Invertible matrix theorem — 12+ equivalent conditions: Exercises

Intuition How to use this page

Each problem below tests one idea from the Invertible Matrix Theorem (IMT). Try it first, then open the collapsible solution. The levels climb from "just recognise the condition" up to "chain several conditions together and prove something new." A quick reminder of the vocabulary before we start:

A matrix is square when it has the same number of rows and columns ( $n \times n$ ). The IMT only speaks about square matrices.
$A$ is invertible if there is a matrix $B$ with $A B = B A = I$ , where $I$ is the identity (1's down the diagonal, 0's elsewhere).
The null space $Nul A$ is the set of all vectors $x$ with $A x = 0$ . If the only such vector is $x = 0$ we say the null space is trivial.
The determinant $det A$ is a single number attached to a square matrix; the IMT says $A$ is invertible exactly when $det A \neq = 0$ .

Throughout, "IMT #k" refers to condition number $k$ in the parent list. Prerequisite ideas live at Determinant, Rank and Nullity, Linear Independence and Span, Eigenvalues and Eigenvectors, Elementary Matrices and Row Reduction, Linear Transformations — Injective and Surjective and Basis and Dimension.

Level 1 — Recognition

Here you only need to spot which condition to apply and read off a yes/no.

Exercise 1.1

For $A = (4623)$ , decide whether $A$ is invertible using the determinant.

Recall Solution 1.1

What we do: compute the determinant of a $2 \times 2$ matrix. $det A = (4) (3) - (2) (6) = 12 - 12 = 0.$ Why this tool: IMT #13 says invertible $⟺ det \neq = 0$ , and for a $2 \times 2$ the determinant is the fastest possible check. Since $det A = 0$ , $A$ is NOT invertible. (Notice row 2 $= 1.5 \times$ row 1 — the rows are dependent, consistent with $det = 0$ .)

Exercise 1.2

Is $A = 100020503$ invertible? Use the fact that it is triangular.

Recall Solution 1.2

What we do: a triangular matrix (all zeros below the diagonal) has determinant equal to the product of its diagonal entries. $det A = 1 \cdot 2 \cdot 3 = 6 \neq = 0.$ Why this tool: IMT #13 again — and for triangular matrices the determinant is just the diagonal product, so no row reduction is needed. $det A = 6 \neq = 0 \Rightarrow$ $A$ IS invertible.

Exercise 1.3

A matrix $A$ is $3 \times 5$ and its 5 columns are linearly independent. Is $A$ invertible?

Recall Solution 1.3

What we do: check the hypothesis of the IMT before anything else. The IMT applies only to square matrices. A $3 \times 5$ matrix is not square, so the word "invertible" (a two-sided inverse) does not even apply. Moreover 5 columns cannot be independent in $R^{3}$ — you cannot fit more than 3 independent vectors into 3-dimensional space (see Basis and Dimension). So the premise is actually impossible. Answer: NOT invertible — the IMT does not apply to non-square matrices.

Level 2 — Application

Now you actually compute something (a determinant, a null vector, a row reduction) to reach the verdict.

Exercise 2.1

Determine invertibility of $A = 210121012$ by computing $det A$ .

Recall Solution 2.1

What we do: expand the $3 \times 3$ determinant along the first row. $det A = 2 det (2112) - 1 det (1012) + 0.$ Compute each $2 \times 2$ : $det (2112) = 4 - 1 = 3$ , and $det (1012) = 2 - 0 = 2$ . $det A = 2 (3) - 1 (2) + 0 = 6 - 2 = 4.$ Why this tool: cofactor expansion turns a $3 \times 3$ into small $2 \times 2$ 's; the zero in position $(1, 3)$ makes the last term vanish, saving work. $det A = 4 \neq = 0 \Rightarrow$ invertible.

Exercise 2.2

For $A = 123246134$ , find a nonzero vector in $Nul A$ and use it to decide invertibility.

Recall Solution 2.2

What we do: look for a linear dependence among the columns. The first two columns are $(1, 2, 3)$ and $(2, 4, 6) = 2 \times$ column 1. So column $2 - 2 \times$ column $1 = 0$ , i.e. the weights $(- 2, 1, 0)$ combine the columns to zero. Why this tool: matrix–vector product $A x$ is a linear combination of columns weighted by $x$ (IMT Link B). A dependence among columns is exactly a null vector. Check directly with $x = (- 2, 1, 0)^{T}$ : $A x = - 2 + 2 + 0 - 4 + 4 + 0 - 6 + 6 + 0 = 000 . ✓$ A nonzero null vector exists $\Rightarrow$ IMT #4 fails $\Rightarrow$ NOT invertible.

Exercise 2.3

Use row reduction to decide whether $A = (1325)$ is invertible, and count the pivots.

