Multivariable-Calculus · Unit 4 · Video 4 · Interactive Practice

One Number Decides: det A, the Inverse, and Why the Origin Is the Only Solution

IKey Formulas

Formula	Name	What it takes
$A^{-1} = \dfrac{1}{\det A}\,\mathrm{adj}(A)$	Inverse from the adjoint	Minors, cofactor signs, transpose — then one division
$A \text{ is invertible} \iff \det A \neq 0$	Invertibility criterion	One number, read off the coefficients
$\det A \neq 0 \;\Rightarrow\; X = A^{-1}B$	Unique solution of $AX = B$	The inverse and the right-hand side
$AX = \mathbf{0}$ is solved by $X = \mathbf{0}$	Trivial solution	Nothing — every plane passes through the origin

Key Insight: The first three steps of the cofactor recipe always run; only the fourth — dividing by $\det A$ — can fail. And when it fails, no other method rescues it: $\det M \cdot \det A = \det I = 1$ is impossible once $\det A = 0$ .

IIVisualization 1 — Where the Recipe Stalls

Minors, cofactors and the transpose always go through; the fourth step divides by $\det A$ .

Bottom-right entry a₃₃ 3

IIIVisualization 2 — Three Planes Through the Origin

The first two planes meet in a line; $\det A$ decides whether the third plane contains it.

Bottom-right entry a₃₃ 3

💡 Scaling keeps a solution a solution: if $AX = \mathbf{0}$ , then $A(2X) = \mathbf{0}$ as well — so a single nonzero solution already brings infinitely many.

IVVisualization 3 — One Solution for Every B

Every right-hand side gets exactly one $X$ when $\det A = 4$ ; $B = \mathbf{0}$ gets the origin.

b₁ 1 b₂ 2 b₃ 3

VQuiz Questions

Problem 1 · Compute the Determinant, Then Decide

Given: $A = \begin{bmatrix} 1 & 2 & 0 \\ 3 & 1 & 2 \\ 0 & 1 & 1 \end{bmatrix}$ — expand along the first row and decide whether $A^{-1}$ exists.

✅ Correct!

\det A = -7 \neq 0

, so the division by

\det A

goes through and

A^{-1} = -\tfrac{1}{7}\mathrm{adj}(A)

❌ Check the cofactor signs. The middle term of a first-row expansion is subtracted:

-2(3 \cdot 1 - 2 \cdot 0) = -6

, not

+6

❌ A term went missing. You have the first minor,

1(1 \cdot 1 - 2 \cdot 1) = -1

; the

a_{12}

term still has to be subtracted.

❌ The determinant is right, the conclusion is not. The criterion is

\det A \neq 0

, not

\det A > 0

;

-7

is nonzero, so

A^{-1}

exists.

❌ Not quite. Expand along the first row with the

+\,-\,+

pattern:

1(1 - 2) - 2(3 - 0) + 0

Show solution

Expanding along the first row, minors first:

\det A = 1\begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} - 2\begin{vmatrix} 3 & 2 \\ 0 & 1 \end{vmatrix} + 0\begin{vmatrix} 3 & 1 \\ 0 & 1 \end{vmatrix}

\det A = 1(1 - 2) - 2(3 - 0) + 0 = -1 - 6 = -7

Since $\det A = -7 \neq 0$ , the matrix is invertible: the adjoint is built as always, and the last step divides by $-7$ instead of stalling. A negative determinant is no obstacle — only a zero determinant is.

Problem 2 · What Exactly Fails When $\det A = 0$

Given: a $3 \times 3$ matrix $A$ with $\det A = 0$ . Which statement is correct?

✅ Correct! Minors, cofactors and the transpose need no division, so

\mathrm{adj}(A)

always exists — and

\det(MA) = \det M \cdot \det A

rules out every inverse, not just this recipe's.

❌ The minors are fine. Each minor is a

2 \times 2

determinant; nothing in steps 1–3 divides by anything.

\mathrm{adj}(A)

comes out even when

\det A = 0

❌ Stronger than that. If some

M

had

MA = I

, then

\det M \cdot \det A = \det I = 1

— impossible with

\det A = 0

. No method can succeed.

❌ The homogeneous system always has one.

X = \mathbf{0}

satisfies

AX = \mathbf{0}

for every

A

; what a zero determinant costs you is uniqueness.

❌ Not quite. Ask which of the four steps needs a division.

Show solution

Steps 1–3 — minors, the checkerboard of signs, the transpose — are built from products and differences of entries. They never divide, so $\mathrm{adj}(A)$ exists for every square matrix.

Step 4 is $A^{-1} = \dfrac{1}{\det A}\mathrm{adj}(A)$ , and with $\det A = 0$ it asks for $\tfrac{1}{0}$ .

And nothing else works either. Suppose $MA = I$ for some $M$ . Taking determinants,

\det M \cdot \det A = \det I = 1

A product equal to $1$ has no zero factor, so $\det A \neq 0$ . Contrapositive: $\det A = 0$ means no inverse exists at all.

