The Inverse of a Matrix and Matrix Pathology

• Division by a matrix is not defined but the equivalent is obtained from the inverse of the matrix.
• If the product of two square matrices, $A*B$, equals to the identity matrix, $I$, $B$ is said to be the inverse of $A$ (and also $A$ is the inverse of $B$).
• Matrices do not commute ($A*B \neq B*A$) on multiplication but inverses are an exception: $A*A^{-1}= A^{-1}*A$.
• To find the inverse of matrix $A$, use an elimination method.
• We augment the $A$ matrix with the identity matrix of the same size and solve. The solution is $A^{-1}$. Example;

$$A=\left[ \begin{array}{rrr} 1 & -1 & 2 \\ 3 & 0 & 1 \\ 1 & 0 & 2 \\ \end{array} \right],$$

$$\left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & 0\\ 3 & 0 & 1 & 0 &1 & 0\\ 1 & 0 & 2 & 0 &0 & 1\\ \end{array} \right],$$

$$\begin{array}{r} \\ R_2-(3/1)R_1 \rightarrow \\ R_3-(1/1)R_1 \rightarrow \\ \end{array}$$

$$\left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & 0\\ 0 & 3 &-5 &-3 &1 & 0\\ 0 & 1 & 0 &-1&0 & 1\\ \end{array} \right],$$

$$\left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & 0\\ 0 & 1 & 0 &-1&0 & 1\\ 0 & 3 &-5 &-3 &1 & 0\\ \end{array}\right],$$

$$\begin{array}{r} \\ \\ R_3-(3/1)R_2 \rightarrow \\ \end{array}$$

• Cont.

  
  

  $$\left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & 0\\ 0 & 1 & 0 &-1&0 & 1\\ 0 & 0 &-5 & 0 &1 &-3\\ \end{array}\right],$$
  
  

  
  

  $$\begin{array}{r} \\ \\ R_3/(-5) \\ \end{array},$$
  
  

  
  

  $$\begin{array}{r} R_1-(2/1)R_3 \rightarrow \\ \\ \\ \end{array},$$
  
  

  
  

  $$\left[ \begin{array}{rrrrrr} 1 & -1 & 0 & 1 &2/5&-6/5\\ 0... ... 0 &-1&0 & 1\\ 0 & 0 &1 & 0 &-1/5&3/5\\ \end{array}\right],$$
  
  

  
  

  $$\begin{array}{r} \\ R_2-(1/-1)R_1 \rightarrow \\ \\ \end{array},$$
  
  

  
  

  $$\left[ \begin{array}{rrrrrr} 1 & 0 & 0 & 0 &2/5&-1/5\\ 0 ... ...& 0 &-1&0 & 1\\ 0 & 0 &1 & 0 &-1/5&3/5\\ \end{array}\right]$$
  
  

• We confirm the fact that we have found the inverse by multiplication:

  
  

  $$\underbrace{\left[ \begin{array}{rrr} 1 & -1 & 2 \\ 3 & 0 & 1 \\ 1 & 0 & 2 \\ \end{array} \right]}_A*$$
  
  

  
  

  $$\underbrace{\left[ \begin{array}{rrr} 0 &2/5&-1/5\\ -1&0 & 1\\ 0 &-1/5&3/5\\ \end{array} \right]}_{A^{-1}}=$$
  
  

  
  

  $$\underbrace{\left[ \begin{array}{rrr} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 &1 \\ \end{array} \right]}_I$$
  
  

• It is more efficient to use Gaussian elimination. We show only the final triangular matrix; we used pivoting:
  
  
  
  $$\left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & 0\\ 3 & ... ... (0.333) & (0) & 1.667 & 0 &-0.333 & 1\\ \end{array} \right]$$
  
  

• After doing the back-substitutions, we get
  
  $$\left[ \begin{array}{rrrrrr} 3 & 0 & 1 & 0 &0.4& -0.2\\ (... ... (0.333) & (0) & 1.667 & 0 &-0.2 & 0.6\\ \end{array} \right]$$
  
  

• If we have the inverse of a matrix, we can use it to solve a set of equations, $Ax=b$,
• because multiplying by $A^{-1}$ gives the answer ($x$):
  
  $$\begin{array}{l} A^{-1}Ax=A^{-1}b \\ x=A^{-1}b\\ \end{array}$$