Inverse matrix

What is the inverse matrix?

Every matrix that isn’t singular has an inverse matrix. Whenever we multiply a matrix by its inverse matrix, we get the identity matrix, I.

Representing the inverse matrix

• The inverse matrix of M can be written as M^{-1}.
• This then makes sense why they multiply to the identity matrix: MM^{-1} = I
  • anything multiplied by its inverse is equal to the identity, or 1 if we’re working with regular scalars.

Finding the inverse matrix

Find the inverse matrix of \begin{bmatrix} 4 & 2 \\ 5 & 3 \end{bmatrix}

• \begin{pmatrix}4 & 2 \\ 5 & 3\end{pmatrix}\begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}
• 4a + 2c = 1
• 4b + 2d = 0
• 5a + 3c = 0
• 5b + 3d = 1
• Solve the equations with a and c simultaneously:
  • 4a+2c=1 (1)
  • 5a+3c=0 (2)
  • Multiply (1) by 3: 12a + 6c = 3 (3)
  • Multiply (2) by -2: -10a - 6c = 0 (4)
  • Solve using elimination:
    • (3) + (4): 2a = 3 \Rightarrow a = \frac{3}{2}
  • Substitute a into (1): 4(\frac{3}{2}) + 2c = 1 \Rightarrow 6 + 2c = 1 \Rightarrow 2c = -5 \Rightarrow c = -\frac{5}{2}
  • a = \frac{3}{2}, c = -\frac{5}{2}

General form of the inverse matrix

• Let M=\begin{pmatrix}a&b\\c&d\end{pmatrix}
• Find the determinant:
  • ad-bc
• Swap around the placement of a and d
• Negate b and c
• Divide the resulting matrix by the determinant, which we said was ad-bc

From that, we can derive the formula of the inverse of a 2x2 matrix:

M^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}

Find the inverse matrix of \begin{bmatrix} 4 & 2 \\ 5 & 3 \end{bmatrix} using the general form

• a=4, b=2, c=5, d=3
• ad-bc = 4\times 3 - 2\times 5 = 12 - 10 = 2
• M^{-1} = \frac{1}{2}\begin{pmatrix}3 & -2 \\ -5 & 4\end{pmatrix} = \begin{pmatrix}\frac{3}{2} & -1 \\ -\frac{5}{2} & 2\end{pmatrix}
• Answer: \begin{bmatrix} \frac{3}{2} & -1 \\ -\frac{5}{2} & 2 \end{bmatrix}

Find the inverse of \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}

• a=1, b=2, c=3, d=4
• ad-bc = 1\times 4 - 2\times 3 = 4 - 6 = -2
• M^{-1} = \frac{1}{-2}\begin{pmatrix}4 & -2 \\ -3 & 1\end{pmatrix} = \begin{pmatrix}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{pmatrix}
• Answer: \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}

Find the inverse of \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

• a=1, b=0, c=0, d=1
• ad-bc = 1\times 1 - 0\times 0 = 1
• M^{-1} = \frac{1}{1}\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}
• Answer: \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

The inverse of a matrix M is \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}. What is M?

• The inverse is reversable - so just like we can find M^{-1} given M by finding the inverse of M, we can also find M given M^{-1} by finding the inverse of M^{-1}.
• M = (M^{-1})^{-1}
• M = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}^{-1}
• a=2, b=3, c=4, d=5
• ad-bc = 2\times 5 - 3\times 4 = 10 - 12 = -2
• M = \frac{1}{-2}\begin{pmatrix}5 & -3 \\ -4 & 2\end{pmatrix} = \begin{pmatrix}-\frac{5}{2} & \frac{3}{2} \\ 2 & -1\end{pmatrix}
• Answer: \begin{bmatrix} -\frac{5}{2} & \frac{3}{2} \\ 2 & -1 \end{bmatrix}

Find the inverse of matrix product AB

• Let the inverse function be X
• ABX=I (because matrix multiplication is associative)
• A^{-1}ABX=A^{-1}I (pre-multiply both sides by A^{-1})
• BX=A^{-1}I
• IBX=A^{-1}I
• BX=A^{-1}
• B^{-1}BX=B^{-1}A^{-1}
• IX=B^{-1}A^{-1}
• X=B^{-1}A{-1}
• Answer: X+B^{-1}A^{-1}

Remember this. THe inverse of AB=B^{-1}A^{-1}.

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