Problem 3 · Tune the Matrix Until the Origin Has Company

Given: the homogeneous system $AX = \mathbf{0}$ with $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & k & 2 \\ 1 & 1 & 3 \end{bmatrix}$ — find $\det A$ in terms of $k$ , then the value of $k$ at which the system gains a nonzero solution.

What is $\det A$ ?

For which $k$ does $AX = \mathbf{0}$ have a nonzero solution?

✅ Correct!

\det A = 2k - 8

vanishes at

k = 4

; for every other

k

the three planes through the origin meet there and nowhere else.

❌ Check the expansion. Along the first row:

1(3k - 2) - 2(6 - 2) + 1(2 - k)

— keep the middle term subtracted and collect the

k

's.

❌ That is only the first term.

1(3k - 2)

is the

a_{11}

cofactor alone; the other two,

-2(6 - 2)

and

+1(2 - k)

, still have to be added in — together they contribute

-8

and

2 - k

❌ The middle term is subtracted, not added. Adding it gives

(3k - 2) + 8 + (2 - k) = 2k + 8

; with the

+\,-\,+

pattern the term is

-2(6 - 2) = -8

❌ Check the sign inside the last minor. It is

2 \cdot 1 - k \cdot 1 = 2 - k

, not

2 + k

; reading it with the wrong sign turns

2k - 8

into

4k - 8

❌ Solve $\det A = 0$ . A nonzero solution appears exactly when the determinant vanishes.

❌ That is the trivial solution.

X = \mathbf{0}

does solve the system for every

k

— the question is when a nonzero solution joins it, and that needs

\det A = 0

❌ Not quite. Compute

\det A

as a function of

k

first, then set it to zero.

Show solution

Step 1 — expand along the first row:

\det A = 1(k \cdot 3 - 2 \cdot 1) - 2(2 \cdot 3 - 2 \cdot 1) + 1(2 \cdot 1 - k \cdot 1)

\det A = (3k - 2) - 2(4) + (2 - k) = 3k - 2 - 8 + 2 - k = 2k - 8

Step 2 — the criterion. With $\det A \neq 0$ the inverse exists and $X = A^{-1}\mathbf{0} = \mathbf{0}$ is the only solution. So a nonzero solution requires

2k - 8 = 0 \quad \Longrightarrow \quad k = 4

At $k = 4$ the three planes still all pass through the origin, but they no longer pin it down alone: the first two meet in a line that now lies inside the third.

Problem 4 · A Claimed Inverse, and What Survives

Given: $\det A = 0$ , yet a classmate claims to have found a matrix $M$ with $MA = I$ . Take determinants of both sides, using $\det(MA) = \det M \cdot \det A$ and $\det I = 1$ .

What do the two sides become?

With $\det A = 0$ , the homogeneous system $AX = \mathbf{0}$ :

✅ Correct!

\det M \cdot 0 = 0

can never equal

1

, so no inverse exists — and the uniqueness argument

X = A^{-1}\mathbf{0} = \mathbf{0}

loses the very

A^{-1}

it runs on, leaving the origin as one solution among possibly many.

❌ Substitute $\det A = 0$ . The left side is

\det M \cdot 0

, whatever

\det M

is; the right side is

\det I = 1

❌ $\det I$ is not zero. The identity has determinant

1

, so the equation reads

0 = 1

— a contradiction, not a consistency.

❌ The rule has no exceptions.

\det(MA) = \det M \cdot \det A

holds for all square

M, A

— which is exactly why

\det A = 0

is fatal.

❌ Check the trivial solution.

A\mathbf{0} = \mathbf{0}

for every

A

; what needs

\det A \neq 0

is the argument that no other solution exists.

❌ Solutions never come in twos. If some

X \neq \mathbf{0}

solves

AX = \mathbf{0}

, then so does

tX

for every scalar

t

— so the count is either exactly one or infinitely many.

❌ Not quite. Take determinants first, then ask which step of the uniqueness argument used

A^{-1}

Show solution

The claim cannot hold. Taking determinants of $MA = I$ :

\det M \cdot \det A = \det I = 1

With $\det A = 0$ the left side is $\det M \cdot 0 = 0$ , so the equation reads $0 = 1$ . No $M$ can do it.

What the homogeneous system keeps. $A\mathbf{0} = \mathbf{0}$ holds for every matrix, so $X = \mathbf{0}$ is still a solution — geometrically, all three planes pass through the origin.

What it loses. Uniqueness came from multiplying $AX = \mathbf{0}$ on the left by $A^{-1}$ :

A^{-1}(AX) = A^{-1}\mathbf{0} \quad \Longrightarrow \quad X = \mathbf{0}

That argument needs $A^{-1}$ , which no longer exists. So the origin remains a solution, but it need not be alone.

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