5.1 Invertible matrices

A square matrix can send different inputs to the same output, or leave some outputs unreachable. An inverse must undo its action completely: every output must come from exactly one input. This requirement connects reversibility to the solution sets of $Ax=b$ . Entries are real unless another field is specified.

Before you start

Use matrix multiplication in its given order and recall that each elementary row operation is reversible. A system $Ax=b$ has a unique solution exactly when it is consistent and every variable column is a pivot column after reduction. Invertibility will guarantee this for every right-hand side of the appropriate dimension.

Left and right inverses

Definition 5.1.1

(Left inverse and right inverse)

Let $A$ be a $p \times q$ matrix.

A $q \times p$ matrix $H$ is a left inverse of $A$ if $HA = I_q$ .
A $q \times p$ matrix $G$ is a right inverse of $A$ if $AG = I_p$ .

For rectangular matrices, a left inverse or a right inverse can exist alone. If both exist, however, they must coincide. The square case will give the stronger conclusion that either one-sided inverse already guarantees the other.

Definition 5.1.2

(Invertible matrix)

Let $A$ be a $p \times p$ square matrix. We say that $A$ is invertible if there exists a $p \times p$ matrix $B$ such that

BA = AB = I_p.

The matrix $B$ is called the inverse of $A$ , and we write $B = A^{-1}$ .

Theorem 5.1.3

(The inverse is unique)

Let $A$ be $p\times q$ and let $B,C$ be $q\times p$ . If $B$ is a left inverse of $A$ and $C$ is a right inverse, then $B=C$ . In particular, an invertible square matrix has exactly one inverse.

Proof

(Why the inverse is unique)

The assumptions are $BA=I_q$ and $AC=I_p$ . Associativity gives

B=BI_p=B(AC)=(BA)C=I_qC=C.

No factors were interchanged. For an invertible square matrix, any two inverses satisfy these hypotheses, so they coincide.

What invertibility means

The equations defining an inverse restore every input vector. Diagonal matrices make that action especially transparent.

Worked example 5.1.4

(A diagonal matrix is easy to invert)

Let

D = \operatorname{diag}(2, -1, 3).

Then

D^{-1} = \operatorname{diag}\left(\tfrac{1}{2}, -1, \tfrac{1}{3}\right).

This works because each diagonal entry is replaced by its reciprocal, and the off-diagonal zeros stay zero. Multiplying $D$ by $D^{-1}$ gives $I_3$ .

Equivalent formulations

Theorem 5.1.5

(Equivalent ways to recognize invertibility)

Let $A$ be a real $p\times p$ matrix. The following conditions are equivalent:

$A$ is invertible.
$A$ is row-equivalent to $I_p$ .
$A$ is a product of elementary row-operation matrices of order $p$ .
$A$ has a left inverse.
$A$ has a right inverse.
$A$ is nonsingular: $Ax=0$ has only the zero solution.
For every $b\in\mathbb R^p$ , the system $Ax=b$ is consistent.
For every $b\in\mathbb R^p$ , the system $Ax=b$ has exactly one solution.

Equivalently, the columns of $A$ are linearly independent, or they span $\mathbb R^p$ . When these conditions hold, the unique solution is $A^{-1}b$ .

Proof

(Where the square hypothesis enters the equivalence proof)

A left inverse $H$ forces the homogeneous solution to be zero: if $Ax=0$ , then $x=HAx=0$ . Conversely, suppose the homogeneous system has only the zero solution. In any RREF of $A$ , a free variable would produce a nonzero homogeneous solution. Thus every column has a pivot. Because $A$ is square, there are also exactly $p$ pivot rows, and the RREF is $I_p$ .

If a row-operation product $E$ satisfies $EA=I_p$ , the reverse chain shows that $E$ is invertible and $A=E^{-1}$ . Hence $A$ is invertible and is itself a product of elementary matrices. Conversely, any such product is invertible, and an inverse is in particular a left inverse. This connects conditions 1–4 and 6 without assuming uniqueness of RREF.

