Invertible matrix theorem — 12+ equivalent conditions: Worked examples

Definition The symbols and words used on this page

$n \times n$ / square matrix: a grid of numbers with the same number of rows and columns ( $n$ of each). The theorem only speaks about square matrices; we meet one non-square matrix on purpose to see the theorem stay silent.
$I$ (identity matrix): the square matrix with $1$ s down the diagonal and $0$ s elsewhere; e.g. $I_{2} = (1001)$ , and $I_{3}$ is its $3 \times 3$ version. It is the "do-nothing" matrix: $A I = I A = A$ . Being able to reach $I$ by row operations is what "invertible" means.
$R^{n}$ : the space of all lists of $n$ real numbers, e.g. $R^{2}$ is the flat plane of all $(x_{1}, x_{2})$ , and $R^{3}$ is ordinary $3$ D space.
$A x = b$ : the matrix $A$ acting on an input vector $x$ to give an output vector $b$ . "Solvable for every $b$ " asks: can the machine hit any target?
$0$ (zero vector) and null space: $0 = (0, \dots, 0)$ . The null space is every input $x$ that the machine crushes to $0$ . If only $x = 0$ is crushed, the null space is trivial — a good sign.
pivot & RREF: when you tidy a matrix with row reduction, each row's first nonzero leading entry (after cleaning up) is a pivot. The fully tidied form is the Reduced Row Echelon Form (RREF). A square matrix with a pivot in every row and column has RREF equal to $I$ — that is exactly invertibility.
rank: the number of pivots = the number of genuinely independent directions in the rows or columns. "Full rank" means rank $= n$ .

The scenario matrix

Every square matrix falls into exactly one of two camps — invertible or not — but the reason it lands there, and the cheapest way to see it, varies. Here is the full grid of cases we will cover. The "Cheapest test" column names the test from the box above.

Cell	Scenario class	What makes it interesting	Cheapest test	Example
A	Tiny $2 \times 2$ , $det \neq = 0$	baseline invertible	[det]	Ex 1
B	Tiny $2 \times 2$ , $det = 0$	baseline singular	[det]	Ex 2
C	Degenerate: a zero row/column	one row is all zeros	spot the zero → [RREF]	Ex 3
D	Degenerate: repeated / proportional rows	dependence hides in plain sight	[null] / [indep]	Ex 4
E	$3 \times 3$ triangular	limiting-easy determinant	diagonal product → [det]	Ex 5
F	Eigenvalue angle — is $0$ an eigenvalue?	connects to spectrum	[eig]	Ex 6
G	Non-square trap	theorem does NOT apply	squareness check first	Ex 7
H	One-sided inverse twist	right inverse $\Rightarrow$ full inverse	[side]	Ex 8
I	Word problem — mixing recipe	real-world solvability of $A x = b$	[solve]	Ex 9
J	Full $3 \times 3$ ( $det$ by cofactors) + parameter twist	non-triangular determinant, then "for which $t$ does it break?"	[det] (cofactor / as a function of $t$ )	Ex 10, Ex 11

The two big camps and the tests that split them are shown in the flowchart below. Its consequence nodes are exactly the eight named tests from the formula box — so the picture and the box speak the same language.

Intuition How to read this flowchart

Start at the top node, "Square n x n matrix". The first fork ("det zero ?") is the single decisive test [det] — an arrow to the left (labelled yes) if $det A = 0$ , to the right (labelled no) otherwise. Once you land on NOT invertible (left) or invertible (right), the arrows below fan out into the consequences.

Every consequence node is one of the eight tests from the box, named the same way:

Left branch (failure): [indep] fails (dependent columns), [null] fails (a nonzero null vector), [eig] fails ( $0$ is an eigenvalue), and rank drops below $n$ ([rank] fails).
Right branch (success): [rank] holds (rank $= n$ ), [indep] holds (independent columns), and [solve] holds (every $b$ reachable).

These are not new facts to learn — they are the same tests you already read in the box, re-drawn as arrows so you can see that one fork decides them all. We keep only three consequence nodes on the success side (rather than every one) because those three — full rank, independent columns, "hits every target" — are the ones the worked examples below actually use; the rest ride along silently. Every example is really just this picture: "check the fork [det], then read the whole branch."

The worked examples

Cell A — baseline invertible $2 \times 2$

Worked example Example 1 —

$det \neq = 0$ , everything works Is $A = (3214)$ invertible?

Forecast: Guess before reading — does a matrix whose rows clearly point in different directions look reversible?

