LINEAR-ALGEBRA · Unit 3 · Video 2 · Interactive Practice

Matrix Inverses: What They Are and When They Don't Exist

IKey Formulas

Formula	Name	When to Use
$A^{-1}A = AA^{-1} = I$	Inverse definition	Defines what "inverse" means
$\det\!\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc$	2×2 Determinant	Quick singularity check
$\det(A) = 0 \iff A$ is singular	Singularity test	Zero determinant means no inverse exists
$Ax = 0$ has nontrivial solution $\iff A$ is singular	Null space test	Non-zero $x$ in null space proves no inverse

IIVisualization 1 — A Matrix Deforms the Plane

Multiplying by $A$ warps the unit square into a parallelogram of area $|\det A|$ — and zero area means no inverse.

a 2 b 1 c 1 d 2

Only whether $\det A$ is zero decides invertibility — its sign merely records whether the transformation flips the plane's orientation.

IIIVisualization 2 — When Columns Span the Plane

The columns of $A$ span the whole plane only when they point in different directions.

col₂ x 4 col₂ y 1

IVVisualization 3 — A Whole Line Maps to One Point

For the singular matrix $A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$ , infinitely many inputs share the single output $(1, 2)$ .

null-space scalar k 0

If $A^{-1}$ existed, $A^{-1}(1,2)$ would have to return one vector — but infinitely many inputs give $(1,2)$ , so no reverse map can exist.

VQuiz Questions

Question 1

Consider the matrix $A = \begin{bmatrix} 3 & 6 \\ 1 & 2 \end{bmatrix}$ .

What is $\det(A)$ , and is $A$ invertible?

✅ Correct! The determinant is zero, so $A$ has no inverse.

❌ Not quite. Compute $ad - bc$ carefully: $(3)(2) - (6)(1) = ?$

Show solution

Solution:

For $A = \begin{bmatrix} 3 & 6 \\ 1 & 2 \end{bmatrix}$ , the determinant is:

$\det(A) = ad - bc = (3)(2) - (6)(1) = 6 - 6 = 0$

Since $\det(A) = 0$ , the matrix is singular (not invertible).

Notice that column 2 $= (6, 2) = 2 \times (3, 1) = 2 \times$ column 1, confirming the columns are linearly dependent.

Question 2

True or False: If $A$ is a square matrix and $Ax = 0$ has only the trivial solution $x = 0$ , then $A$ is invertible.

✅ Correct! Only the trivial null space means the columns are independent and $A$ is invertible.

❌ Not quite. Recall from the video: a non-trivial null vector is 'the exact dividing line' for singularity.

Show solution

Solution: True

This is a fundamental equivalence in linear algebra. For a square matrix $A$ :

$Ax = 0 \text{ has only } x = 0 \;\;\iff\;\; \text{columns of } A \text{ are linearly independent} \;\;\iff\;\; A \text{ is invertible}$

The video showed the converse direction: if a non-trivial solution exists (like $x = (3, -1)$ for the singular matrix), then $A$ cannot be invertible — proved by contradiction.

The contrapositive gives us this result: no non-trivial null space vector means $A$ is invertible.

Question 3

In the proof by contradiction from the video, we assume $A^{-1}$ exists for $A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$ , knowing that $Ax = 0$ where $x = (3, -1) \neq 0$ .

What contradiction do we arrive at?

✅ Correct! The assumption forces $x = 0$ , contradicting $x = (3, -1) \neq 0$ .

❌ Not quite. Trace the algebra: multiply $Ax = 0$ on the left by $A^{-1}$ and simplify step by step.

Show solution

Solution:

The proof proceeds step by step:

Assume $A^{-1}$ exists.
Given: $Ax = 0$ where $x = (3, -1) \neq 0$ .
Multiply both sides on the left by $A^{-1}$ :

$A^{-1}(Ax) = A^{-1} \cdot 0$

By associativity: $(A^{-1}A)x = 0$
By definition of inverse: $Ix = 0$
Therefore: $x = 0$

But we chose $x = (3, -1) \neq 0$ . This is a contradiction! ⚡

Therefore, our assumption was wrong — $A^{-1}$ does not exist.

Question 4

Which of the following matrices is invertible?

✅ Correct! $B$ has determinant $-2 \neq 0$ , making it the only invertible option.

❌ Not quite. Compute $ad - bc$ for each matrix. Which one gives a nonzero result?

Show solution

Solution:

Compute the determinant $ad - bc$ for each:

Matrix	Determinant	Invertible?
$A = \begin{bmatrix} 2 & 4 \\ 3 & 6 \end{bmatrix}$	$(2)(6) - (4)(3) = 12 - 12 = 0$	❌ Singular
$B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$	$(1)(4) - (2)(3) = 4 - 6 = -2$	✅ Invertible
$C = \begin{bmatrix} 5 & 10 \\ 1 & 2 \end{bmatrix}$	$(5)(2) - (10)(1) = 10 - 10 = 0$	❌ Singular
$D = \begin{bmatrix} 0 & 0 \\ 1 & 2 \end{bmatrix}$	$(0)(2) - (0)(1) = 0$	❌ Singular

Only $B$ has a nonzero determinant ( $-2$ ), so it is the only invertible matrix. Note: the sign of the determinant doesn't matter — only whether it equals zero or not.

Solved: 0 / 4