Recall Solution 2.3

What we do: apply elementary row operations (see Elementary Matrices and Row Reduction). $(1325) R_{2} \to R_{2} - 3 R_{1} (10 2 - 1) R_{2} \to - R_{2} (1021) R_{1} \to R_{1} - 2 R_{2} (1001) = I .$ Why this tool: IMT #2/#3 — reaching $I$ with a pivot in every row and column is a direct certificate of invertibility. Two pivots, RREF $= I \Rightarrow$ invertible. Cross-check: $det = 1 \cdot 5 - 2 \cdot 3 = - 1 \neq = 0. ✓$

Level 3 — Analysis

Reason about a family of matrices: for which parameter values is it invertible?

Exercise 3.1

For which real numbers $k$ is $A = (1 k k 4)$ invertible?

Recall Solution 3.1

What we do: invertibility fails exactly when $det A = 0$ , so solve $det A = 0$ and exclude those $k$ . $det A = (1) (4) - (k) (k) = 4 - k^{2} .$ Why this tool: IMT #13. Treating the determinant as a function of $k$ turns "invertible?" into a simple equation. Set $4 - k^{2} = 0 \Rightarrow k^{2} = 4 \Rightarrow k = \pm 2$ . So $A$ is NOT invertible for $k = 2$ or $k = - 2$ , and invertible for all other $k$ , i.e. $k \in R ∖ {- 2, 2}$ .

Exercise 3.2

$A = 111124 13 t$ . Find every value of $t$ for which $A$ is singular (not invertible).

Recall Solution 3.2

What we do: compute $det A$ as a function of $t$ , then find its roots. Expand along the first row: $det A = 1 det (24 3 t) - 1 det (11 3 t) + 1 det (1124) .$ Each minor: $det (24 3 t) = 2 t - 12$ , $det (11 3 t) = t - 3$ , $det (1124) = 4 - 2 = 2$ . $det A = (2 t - 12) - (t - 3) + 2 = 2 t - 12 - t + 3 + 2 = t - 7.$ Why this tool: IMT #13; writing the determinant as the linear expression $t - 7$ makes the singular case obvious. $det A = 0 ⟺ t = 7$ . So $A$ is singular exactly when $t = 7$ , and invertible for every other $t$ .

Exercise 3.3

Suppose $A$ is $3 \times 3$ with eigenvalues $2, - 1, 0$ . Is $A$ invertible? What is $det A$ ?

Recall Solution 3.3

What we do: use the link between eigenvalues and invertibility (see Eigenvalues and Eigenvectors). Why this tool: IMT #14 — $A$ is invertible $⟺ 0$ is not an eigenvalue. Here $0$ is listed as an eigenvalue. Therefore $A$ is NOT invertible. Confirming via $det =$ product of eigenvalues: $det A = 2 \cdot (- 1) \cdot 0 = 0,$ consistent with IMT #13. A zero eigenvalue $λ = 0$ gives a nonzero $x$ with $A x = 0 x = 0$ — a nontrivial null vector, so IMT #4 also fails.

Level 4 — Synthesis

Chain several IMT conditions and prove statements about general matrices.

Exercise 4.1

$A$ is $n \times n$ and there is a matrix $D$ with $A D = I$ (a right inverse). Prove $A$ is invertible, and hence $D A = I$ too.

Recall Solution 4.1

What we do: show the map $x \mapsto A x$ is onto, then use squareness to jump to full invertibility. Step 1 (surjective). For any $b \in R^{n}$ , set $x = D b$ . Then $A x = A (D b) = (A D) b = I b = b$ . So every $b$ has a preimage — the map is onto (IMT #9), equivalently $A x = b$ solvable for all $b$ (IMT #7). Why this step: $A D = I$ literally hands us a formula $x = D b$ that solves $A x = b$ . Step 2 (climb the IMT). For a square matrix, onto (#9) is in the equivalence loop, so all conditions hold — in particular $A$ is invertible (#1). Invertibility gives a genuine $A^{- 1}$ with $A^{- 1} A = A A^{- 1} = I$ . Step 3 (identify $D$ ). Multiply $A D = I$ on the left by $A^{- 1}$ : $A^{- 1} A D = A^{- 1} \Rightarrow D = A^{- 1}$ . Hence $D A = A^{- 1} A = I$ . $■$ Key point: squareness is what lets a one-sided inverse become two-sided (IMT #10, #11).

Exercise 4.2

$A$ and $B$ are $n \times n$ and their product $A B$ is invertible. Prove that both $A$ and $B$ are invertible.

Recall Solution 4.2

What we do: exhibit one-sided inverses for $A$ and for $B$ , then apply Exercise 4.1's principle. Let $M = (A B)^{- 1}$ , so $(A B) M = I$ and $M (A B) = I$ . $B$ is invertible: from $M (A B) = I$ we read $(M A) B = I$ , a left inverse for $B$ . For square $B$ , IMT #10 $\Rightarrow B$ invertible. $A$ is invertible: from $(A B) M = I$ we read $A (B M) = I$ , a right inverse for $A$ . For square $A$ , IMT #11 $\Rightarrow A$ invertible. Why this tool: we never computed inverses; we just found one-sided inverses and let the IMT (via Ex. 4.1) upgrade them. $■$

Exercise 4.3

Show that if $A$ is invertible, then $A^{T}$ is invertible, using the determinant.

