Inverse Matrices
By Catalin David
A matrix is invertible if its determinant is not 0. If A is an invertible matrix, then its inverse is $A^{-1}$ where $A^{-1}=\frac{1}{\left|A\right|} \cdot adj(A)$. $adj(A)$ is the adjugate of the matrix.
Determining the Inverse of a Matrix
- We calculate the determinant of the matrix.
- We write the transpose of the matrix.
- Every element of the transpose is replaced with its cofactor. The resulting matrix is the adjugate.
- We calculate the inverse.
Example 46
$A=\begin{pmatrix}
1 & 3\\
2 & 5
\end{pmatrix}$
$\left|A\right|=1\cdot 5-6=-1$
The matrix is invertible, so we can calculate its inverse.
$ A^{T}= \begin{pmatrix} 1 & 2\\ 3 & 5 \end{pmatrix}$
We replace the elements of the transpose with their cofactors.
$1\longrightarrow (-1)^{1+1}\cdot \Delta_{1,1}=(-1)^{2}\cdot5 = 5$
$2\longrightarrow (-1)^{1+2}\cdot \Delta_{1,2}=(-1)^{3}\cdot3 = -3$
$3\longrightarrow (-1)^{2+1}\cdot \Delta_{2,1}=(-1)^{3}\cdot2 = -2$
$5\longrightarrow (-1)^{2+2}\cdot \Delta_{2,2}=(-1)^{4}\cdot1 = 1$
$adj(A)= \begin{pmatrix} 5 & -3\\ -2 & 1\\ \end{pmatrix}$
$A^{-1}=- \begin{pmatrix} 5 & -3\\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -5 & 3\\ 2 & -1 \end{pmatrix}$
Example 47
$B=\begin{pmatrix}
2 & -7\\
-1 & 6
\end{pmatrix}$
$\left|B\right|=2\cdot 6-(-7)\cdot (-1) = 5$
The matrix invertible, so we can calculate its inverse.
$A^{T}=
\begin{pmatrix}
2 & -1\\
-7 & 6
\end{pmatrix}$
We replace the elements of the transpose with their cofactors.
$2\longrightarrow (-1)^{1+1}\cdot \Delta_{1,1}=(-1)^{2}\cdot6 = 6$
$-1\longrightarrow (-1)^{1+2}\cdot \Delta_{1,2}=(-1)^{3}\cdot(-7) = 7$
$-7\longrightarrow (-1)^{2+1}\cdot \Delta_{2,1}=(-1)^{3}\cdot(-1) = 1$
$6\longrightarrow (-1)^{2+2}\cdot \Delta_{2,2}=(-1)^{4}\cdot2 = 2$
$adj(A)= \begin{pmatrix} 6 & 7\\ 1 & 2 \end{pmatrix}$
$A^{-1}=\frac{1}{5} \begin{pmatrix} 6 & 7\\ 1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{6}{5} & \frac{7}{5}\\ \frac{1}{5} & \frac{2}{5} \end{pmatrix}$
Example 48
$C=\begin{pmatrix}
1 & 3 & 2\\
4 & 1 & 1\\
1 & 2 & 3\\
\end{pmatrix}$
By applying the formula to calculate the determinant we obtain $\left|B\right|=-18$.
The matrix is invertible, so we can calculate its inverse.
$C^{T}=\begin{pmatrix}
1 & 4 & 1\\
3 & 1 & 2\\
2 & 1 & 3
\end{pmatrix}$
We replace every element of the transpose with its cofactor.
$ 1\longrightarrow (-1)^{1+1}\cdot \Delta_{1,1}=(-1)^{2}\cdot
\begin{vmatrix}
1 & 2\\
1 & 3
\end{vmatrix} = 3 - 2 = 1$
$4\longrightarrow (-1)^{1+2}\cdot \Delta_{1,2}=(-1)^{3}\cdot \begin{vmatrix} 3 & 2\\ 2 & 3 \end{vmatrix} = -(9-4)=-5$
$1\longrightarrow (-1)^{1+3}\cdot \Delta_{1,3}=(-1)^{4}\cdot \begin{vmatrix} 3 & 1\\ 2 & 1 \end{vmatrix} = 3-2=1$
$3\longrightarrow (-1)^{2+1}\cdot \Delta_{2,1}=(-1)^{3}\cdot \begin{vmatrix} 4 & 1\\ 1 & 3\\ \end{vmatrix} = -(12-1)=-11$
$1\longrightarrow (-1)^{2+2}\cdot \Delta_{2,2}=$ $(-1)^{4}\cdot\begin{vmatrix} 1 & 1\\ 2 & 3\\ \end{vmatrix}=3-2=1$
$2\longrightarrow (-1)^{1+3}\cdot \Delta_{2,3}=$ $(-1)^{5}\cdot\begin{vmatrix} 1 & 4\\ 2 & 1 \end{vmatrix}= -(1-8)=7$
$2\longrightarrow (-1)^{3+1}\cdot \Delta_{3,1}=$ $(-1)^{4}\cdot\begin{vmatrix} 4 & 1\\ 1 & 2 \end{vmatrix}=8-1=7$
$1\longrightarrow (-1)^{3+2}\cdot \Delta_{3,2}=$ $(-1)^{5}\cdot \begin{vmatrix} 1 & 1\\ 3 & 2 \end{vmatrix}=-(2-3)=1$
$3\longrightarrow (-1)^{3+3}\cdot \Delta_{3,3}=$ $(-1)^{6}\cdot\begin{vmatrix} 1 & 4\\ 3 & 1 \end{vmatrix}=1-12=-11$
$adj(A)= \begin{pmatrix} 1 & -5 & 1\\ -11 & 1 & 7\\ 7 & 1 & -11 \end{pmatrix}$
$A^{-1} = - \frac{1}{18}\cdot \begin{pmatrix} 1 & -5 & 1\\ -11 & 1 & 7\\ 7 & 1 & -11 \end{pmatrix} =$ $\begin{pmatrix} - \frac{1}{18} & \frac{5}{18} & -\frac{1}{18}\\ \frac{11}{18} & -\frac{1}{18} & -\frac{7}{18}\\ -\frac{7}{18} & -\frac{1}{18} & \frac{11}{18} \end{pmatrix}$
Properties of the Inverse of a Matrix
If A is an invertible matrix, then:
$A\cdot A^{-1} = A^{-1}\cdot A=I_{n}$
Example 49
$A=\begin{pmatrix}
1 & 3\\
2 & 5
\end{pmatrix}$
$A^{-1}= \begin{pmatrix} -5 & 3\\ 2 & -1 \end{pmatrix}$
$A\cdot A^{-1}= \begin{pmatrix} 1 & 3\\ 2 & 5 \end{pmatrix} \begin{pmatrix} -5 & 3\\ 2 & -1 \end{pmatrix}=$ $\begin{pmatrix} 1\cdot(-5)+3\cdot2 & 1\cdot3 + 3\cdot(-1)\\ 2\cdot(-5)+5\cdot2 & 2\cdot3 +5\cdot(-1) \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}= I_{2}$
$A^{-1}\cdot A= \begin{pmatrix} -5 & 3\\ 2 & -1 \end{pmatrix} \begin{pmatrix} 1 & 3\\ 2 & 5 \end{pmatrix}=$ $\begin{pmatrix} -5\cdot1 + 3\cdot2 & -5\cdot3 + 3\cdot 5\\ 2\cdot1 +(-1)\cdot2 & 2\cdot3 +(-1)\cdot5 \end{pmatrix}= \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}=I_{2}$