Multivariable-Calculus · Unit 4 · Video 3 · Interactive Practice

Three Planes, One Point: Why AX = B Usually Has Exactly One Solution, and When It Does Not

IKey Formulas

Formula	Name	What you need
$AX = B$	Matrix form of a $3 \times 3$ system	Coefficients in $A$ , with a $0$ for every missing variable
$X = A^{-1}B$	Solution of the system	$A^{-1}$ exists — and the inverse multiplies on the left
$\mathcal{P}_1 \cap \mathcal{P}_2$ is a line along $\mathbf{N}_1 \times \mathbf{N}_2$	Line of the first two planes	Two non-parallel normals
$\operatorname{dist}(O, \mathcal{P}) = \dfrac{\|d\|}{\|\mathbf{N}\|}$	Distance from the origin to $ax + by + cz = d$	$\mathbf{N} = \langle a, b, c \rangle$

Key Insight: Exactly two solutions is impossible — if $P$ and $Q$ both satisfy every equation, so does every point of the line through them. The solution set is empty, a single point, a line, or a plane, and nothing else.

IITwo Planes Fix a Line, the Third Picks a Point

Every point of $\mathcal{P}_1 \cap \mathcal{P}_2$ satisfies the first two equations; the third keeps exactly one of them.

💡 The picture assumes $\mathcal{P}_1$ and $\mathcal{P}_2$ really cross. Were they parallel but distinct there would be no line to start from and no solution whatever $\mathcal{P}_3$ does; were they the same plane the line would widen into that plane, and three copies of one equation make the solution set that whole plane.

IIIWhen the Third Plane Misses the Line

Tilt or shift $\mathcal{P}_3$ and the single point becomes no solution — or the whole line.

P₃ tilt (k) 3 P₃ shift (d) 3

💡 The coefficient of $t$ in that substitution is $\mathbf{N}_3 \cdot (\mathbf{N}_1 \times \mathbf{N}_2) = \det A$ , here $1 + k$ : that one number decides whether the line pierces $\mathcal{P}_3$ at all, and when it vanishes the constant $d$ settles which degenerate case you are in — the whole line at $d = 3$ , nothing otherwise. Reading it off $A$ is the next video.

IVWhy the Inverse Goes on the Left

Left-multiplying $AX = B$ by $A^{-1}$ collapses the left side to $X$ ; the reversed product does not exist.

Step 1

Step 1 — Three equations, one matrix equation

\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 2 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}

Row 1 of

A

dotted with

X

reproduces

x + z = 1

; the

0

holds the place of the missing

y

Step 2 — Left-multiply both sides by $A^{-1}$

A^{-1}(AX) = A^{-1}B

Step 3 — The pair $A^{-1}A$ is the identity

(A^{-1}A)X = IX = X

X = A^{-1}B \ \checkmark

Step 4 — In this order the sizes agree

\underbrace{(3 \times 3)}_{A^{-1}}\ \underbrace{(3 \times 1)}_{B} = 3 \times 1

The inner sizes are both

3

, and the product is a column — which is what

X

is.

Step 5 — In the other order they do not

BA^{-1}: \quad \underbrace{(3 \times 1)}_{B}\ \underbrace{(3 \times 3)}_{A^{-1}}

The inner sizes

1

and

3

disagree, so

BA^{-1}

is undefined. Here

X = \langle 1, 1, 0 \rangle

\ 1 + 0 = 1

\ 1 + 1 = 2

\ 1 + 2 + 0 = 3 \ \checkmark

VQuiz Questions

Problem 1 · Reading Off the Coefficient Matrix

Given: the system $x + z = 1$ , $\ x + y = 2$ , $\ x + 2y + 3z = 3$ written as $AX = B$ — find the second row of $A$ .

✅ Correct!

x + y = 2

carries no

z

, so a

0

holds that column.

❌ That is the first row.

x + z = 1

gives

\begin{pmatrix} 1 & 0 & 1 \end{pmatrix}

; the second equation is

x + y = 2

❌ A row needs one entry per variable. A missing variable contributes

0

, not a missing entry — otherwise

A

would not be

3 \times 3

and

AX

would not exist.

❌ That is the third row.

x + 2y + 3z = 3

gives

\begin{pmatrix} 1 & 2 & 3 \end{pmatrix}

❌ Not quite. Read the coefficients of

x

y

z

— in that order — out of

x + y = 2

Show solution

Write every equation with all three variables showing:

\begin{aligned} x + 0y + \ \ z &= 1 \\ x + \ \ y + 0z &= 2 \\ x + 2y + 3z &= 3 \end{aligned}

The coefficients, row by row, are the rows of $A$ :

A = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 2 & 3 \end{pmatrix}, \qquad B = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}

The second row is $\begin{pmatrix} 1 & 1 & 0 \end{pmatrix}$ , and row 2 of $A$ dotted with $X$ returns $x + y = 2$ .