Recall Solution 4.3

What we do: transfer invertibility across the transpose using two determinant facts. Fact: $det (A^{T}) = det A$ (transposing does not change the determinant). Since $A$ invertible $\Rightarrow det A \neq = 0$ (IMT #13), we get $det (A^{T}) = det A \neq = 0$ . By IMT #13 applied to $A^{T}$ , $A^{T}$ is invertible (this is IMT #12). $■$ Why this tool: the determinant is a single number that "sees through" the transpose, so the whole IMT machinery carries over instantly.

Level 5 — Mastery

Full proofs weaving geometry, rank and the equivalence structure.

Exercise 5.1

Let $A$ be $n \times n$ . Prove directly that " $A x = 0$ has only $x = 0$ " is equivalent to " $A$ has rank $n$ ", and interpret geometrically for $n = 2$ .

Recall Solution 5.1

What we do: connect null space, pivots and rank (see Rank and Nullity). ( $\Rightarrow$ ) Suppose $A x = 0$ forces $x = 0$ . Every non-pivot (free) column would create a free variable and hence a nonzero solution; since none exist, there are no free columns. So all $n$ columns are pivot columns $\Rightarrow$ rank $=$ number of pivots $= n$ . ( $\Leftarrow$ ) Suppose rank $= n$ . Then there are $n$ pivots, one per column, so there are no free variables. A homogeneous system with no free variables has the unique solution $x = 0$ . Rank–Nullity check: $dim Nul A = n - rank A = n - n = 0$ , matching "trivial null space" (IMT #17). Geometry ( $n = 2$ ): the columns of $A$ are two vectors in the plane.

Figure — Invertible matrix theorem — 12+ equivalent conditions

When they are independent (rank 2), they point in different directions and their linear combinations sweep out the whole plane — only the weights

(0, 0)

land on the origin, so the null space is just the origin. When they are dependent (rank 1), both lie on one line; a nonzero combination returns to the origin, giving a nontrivial null vector.

■

Exercise 5.2

Prove: for a square $A$ , "columns span $R^{n}$ " implies "columns are linearly independent." (This is the surprise the IMT hides — for square matrices spanning forces independence.)

Recall Solution 5.2

What we do: count pivots, using that a square matrix has exactly $n$ rows and $n$ columns. Spanning $\Rightarrow$ pivot in every row. If the columns span $R^{n}$ , the map $x \mapsto A x$ is onto (IMT #9), which needs a pivot in every row — otherwise some $b$ would be unreachable. That is $n$ pivots (one per row). $n$ pivots $\Rightarrow$ pivot in every column. With $n$ pivots spread over $n$ columns, each column holds exactly one pivot; there are no free columns. No free columns $\Rightarrow$ independent (IMT #5, via Link B). $■$ Why it only works for squares: the argument uses " $n$ pivots fill both all rows AND all columns" — possible only when rows $=$ columns. For a $2 \times 3$ matrix the columns can span $R^{2}$ with just 2 pivots, leaving a free column, so they are not independent.

Exercise 5.3

Let $A = (1224)$ . (a) Show it is singular three different ways (determinant, null vector, rank). (b) Sketch what the map $x \mapsto A x$ does to the plane.

Recall Solution 5.3

(a) Three certificates of singularity:

Determinant (IMT #13): $det A = (1) (4) - (2) (2) = 4 - 4 = 0.$
Null vector (IMT #4): column $2 = 2 \times$ column $1$ , so $x = (2, - 1)^{T}$ gives $A x = (1 \cdot 2 + 2 \cdot (- 1), 2 \cdot 2 + 4 \cdot (- 1))^{T} = (0, 0)^{T}$ .
Rank (IMT #15): row 2 $= 2 \times$ row 1, so after $R_{2} \to R_{2} - 2 R_{1}$ we get $(1020)$ — one pivot, so rank $= 1 < 2$ . Why three ways: the whole point of the IMT is that these are the same fact; here we watch the single failure appear in three costumes. (b) Geometry:
Both columns $(1, 2)$ and $(2, 4)$ lie on the same line $y = 2 x$ . Every input vector is mapped onto that one line — the map collapses the 2D plane onto a 1D line. The entire perpendicular direction $(2, - 1)$ is squashed to the origin (that's the null vector). A collapse destroys information, so the transformation cannot be undone: not invertible. $■$

Recall One-sentence summary

Every exercise here reduces to a pivot count — determinant, rank, null space, eigenvalues, spanning and independence are just different lenses on "does $A$ have $n$ pivots?"