Step 1. Compute the determinant $det A = (3) (4) - (1) (2) = 12 - 2 = 10$ . Why this step? For a $2 \times 2$ , test [det] is the single cheapest condition — one multiplication, one subtraction. And $det \neq = 0$ instantly triggers the entire theorem.

Step 2. $det A = 10 \neq = 0 \Rightarrow$ invertible. Read off the rest for free. Why this step? Because all the tests in the box are equivalent, confirming one ([det]) automatically confirms every other: columns linearly independent, rank $= 2$ , null space $= {0}$ , $0$ is not an eigenvalue. No extra work needed — that is the whole point of the theorem.

Step 3. Since it is invertible we can write the inverse using the $2 \times 2$ formula $A^{- 1} = \frac{1}{d e t A} (d - c - b a)$ : $A^{- 1} = \frac{1}{10} (4 - 2 - 1 3) .$ Why this step? Having decided invertibility, it is worth exhibiting the actual inverse so we can sanity-check it directly against $I$ in the next line.

Verify: $A A^{- 1} = \frac{1}{10} (3214) (4 - 2 - 1 3) = \frac{1}{10} (100010) = I$ . ✔

Cell B — baseline singular $2 \times 2$

Worked example Example 2 —

$det = 0$ , everything fails together Is $A = (2163)$ invertible?

Forecast: Look at the two rows. Do you notice a hidden relationship before computing anything?

Step 1. Compute $det A = (2) (3) - (6) (1) = 6 - 6 = 0$ . Why this step? Test [det] again — cheapest for $2 \times 2$ , and $det = 0$ is the alarm bell for the whole failing side of the theorem.

Step 2. $det A = 0 \Rightarrow$ not invertible. Every equivalent condition fails: columns dependent, rank $= 1 < 2$ , a nonzero null vector exists, $0$ is an eigenvalue. Why this step? Equivalence runs both ways: if the one cheap test ([det]) fails, all the others must fail too. So a single zero determinant condemns the whole list at once.

Step 3. Find the dependence: row 1 $= 2 \times$ row 2. So column 2 $= 3 \times$ column 1 (since $6 = 3 \cdot 2$ , $3 = 3 \cdot 1$ ). The null vector satisfies $x_{1} (col_{1}) + x_{2} (col_{2}) = 0$ , i.e. $3 (col_{1}) - (col_{2}) = 0$ , so $x = (3, - 1)^{T}$ . Why this step? Producing the null vector proves test [null] fails by direct witness — a concrete nonzero input the machine crushes to $0$ — so we do not rely on the determinant alone.

Verify: $A (3 - 1) = (2 \cdot 3 + 6 \cdot (- 1) 1 \cdot 3 + 3 \cdot (- 1)) = (00)$ . ✔ A nonzero input squashed to zero — the machine destroyed information.

Cell C — degenerate: a zero row

Worked example Example 3 — a whole row of zeros

Is $A = 501703209$ invertible?

Forecast: A row of all zeros — can such a matrix ever be reversed?

Step 1. Spot the zero row (row 2). A zero row means row reduction can never place a pivot (a leading nonzero entry) in that row. Why this step? Test [RREF] needs a pivot in every row. A zero row forbids that immediately — no arithmetic required.

Step 2. With fewer than $3$ pivots, the RREF (the fully tidied form) cannot be $I_{3}$ , so not invertible. Why this step? [RREF] says invertible $⟺$ the tidied form equals $I_{3}$ . A missing pivot means the tidied form still has a zero row, so it can never match the identity — condemning invertibility.

Step 3. Cross-check with the determinant: expanding along the zero row gives $det A = 0$ (every term multiplies a zero entry). Why this step? Running an independent test ([det]) that must agree with [RREF] guards against a slip — two equivalent conditions giving the same verdict raises our confidence.

Verify: $det A = 0$ (cofactor expansion along row 2 — all terms vanish). ✔ Both tests agree.

Cell D — degenerate: proportional rows hiding in a $3 \times 3$

Worked example Example 4 — dependence you must hunt for

Is $A = 120241365$ invertible?

Forecast: Two of these rows are secretly the same direction. Which two?

Step 1. Notice row 2 $= 2 \times$ row 1 exactly. So the rows are dependent (one is a scaled copy of another). Why this step? Spotting proportionality by eye is the cheapest possible route — no determinant, no reduction. Dependent rows drop the rank below $n$ .