Now suppose $AG=I_p$ . Then $A$ is a left inverse of the square matrix $G$ . The already proved left-inverse criterion makes $G$ invertible, and $A=A(GG^{-1})=(AG)G^{-1}=G^{-1}$ is invertible. This proves the right-inverse criterion without silently reversing a product.

If every $Ax=b$ is consistent, choose solutions $u_j$ of $Au_j=e_j$ , where $e_j$ is column $j$ of $I_p$ . Their matrix $U=[u_1\ \cdots\ u_p]$ satisfies $AU=I_p$ , so the right-inverse criterion applies. Finally, when $A$ is invertible, $A(A^{-1}b)=b$ proves existence, and any solution $x$ satisfies $x=A^{-1}Ax=A^{-1}b$ , proving uniqueness. Uniqueness for every right-hand side implies consistency for every right-hand side.

The column statements are translations: independence means that $Ax=0$ has only the zero coefficient vector, and spanning means that every $b$ has a representation $Ax=b$ . Thus no additional assumption about the columns is being introduced.

Row reduction and the inverse

Each elementary row-operation matrix is invertible; its inverse encodes the reverse operation, as proved in 3.5 Elementary row-operation matrices. If the chronological operations have matrices $E_1,\ldots,E_k$ , their combined left multiplier is $E=E_k\cdots E_1$ .

Apply every operation across the entire augmented matrix. Then

E[A\mid I_p]=[EA\mid E].

If the left block becomes $I_p$ , then $EA=I_p$ . Since $E$ is invertible, $A=E^{-1}$ and the right block $E$ is $A^{-1}$ . If reduction leaves a missing pivot in the left block, the square matrix fails the equivalent conditions above. The method tests invertibility as well as computing an inverse; a complicated right block cannot compensate for a missing left pivot.

Equivalent tests for invertibility

Follow the row-reduction dictionary behind inverse extraction: reversible row operations turn A into I exactly when the right block becomes A^{-1}.

Reversible row operations
Every elementary row operation is left multiplication by an invertible matrix, so a full reduction records an accumulated product.
Reaching the identity
If E_k...E_1 A=I, then the accumulated product E_k...E_1 is A^{-1}.
Augmented inverse method
Applying the same row operations to [A | I] gives [I | A^{-1}] when the left block reaches I.
Failure mode
If a pivot is missing, the left block cannot become I. Equivalently, Ax=0 has a nonzero solution and A is singular.
Equivalent dictionary
Invertible, nonsingular, row-equivalent to I, trivial null space, and unique solutions for every b are one package.

The row-reduction test, the inverse formula, the null-space test, and unique solvability all say the same thing for a square matrix.

Interactive · Step throughStep 1 of 7

Follow one inverse-by-row-reduction example

Apply each row operation to the whole augmented matrix $[A \mid I]$ . When the left block becomes I, read $A^{-1}$ from the right block.

[A | I]

1	2	1	1	0	0
0	1	1	0	1	0
2	3	4	0	0	1

1pivot used by this operation0changed in this stepleft of the bar: A; right: the same operations applied to I

Start with A = [[1,2,1],[0,1,1],[2,3,4]] and augment it by I. Apply every row operation to all six entries of the row; the right block records the same operations applied to I.

Pivot in focus: Row 1, column 1: 1
Changed in this step: None (starting matrix)

Left block not yet in row echelon form

Row 3 of the left block still has 2 in column 1, which is not to the right of the pivot in row 2. Whether A is invertible is not decided yet.

NoticeA row operation acts on the whole row of [A | I], right block included; otherwise the right block no longer records what was done to I.

Interactive · Step throughStep 1 of 5

Trace one full row-reduction path

Each step applies one group of elementary row operations. Watch the entries above and below each pivot become 0.

Choose the first pivot in column 1.

1	2	2	4
1	3	3	5
2	6	5	6

1pivot in focus0changed in this stepright of the bar: constants

Start with the augmented matrix. The first pivot should help us clear the entries underneath it.

