  • Method for Finding 3×3 Matrix Inverse:
    • Augment the original matrix with the identity matrix
    • Perform row operations to transform the left side into the identity matrix
    • The right side becomes the inverse matrix
  • Steps to Find 3×3 Matrix Inverse:
    • Write the augmented matrix [latex][A|I][/latex]
    • Use elementary row operations to transform [latex]A[/latex] into [latex]I[/latex]
    • The resulting right side is [latex]A^{-1}[/latex]
  • Singularity:
    • Not all 3×3 matrices have inverses
    • A matrix with no inverse is called singular
    • If row operations result in a row of zeros, the matrix is singular
  • Verification:
    • Check that [latex]AA^{-1} = A^{-1}A = I[/latex]
Find the inverse of the [latex]3\times 3[/latex] matrix.

[latex]A=\left[\begin{array}{ccc}\hfill 2&\hfill -17& \hfill 11\\ \hfill -1& \hfill 11& \hfill -7\\ \hfill 0& \hfill 3& \hfill -2\end{array}\right][/latex]

Solving a System of Linear Equations Using the Inverse of a Matrix

The Main Idea

  • Matrix Representation of Systems:
    • [latex]AX = B[/latex], where:
      • [latex]A[/latex]: Coefficient matrix
      • [latex]X[/latex]: Variable matrix
      • [latex]B[/latex]: Constant matrix
  • Solution Method:
    • Multiply both sides by [latex]A^{-1}[/latex]: [latex]A^{-1}AX = A^{-1}B[/latex]
    • Simplify: [latex]IX = A^{-1}B[/latex]
    • Solution: [latex]X = A^{-1}B[/latex]
  • Matrix Inverse Properties:
    • [latex]A^{-1}A = AA^{-1} = I[/latex]
    • Not all matrices have inverses (singular matrices)
  • Order Matters:
    • [latex]A^{-1}B \neq BA^{-1}[/latex] (matrix multiplication is not commutative)
Solve the system using the inverse of the coefficient matrix.

[latex]\begin{array}{l}\text{ }2x - 17y+11z=0\hfill \\ \text{ }-x+11y - 7z=8\hfill \\ \text{ }3y - 2z=-2\hfill \end{array}[/latex]

You can view the transcript for “Using a Matrix Equation to Solve a System of Equations” here (opens in new window).