3.3 The Inverse Matrix and the Determinant
Introduction
To solve the scalar equation $ax = b$, you multiply both sides by $a^{-1} = \frac{1}{a}$ (provided $a \neq 0$) and get $x = a^{-1}b$. Could we do the same thing with a matrix equation $A\vec{x} = \vec{b}$? If there exists a “matrix version of the reciprocal” for $A$, we could multiply both sides by it and immediately find $\vec{x}$. That matrix is the inverse matrix, and the idea is exactly this simple.
But just as $\frac{1}{0}$ does not exist for scalars, some matrices have no inverse either. The tool that tells us whether an inverse exists is the determinant.
Definition of the Inverse Matrix
Definition
For an $n \times n$ square matrix $A$, the inverse matrix $A^{-1}$ is the matrix satisfying
\[A A^{-1} = A^{-1} A = I\]A matrix that has an inverse is called invertible (or non-singular). A matrix without an inverse is called singular.
Uniqueness
If an inverse exists, it is unique. Suppose both $B$ and $C$ are inverses of $A$. Then:
\[B = BI = B(AC) = (BA)C = IC = C\]Properties of the Inverse
\((A^{-1})^{-1} = A\) \((AB)^{-1} = B^{-1}A^{-1} \quad \text{(note the reversal!)}\) \((A^T)^{-1} = (A^{-1})^T\) \((cA)^{-1} = \frac{1}{c}A^{-1} \quad (c \neq 0)\)
The 2x2 Inverse Formula
Formula
\[A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \implies A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\]This works only when $ad - bc \neq 0$.
How to Remember It
Swap the diagonal entries ($a \leftrightarrow d$), flip the signs of the off-diagonal entries ($b \to -b$, $c \to -c$), and divide by $ad - bc$.
Example
\[A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\] \[\det A = 3 \cdot 2 - 1 \cdot 5 = 6 - 5 = 1\] \[A^{-1} = \frac{1}{1}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\]Verification: $AA^{-1} = \begin{pmatrix}3&1\5&2\end{pmatrix}\begin{pmatrix}2&-1\-5&3\end{pmatrix} = \begin{pmatrix}1&0\0&1\end{pmatrix} = I$ $\checkmark$
The Determinant
The 2x2 Determinant
\[\det A = \det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc\]| This is also written as $ | A | $. |
Geometric Meaning – Area Scaling Factor
| In two dimensions, the absolute value of the determinant equals the area of the parallelogram spanned by the column vectors of $A$. More generally, $A$ scales the area of every shape by a factor of $ | \det A | $. |
$ \det A > 1$: areas are enlarged. $ \det A = 1$: areas are preserved (an area-preserving transformation). $ \det A = 0$: areas collapse to zero – the plane is squished onto a line (or a point). - $\det A < 0$: the orientation is flipped.
The 3x3 Determinant
\[\det\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = a_{11}(a_{22}a_{33}-a_{23}a_{32}) - a_{12}(a_{21}a_{33}-a_{23}a_{31}) + a_{13}(a_{21}a_{32}-a_{22}a_{31})\]This is called the Laplace expansion (or cofactor expansion) along the first row.
Properties of the Determinant
Determinant of a Product
\[\det(AB) = \det(A) \cdot \det(B)\]This is an extremely powerful property. The determinant of a product equals the product of the determinants.
Determinant of the Transpose
\[\det(A^T) = \det(A)\]Determinant of the Inverse
\[\det(A^{-1}) = \frac{1}{\det(A)}\]This follows from $\det(AA^{-1}) = \det(I) = 1$, so $\det(A)\det(A^{-1}) = 1$.
Determinant of a Scalar Multiple
\[\det(cA) = c^n \det(A) \quad (n \times n \text{ matrix})\]The Invertibility Condition
\[\boxed{A^{-1} \text{ exists} \quad \Leftrightarrow \quad \det A \neq 0}\]This is the fundamental test. If $\det A = 0$, then $A$ is singular and has no inverse.
Geometrically, $\det A = 0$ means the transformation collapses space into a lower dimension. Information is lost, so there is no way to reverse the transformation and recover the original vectors.
\[A\vec{x} = \vec{b} \text{ has a unique solution} \quad \Leftrightarrow \quad \det A \neq 0 \quad \Leftrightarrow \quad \vec{x} = A^{-1}\vec{b}\]Key Takeaways
| Concept | Description / Formula | ||
|---|---|---|---|
| Inverse definition | $AA^{-1} = A^{-1}A = I$ | ||
| 2x2 inverse | $A^{-1} = \frac{1}{ad-bc}\begin{pmatrix}d&-b\-c&a\end{pmatrix}$ | ||
| $(AB)^{-1}$ | $B^{-1}A^{-1}$ (order reversal) | ||
| 2x2 determinant | $\det A = ad - bc$ | ||
| Geometric meaning | Area (volume) scaling factor $= | \det A | $ |
| Determinant of product | $\det(AB) = \det A \cdot \det B$ | ||
| Invertibility condition | $\det A \neq 0$ | ||
| Solving linear systems | If $\det A \neq 0$, then $\vec{x} = A^{-1}\vec{b}$ |
In the next post, we explore linear transformations – how matrices transform vector spaces, and what that looks like geometrically.