Pivot: Row 1, column 1: 1
Changed in this step: None (starting matrix)

Not yet in row echelon form

Row 2 still has 1 in column 1, which is not to the right of the pivot in row 1.

NoticeColumn 1 already has a convenient pivot 1 in the first row, so we do not need a row swap.

How to use the invertibility dictionary

Choose the criterion matching the given evidence: pivots, a one-sided identity, a nonzero null-space vector, or information about all right-hand sides.

Worked example 5.1.6

(A nonzero null-space vector proves non-invertibility)

Suppose $A$ is a square matrix and

v = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \ne 0, \qquad Av = 0.

Then $A$ cannot be invertible. Indeed, if $A^{-1}$ existed, multiplying $Av = 0$ on the left by $A^{-1}$ would give

A^{-1}Av = A^{-1}0.

The left side is $v$ , while the right side is $0$ . This would force $v = 0$ , contradicting the given vector. Therefore the existence of one nonzero solution of $Ax = 0$ already proves that $A$ is singular.

Worked example 5.1.7

(Use an inverse to solve a system)

Suppose

A^{-1} = \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix}, \qquad b = \begin{bmatrix} 5 \\ -1 \end{bmatrix}.

If $Ax = b$ , then multiply both sides by $A^{-1}$ :

x = A^{-1}b = \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 5 \\ -1 \end{bmatrix} = \begin{bmatrix} 11 \\ -3 \end{bmatrix}.

The inverse is not just a symbol for undoing $A$ ; it is an explicit formula for the unique solution.

Row-equivalence through invertible matrices

Theorem 5.1.8

(Row-equivalence is left multiplication by an invertible matrix)

Suppose $A$ and $B$ are matrices with $p$ rows. Then the following are equivalent:

$A$ and $B$ are row-equivalent.
There exists an invertible $p \times p$ matrix $G$ such that

B = GA.

Moreover, once $B = GA$ , we also have

A = G^{-1}B.

A row-operation chain supplies such a product $G$ . Conversely, the dictionary expresses any invertible $G$ as a product of elementary matrices, so $GA$ is obtained from $A$ by those row operations. Multiplying by $G^{-1}$ recovers $A$ .

Worked example 5.1.9

(Reading a row-equivalence as one matrix equality)

Let

A = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix}, \qquad G = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.

The matrix $G$ is the row-operation matrix for the move

R_2 \leftarrow R_2 - R_1.

GA = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 2 \\ 0 & 1 & 1 \end{bmatrix}.

If we call this new matrix $B$ , then $B = GA$ . That single equation records the entire row operation. Since $G$ is invertible, $A$ and $B$ are row-equivalent.

Theorem 5.1.10

(Row operations preserve linear relations among corresponding columns)

Let $A$ and $B$ be row-equivalent $p \times q$ matrices, and write their columns as

A = [a_1 \; a_2 \; \cdots \; a_q], \qquad B = [b_1 \; b_2 \; \cdots \; b_q].

a_j = \alpha_1 a_{k_1} + \alpha_2 a_{k_2} + \cdots + \alpha_n a_{k_n},

then

b_j = \alpha_1 b_{k_1} + \alpha_2 b_{k_2} + \cdots + \alpha_n b_{k_n}.

In particular, linear dependence and linear independence among corresponding columns are preserved by row-equivalence.

Multiplying by $G$ carries each relation forward; multiplying by $G^{-1}$ recovers it. Thus the same coefficient relations hold in both directions, which also proves preservation of dependence and independence. The columns themselves generally change. The full RREF uniqueness argument and its role in rank are developed in 5.2 Rank and uniqueness of RREF.

Products and inverse order

Theorem 5.1.11

(A square product is invertible exactly when its factors are)

Let $A,B$ be real $p\times p$ matrices. Then $AB$ is invertible if and only if both $A$ and $B$ are invertible. In that case,

(AB)^{-1}=B^{-1}A^{-1}.

Proof

(Construct the inverse before using its name)

First assume the factors are invertible and set $C=B^{-1}A^{-1}$ . Check both sides using associativity:

(AB)C=A(BB^{-1})A^{-1}=I_p,\qquad C(AB)=B^{-1}(A^{-1}A)B=I_p.