Step 2. Dependent rows mean rank $\leq 2 < 3$ , so test [rank] (full rank $= 3$ ) fails, hence not invertible. Why this step? Here we use a fact from rank theory: the number of independent rows always equals the number of independent columns (row rank $=$ column rank). So dependent rows guarantee dependent columns too, which is exactly what [rank]/[indep] forbid. (This equality is proved in the Rank and Nullity note; we simply cite it here.)

Step 3. Concretely, dependent columns means test [null] also fails — some nonzero $x$ has $A x = 0$ . Why this step? We name the specific consequence on the failing branch so the reader sees the flowchart's left side come alive, not just an abstract "rank $< 3$ ".

Verify: $det A = 0$ . (Two proportional rows always force $det = 0$ , since scaling a row scales the determinant, and two equal rows give $det = 0$ .) ✔

Cell E — limiting-easy: a triangular matrix

Worked example Example 5 — triangular determinant is just the diagonal

Is $A = 400950 - 1 73$ invertible?

Forecast: All the entries below the diagonal are zero. Does that make the invertibility question trivially easy?

Step 1. For an upper-triangular matrix, $det A =$ product of diagonal entries $= 4 \cdot 5 \cdot 3 = 60$ . Why this step? Triangular matrices are the limiting-easy case for [det]: the determinant collapses to the diagonal product, no expansion needed. This is the cheapest possible determinant.

Step 2. $det A = 60 \neq = 0 \Rightarrow$ invertible, hence full rank $3$ and no zero eigenvalue. Why this step? One passing test ([det]) drags every equivalent condition along with it — we simply report the ones worth naming ([rank], [eig]).

Step 3. Bonus insight: the eigenvalues of a triangular matrix are its diagonal entries $4, 5, 3$ . None is $0$ , so test [eig] confirms invertibility independently. Why this step? A second, independent route ([eig]) reaching the same verdict is a free reliability check — and it previews the eigenvalue angle of the next example.

Verify: $det A = 60$ and eigenvalues $= {4, 5, 3}$ , none zero. ✔

Cell F — the eigenvalue angle: is $0$ an eigenvalue?

Worked example Example 6 — invertibility as "no zero eigenvalue"

Is $A = (2123)$ invertible? Decide using eigenvalues only.

Forecast: An eigenvalue is a number $λ$ with $A x = λ x$ for some nonzero $x$ . If one of them is $0$ , test [eig] says the matrix is singular. Guess: will $0$ show up?

Step 1. Eigenvalues solve $det (A - λ I) = 0$ . Compute the characteristic polynomial: $det (2 - λ 1 2 3 - λ) = (2 - λ) (3 - λ) - 2 = λ^{2} - 5 λ + 4.$ Why this step? Test [eig] asks a single question: "is $0$ an eigenvalue?" By definition $λ$ is an eigenvalue exactly when $A - λ I$ is singular, i.e. when $det (A - λ I) = 0$ . So the cheapest way to hunt for a zero eigenvalue is to build this one polynomial in $λ$ and check whether $λ = 0$ is among its roots — that is far less work than searching for eigenvectors directly.

Step 2. Factor: $λ^{2} - 5 λ + 4 = (λ - 1) (λ - 4)$ , so the eigenvalues are $λ = 1$ and $λ = 4$ . Why this step? Factoring turns the polynomial into a product that is zero exactly at its roots, reading off both eigenvalues at a glance so we can check for a $0$ among them.

Step 3. Neither eigenvalue is $0$ $\Rightarrow$ invertible. (Equivalently, the constant term $4$ equals $det A$ , and $det A = 4 \neq = 0$ .) Why this step? Connecting the constant term to [det] shows the two tests are the same coin — the product of the eigenvalues is the determinant, so "no zero eigenvalue" and " $det \neq = 0$ " are literally the same statement.

Verify: eigenvalues ${1, 4}$ , product $= 4 = det A \neq = 0$ . ✔ The product-of-eigenvalues rule from the parent note holds.

Cell G — the non-square trap

Worked example Example 7 — the theorem stays silent

The matrix $A = 101011$ ( $3 \times 2$ ) has independent columns. Is it invertible?

Forecast: Independent columns felt like a guarantee of invertibility. Is $A$ square?

Step 1. Count shape: $3$ rows, $2$ columns. Not square. Why this step? The parent note's crucial caveat: the theorem applies only to $n \times n$ matrices. Before any test in the box, check squareness — otherwise every conclusion is void.

Step 2. Because $A$ is not square, "invertible" (a two-sided inverse) is impossible. Its map $x \mapsto A x$ sends $R^{2} \to R^{3}$ ; it can be injective (columns independent) but can never hit all of $R^{3}$ . Why this step? Two columns can span at most a $2$ -dimensional slab, and you need at least $3$ vectors for a basis of the $3$ -dimensional space $R^{3}$ . So test [solve] must fail no matter what — independence and invertibility part ways here.

