Topic 4.11 Notes – The Inverse and Determinant of a Matrix

Identity Matrices, Inverses, and Determinants

For $2\times2$ matrices, the identity matrix is

$$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$

It acts like the number 1 in regular multiplication. If $A$ is any $2\times2$ matrix, then

$$AI=IA=A.$$

An inverse of a matrix $A$ is written $A^{-1}$. It is the matrix that undoes $A$, which means

$$AA^{-1}=A^{-1}A=I.$$

That does not mean “take reciprocals of each entry.” Matrix inverse is a whole-matrix operation.

For $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the determinant is

$$\det(A)=ad-bc.$$

You may also see determinant written with vertical bars:

$$\left|\begin{matrix}a&b\\c&d\end{matrix}\right|=ad-bc.$$

The one connection you want locked in is this:

• $A^{-1}$ exists exactly when $\det(A)\neq 0$

Finding the Inverse of a $2\times2$ Matrix

When $\det(A)\neq 0$, the inverse formula is

$$A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.$$

Here is the hand process in order:

1. Find the determinant $ad-bc$.
2. If it equals 0, stop. No inverse exists.
3. Swap the main diagonal entries $a$ and $d$.
4. Negate both off-diagonal entries $b$ and $c$.
5. Multiply by $\dfrac{1}{\det(A)}$.

Example with $A=\begin{bmatrix}3&1\\2&4\end{bmatrix}$

• $\det(A)=3(4)-1(2)=10$
• So $$A^{-1}=\frac{1}{10}\begin{bmatrix}4&-1\\-2&3\end{bmatrix}$$

Parameters

If a matrix has a variable in it, use the determinant to find when the inverse breaks.

Example
$M=\begin{bmatrix}k&2\\3&k-1\end{bmatrix}$

$$\det(M)=k(k-1)-6=k^2-k-6=(k-3)(k+2)$$

So $M$ is not invertible when $k=3$ or $k=-2$. For all other $k$,

$$M^{-1}=\frac{1}{k^2-k-6}\begin{bmatrix}k-1&-2\\-3&k\end{bmatrix}$$

Verifying an inverse

Check by matrix multiplication, not by multiplying matching entries.

• Compute $AA^{-1}$ or $A^{-1}A$
• You should get $I$

Determinants as Geometry and Invertibility

A $2\times2$ matrix can be seen as two row vectors or two column vectors in $\mathbb R^2$. Those two vectors form a parallelogram. Here, the columns of $A=\begin{bmatrix}3&1\\2&4\end{bmatrix}$ make a parallelogram with area $10$.

Parallelogram from the columns of a $2\times2$ matrix

The geometric fact is:

• $|\det(A)|$ = area of the parallelogram
• If $\det(A)=0$, the vectors are parallel and the shape collapses to area 0

Illustrative examples that all point to the same idea:

• column vectors form the parallelogram
• row vectors can also form it
• parallelogram area is $|\det(A)|$
• parallel vectors when the determinant is 0 means no area

A negative determinant is still fine. The sign can change if you switch vector order, but the area stays the same because area uses absolute value.

Equivalent ways to spot invertibility:

• $\det(A)\neq 0$
• inverse exists
• row vectors are not parallel
• column vectors are not parallel
• parallelogram area is positive

Equivalent ways to spot noninvertibility:

• $\det(A)=0$
• inverse does not exist
• row or column vectors are parallel
• parallelogram has area 0

When to Use Each Idea

• Use the determinant first if asked whether a matrix is invertible.
• Use the determinant first before using the inverse formula, because division by 0 would make the formula invalid.
• Use the inverse formula when asked for exact $A^{-1}$.
• Use determinant reasoning in vector questions about area or parallelism.
• With parameters, solve $\det(A)=0$ to find the values where invertibility changes.
• Technology can find $\det(A)$ or $A^{-1}$, but you still need to interpret the result and state restrictions.

Common Mistakes and Fast Checks

• Mixing up $ad-bc$ with $ad+bc$
• Forgetting that determinant 0 means no inverse
• Swapping the wrong entries or forgetting to negate both off-diagonal entries
• Treating $A^{-1}$ like entry-by-entry reciprocals
• Using $\det(A)$ instead of $|\det(A)|$ for area
• Missing parameter values that make the denominator 0
• After finding an inverse, multiply and make sure you get exactly $I$