The inverse of a matrix $A$, denoted as $A^{-1}$, is a matrix that, when multiplied by $A$, results in the identity matrix $\mathbf I$. Mathematically: $A \cdot A^{-1}= A^{-1} \cdot A=\mathbf I$ ^81f95f **Definition:** For a $2 \times 2$ matrix, the inverse is given by the following $ A^{-1} = \frac{1}{\det (A)} \begin{bmatrix} \phantom{-}d & -b \\-c & \phantom{-}a\end{bmatrix} $ where: $ A=\begin{bmatrix} \phantom{-}d & -b \\-c & \phantom{-}a\end{bmatrix}, \quad \det(A)= ad-bc$ For larger matrices, the inverse is computed using methods like *Gaussian Elimination*, reducing the matrix to the identity form while applying the same row operations to an augmented matrix. **Conditions:** Not all matrices have an inverse. For $A$ to be invertible, it must satisfy the following: - *Square Matrix:* $A$ must have the same number of rows and columns. - *Non-Singular Matrix:* The [[Matrix Determinant|Determinant]] of $A$ must be non-zero $\det(A) \neq 0$. ## Solving Systems of Equations Using the Inverse Matrix Consider the following system of equations. $ \begin{align} a+b+3c &= 15 \\ a+2b+4c &= 21 \\ a+b+2c &= 13 \end{align} $ We can represent these equations in matrix form, where the coefficients on the left-hand side form a matrix $\mathbf A$, the variables form a column vector $r$, and the constants on the right-hand side form a column vector $s$: $ \underbrace{\begin{bmatrix} 1 & 1 & 3 \\ 1 & 2 & 4 \\ 1 & 1 & 2 \end{bmatrix}}_{\mathbf A} \cdot \underbrace{\begin{bmatrix} a \\ b \\ c \end{bmatrix}}_{r} = \underbrace{\begin{bmatrix} 15 \\ 21 \\13 \end{bmatrix}}_{s} $ **Using the Inverse Matrix:** To solve for $r$, we multiply both sides of the matrix equation by the inverse of $\mathbf A$, denoted as $\mathbf A^{-1}$. We use the property that $\mathbf A \cdot \mathbf A^{-1} = \mathbf I$, where $\mathbf I$ is the identity matrix. $ \begin{align} A \cdot r&=s \\ A^{-1}(A \cdot r) &= A^{-1} \cdot s \\ \mathbf I \cdot r &= A^{-1} \cdot s\\ r &= A^{-1} \cdot s \end{align} $ **Gaussian Elimination:** We use row-wise elimination, to make $A$ on the left-hand side equal to the identity matrix. $ \begin{align} &\overbrace{\begin{bmatrix} 1 & 1 & 3 \\ 1 & 2 & 4 \\ 1 & 1 & 2 \end{bmatrix}}^{A} \, A^{-1} &= \overbrace{\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}}^{\mathbf I} \\[12pt] &\begin{bmatrix} 1 & 1 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \, A^{-1} &= \begin{bmatrix} \phantom{-}1 & \phantom{-}0 & \phantom{-}0 \\ -1 & \phantom{-}1 & \phantom{-}0 \\ \phantom{-}1 & \phantom{-}0 & -1 \end{bmatrix} \\[12pt] &\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \, A^{-1} &= \begin{bmatrix} -2 & \phantom{-}0 & \phantom{-}3\\ -2 & \phantom{-}1&\phantom{-}1 \\ \phantom{-}1 & \phantom{-}0 & -1 \end{bmatrix} \\[12pt] &\underbrace{\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}}_{\mathbf I} \, A^{-1} &= \begin{bmatrix} \phantom{-}0 &-1&\phantom{-}2 \\ -2&\phantom{-}1&\phantom{-}1 \\ \phantom{-}1&\phantom{-}0 & -1 \end{bmatrix} \end{align} $ Once this is done the left-hand side is equal to $\mathbf I A^{-1}$ which is equal to $A^{-1}$ itself (right-hand side). Finally we can compute $r$ from the equation above $r=s*A^{-1}$. $ r= \begin{bmatrix} \phantom{-}0 &-1&\phantom{-}2 \\ -2&\phantom{-}1&\phantom{-}1 \\ \phantom{-}1&\phantom{-}0 & -1 \end{bmatrix} * \begin{bmatrix} 15\\21\\13 \end{bmatrix} = \begin{bmatrix} 5\\4\\2 \end{bmatrix} $