Step 3. Conclusion: the columns are independent, yet the matrix is not invertible. Why this step? Stating the paradox plainly cements the lesson — "independent columns $\Rightarrow$ invertible" is only true after the squareness gate.

Verify: Columns independent (rank $= 2$ ) but rank $\neq = 3$ , so the columns cannot span $R^{3}$ : $2 < 3$ . ✔

Cell H — the one-sided inverse twist

Worked example Example 8 — exam-style: a right inverse is enough

Exam claim: " $A$ is $2 \times 2$ and there exists $D$ with $A D = I$ . But we were never told $D A = I$ . Can we conclude $A$ is invertible?"

Forecast: $A D = I$ looks weaker than $A D = D A = I$ . Is a one-sided inverse a real inverse for a square matrix?

Step 1. $A$ is square, and test [side] in the box says a one-sided inverse alone suffices. Why this step? For square matrices, $A D = I$ forces $D A = I$ automatically — this is the content of the parent's "one-sided inverse" mistake box, so a right inverse triggers the whole theorem.

Step 2. Concrete instance: let $A = (1011)$ and propose $D = (10 - 1 1)$ . Why this step? An explicit pair lets us actually multiply and see the claim in action rather than trust it abstractly.

Step 3. $A D = (1011) (10 - 1 1) = (1001) = I$ . So $D$ is a right inverse $\Rightarrow$ invertible, and the "missing" $D A = I$ must also hold. Why this step? Computing $A D$ verifies the hypothesis of [side]; the theorem then hands us $D A = I$ for free, which we confirm below.

Verify: $D A = (10 - 1 1) (1011) = (1001) = I$ . ✔ The other side came for free.

Cell I — the word problem

Worked example Example 9 — the paint-mixing shop

A shop mixes two base paints, Red base and Blue base, into custom colours. Mixing $x_{1}$ litres of Red base and $x_{2}$ litres of Blue base produces a paint whose (redness, blueness) reading is $(redness blueness) = (2112) (x_{1} x_{2}) .$ Question: Can the shop hit any target reading $b \in R^{2}$ , and is the recipe unique?

Forecast: If the two bases point in "different enough" directions, they should span every colour. Guess yes or no.

Step 1. Let $A = (2112)$ . Compute $det A = (2) (2) - (1) (1) = 4 - 1 = 3$ . Why this step? "Solvable for every $b$ " is test [solve], equivalent to [det]. One number settles both existence (can we reach every colour?) and uniqueness (is the recipe forced?).

Step 2. $det A = 3 \neq = 0 \Rightarrow$ invertible. So the columns span $R^{2}$ (every reading is achievable) and the map is injective (exactly one recipe per target — no two different mixes give the same colour). Why this step? [solve] passing tells us the columns span the plane, and the same invertibility makes the map injective — that is why for a square matrix "hits every target" and "unique recipe" come together.

Step 3. To hit a target $b = (4, 5)^{T}$ , solve $x = A^{- 1} b$ with $A^{- 1} = \frac{1}{3} (2 - 1 - 1 2)$ : $x = \frac{1}{3} (2 - 1 - 1 2) (45) = \frac{1}{3} (36) = (12) .$ Why this step? Since $A$ is invertible, the unique recipe is literally $A^{- 1} b$ — applying the inverse undoes the mixing and recovers the litres needed.

Verify: $A (1, 2)^{T} = (2 \cdot 1 + 1 \cdot 2, 1 \cdot 1 + 2 \cdot 2)^{T} = (4, 5)^{T} = b$ . ✔ Units: litres in, colour-reading out, the recipe reproduces the target exactly.

Cell J (part 1) — full $3 \times 3$ determinant by cofactors

Worked example Example 10 — a genuinely

$3 \times 3$ non-triangular determinant Is $A = 211130122$ invertible? None of the shortcuts (zero row, proportional rows, triangular shape) apply, so we must actually compute a $3 \times 3$ determinant.

Forecast: No obvious dependence jumps out. Do you expect the determinant to land on $0$ or not?

Step 1. Expand $det A$ along the first row using cofactors: $det A = 2 det (3022) - 1 det (1122) + 1 det (1130) .$ Why this step? Test [det] is still the cheapest single verdict, but for a full $3 \times 3$ we cannot read $det$ off the diagonal — we break it into three $2 \times 2$ determinants we already know how to do. The $+, -, +$ sign pattern is the standard cofactor checkerboard along row 1.