Problem 2 · Which Side Does the Inverse Go On?

Given: $A$ is an invertible $3 \times 3$ matrix and $B$ is the $3 \times 1$ column of constants in $AX = B$ — find the expression that gives $X$ .

✅ Correct! Left-multiplying by

A^{-1}

turns

AX

into

IX = X

, and

(3 \times 3)(3 \times 1)

returns the column

X

❌ The sizes forbid that order.

BA^{-1}

(3 \times 1)(3 \times 3)

: the inner sizes

1

and

3

disagree, so the product does not exist.

❌ $AB$ is a legal product, but the wrong one. Multiplying by

A

again does not undo

A

— only

A^{-1}A = I

does.

❌ $B$ has no inverse. It is

3 \times 1

, not square, so

B^{-1}

is not defined.

❌ Not quite. Undo

A

by left-multiplying both sides of

AX = B

A^{-1}

Show solution

Left-multiply both sides of $AX = B$ by $A^{-1}$ and use associativity:

A^{-1}(AX) = A^{-1}B \quad\Longrightarrow\quad (A^{-1}A)X = A^{-1}B \quad\Longrightarrow\quad IX = X = A^{-1}B

The sizes decide the side. $A^{-1}B$ is $(3 \times 3)(3 \times 1)$ : inner sizes $3$ and $3$ agree and the product is the $3 \times 1$ column $X$ . Reversed, $BA^{-1}$ is $(3 \times 1)(3 \times 3)$ : the inner sizes are $1$ and $3$ , so no product exists — matrix multiplication is not commutative, and here the other order is not even defined.

Problem 3 · Parallel, or Contained?

Given: $x + z = 1$ and $x + y = 2$ meet in the line $(1 - t,\ 1 + t,\ t)$ , and the third equation is $x + 2y - z = d$ — find what the third equation becomes on that line, and then the value of $d$ that makes the system have infinitely many solutions.

Substituting the line into $x + 2y - z$ gives

So the system has infinitely many solutions when

✅ Correct! The

t

terms cancel, so the third equation reads

3 = d

: at

d = 3

it is

0 = 0

and the whole line solves the system; at any other

d

it is a contradiction and nothing does.

❌ Check the substitution.

(1-t) + 2(1+t) - t

: the

t

terms are

-t + 2t - t

❌ Compare the two sides. The left side is the constant

3

for every point of the line, so the equation can only hold if

d

is that same constant.

Show solution

Put $x = 1 - t$ , $y = 1 + t$ , $z = t$ into the left side of the third equation:

(1 - t) + 2(1 + t) - t = 1 - t + 2 + 2t - t = 3

Every $t$ cancels, so the third equation reads $3 = d$ at every point of the line — the line is parallel to $\mathcal{P}_3$ , which the normals confirm: $\langle 1, 2, -1 \rangle \cdot \langle -1, 1, 1 \rangle = -1 + 2 - 1 = 0$ .

$d = 3$ : the third equation says $3 = 3$ , that is $0 = 0$ . It adds nothing, every point of the line solves the system, and there are infinitely many solutions — the line lies inside $\mathcal{P}_3$ .
$d \ne 3$ , say $d = 5$ : the third equation says $3 = 5$ , a contradiction of the $1 = 2$ kind. The line runs alongside $\mathcal{P}_3$ and never meets it, so there is no solution.

Note what cannot happen: two solutions. Two distinct solutions would force every point of the line through them to be a solution as well.

Problem 4 · What the Constant Measures

Given: the plane $x + y + z = 4$ — find its distance from the origin.

✅ Correct!

\frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx 2.31

, the constant measured in units of

|\mathbf{N}|

❌ The constant is the distance only for a unit normal. Here

\mathbf{N} = \langle 1, 1, 1 \rangle

has length

\sqrt{3}

, not

1

❌ That divides by $|\mathbf{N}|^2 = 3$ . The distance divides by the length

|\mathbf{N}| = \sqrt{3}

❌ Multiplied instead of divided. A longer normal means the same constant pushes the plane less far from the origin.

❌ Not quite. Divide the constant by the length of the normal vector

\langle 1, 1, 1 \rangle

Show solution

The normal vector is $\mathbf{N} = \langle 1, 1, 1 \rangle$ , so $|\mathbf{N}| = \sqrt{1 + 1 + 1} = \sqrt{3}$ , and the constant is $d = 4$ :

\operatorname{dist}(O, \mathcal{P}) = \frac{|d|}{|\mathbf{N}|} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx 2.31

Changing the constant slides the plane to a parallel one, since the normal does not change: $x + y + z = 0$ passes through the origin, and $x + y + z = 4$ has been pushed $4$ units out along $\mathbf{N}$ — but those units are $|\mathbf{N}| = \sqrt{3}$ long each, which is why the distance is $4/\sqrt{3}$ rather than $4$ .

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