Step 2. Evaluate each $2 \times 2$ minor: $det (3022) = 6 - 0 = 6$ , $det (1122) = 2 - 2 = 0$ , $det (1130) = 0 - 3 = - 3$ . Why this step? Reducing the big determinant to small ones is exactly why cofactor expansion works — each minor is a baseline $2 \times 2$ we can do in one line.

Step 3. Combine: $det A = 2 (6) - 1 (0) + 1 (- 3) = 12 - 0 - 3 = 9$ . Since $det A = 9 \neq = 0$ , $A$ is invertible — hence full rank $3$ , independent columns, no zero eigenvalue. Why this step? Putting the pieces back with their signs gives the one number [det] needs; being nonzero flips on the entire success branch of the flowchart.

Verify: cofactor answer $det A = 9 \neq = 0$ . ✔

Cell J (part 2) — the parameter / limiting case

Worked example Example 11 — for which

$t$ does invertibility break? Consider $A (t) = (1 t t 1)$ . For which real values of $t$ is $A (t)$ not invertible?

Forecast: As $t$ grows, the two rows tilt toward each other. At what tilt do they line up and collapse the matrix?

Step 1. Treat the determinant as a function of $t$ : $det A (t) = (1) (1) - (t) (t) = 1 - t^{2}$ . Why this step? Test [det] turns the whole question into "where does $1 - t^{2} = 0$ ?" — solve one equation instead of testing infinitely many matrices.

Step 2. $1 - t^{2} = 0 ⟺ t^{2} = 1 ⟺ t = 1 or t = - 1$ . So $A (t)$ is singular exactly at $t = \pm 1$ , and invertible for every other $t$ . Why this step? The roots of the determinant function are precisely the parameter values that fail [det]; away from them the determinant is nonzero, so invertibility holds.

Step 3. Read the geometry across the whole range (see the figure below):

$- 1 < t < 1$ : $det > 0$ , invertible — the two rows point in clearly different directions.
$t = 1$ : $det = 0$ , both rows become $(1, 1)$ — identical, so they collapse onto one line; the null vector $(1, - 1)^{T}$ appears since $A (1) (1, - 1)^{T} = 0$ .
$t = - 1$ : $det = 0$ , rows become $(1, - 1)$ and $(- 1, 1)$ — negatives of each other, again on one line; the null vector is $(1, 1)^{T}$ since $A (- 1) (1, 1)^{T} = 0$ .
$∣ t ∣ > 1$ : $det < 0$ , invertible again — the map now flips orientation (negative determinant), but reversibility is untouched, because only $det \neq = 0$ matters, not its sign. Why this step? Covering every interval and both signs of the determinant is the "all cases" requirement — the reader never meets a value of $t$ we did not classify, and sees why the collapse happens (rows landing on the same line) exactly at $t = \pm 1$ .

Figure 1 — The determinant curve $det A (t) = 1 - t^{2}$ plotted against the parameter $t$ . The curve crosses zero (pink dots) exactly at $t = - 1$ and $t = + 1$ , the only two singular cases. The blue-shaded hump ( $- 1 < t < 1$ ) is where $det > 0$ and the matrix is invertible; the yellow-shaded tails ( $∣ t ∣ > 1$ ) are where $det < 0$ and it is still invertible. Only the two crossings are singular.

Verify: $det A (1) = 0$ , $det A (- 1) = 0$ , and $det A (0) = 1 \neq = 0$ , $det A (2) = - 3 \neq = 0$ . ✔ Singular precisely at $t = \pm 1$ .

Recall Quick self-test

Cheapest test for a $2 \times 2$ ? ::: The determinant $a d - b c$ (test [det]) — one line settles the whole theorem. How do you get $det$ of a full non-triangular $3 \times 3$ ? ::: Cofactor expansion along a row — three $2 \times 2$ minors with the $+, -, +$ sign pattern. A $3 \times 3$ has a row of all zeros. Invertible? ::: No — that row can never hold a pivot, so it fails [RREF]/[rank]. Independent columns on a $3 \times 2$ matrix — invertible? ::: The theorem does not apply; non-square matrices are never invertible. Square $A$ with $A D = I$ only — is $A$ invertible? ::: Yes; by [side], for square matrices a right inverse is automatically a full inverse. For $(1 t t 1)$ , singular when? ::: $t = \pm 1$ , where $det = 1 - t^{2} = 0$ and the two rows land on one line.