Only now may we conclude that $C=(AB)^{-1}$ . The reversed order places the inverse pairs next to each other; no commutativity is used.

For the converse, assume $AB$ is invertible and write $C=(AB)^{-1}$ . We cannot yet use $A^{-1}$ or $B^{-1}$ . Instead,

A(BC)=I_p,\qquad (CA)B=I_p.

Thus $A$ has a right inverse and $B$ has a left inverse. Both factors are square of the same order, so the previously proved one-sided criteria make them invertible. This use of the dictionary is essential to the converse.

By induction, a finite product of square matrices of the same order is invertible exactly when every factor is invertible, and

(A_1\cdots A_k)^{-1}=A_k^{-1}\cdots A_1^{-1}.

Why the transpose also matters

Theorem 5.1.12

(Transpose, scalar multiples, and powers)

If $A$ is invertible, then $(A^{-1})^{-1}=A$ and $(A^T)^{-1}=(A^{-1})^T$ . For a nonzero real scalar $c$ , $(cA)^{-1}=c^{-1}A^{-1}$ .

Every integer power $A^n$ is invertible, with $(A^n)^{-1}=(A^{-1})^n$ . Here $A^0=I_p$ , and for a positive integer $m$ , $A^{-m}=(A^{-1})^m$ .

The transpose formula follows by transposing both inverse identities, which reverses their factor order. The scalar formula follows by cancelling $c$ and $c^{-1}$ in both products. Positive powers follow from the product theorem; negative powers apply it to $A^{-1}$ , and the zero power is the identity. For a positive integer $n$ , invertibility of $A^n$ also implies invertibility of $A$ . That converse is false for $n=0$ , since $A^0=I_p$ even for singular square $A$ .

Products that equal the identity

Theorem 5.1.13

(A one-sided identity is enough in the square case)

Let $P$ and $Q$ be square matrices of the same size. If

PQ=I,

then both $P$ and $Q$ are invertible, and

Q=P^{-1},\qquad QP=I.

Worked example 5.1.14

(Cyclic identities from $ABCD=I$ )

Suppose $A,B,C,D$ are $5 \times 5$ matrices and

ABCD=I_5.

Group the product as

(ABC)D=I_5.

By the one-sided identity theorem, $ABC$ and $D$ are inverses of each other. Therefore

DABC=I_5.

Next group the same original product cyclically:

(DAB)C=I_5 \qquad\text{and}\qquad (CDA)B=I_5,

after using the already justified cyclic identities. This gives

CDAB=I_5, \qquad BCDA=I_5.

So $BCDA$ , $CDAB$ , and $DABC$ are forced to equal $I_5$ . Arbitrary reorderings such as $DCBA$ or $DBAC$ are not forced by $ABCD=I_5$ ; matrix multiplication still does not commute.

Compute an inverse by row reduction

Worked example 5.1.15

(Find an inverse by row reduction)

Let

A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}.

Start from $[A\mid I_2]$ :

\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 3 & 5 & 0 & 1 \end{array}\right].

Eliminate the entry below the first pivot:

\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -1 & -3 & 1 \end{array}\right].

Now scale the second row and clear the entry above the second pivot:

\left[\begin{array}{cc|cc} 1 & 0 & -5 & 2 \\ 0 & 1 & 3 & -1 \end{array}\right].

A^{-1} = \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix}.

A family of inverse matrices

Worked example 5.1.16

(A parameterized inverse from row reduction)

For a real number $\alpha$ , let

A_\alpha = \begin{bmatrix} 1 & 2 & 1 & 2 \\ 1 & 3 & 1 & 2 \\ 0 & 0 & 1 & \alpha \\ 1 & 2 & 1 & 3 \end{bmatrix}.

Begin with $[A_\alpha\mid I_4]$ . Apply $R_2\leftarrow R_2-R_1$ and $R_4\leftarrow R_4-R_1$ to both halves. The result is

\left[\begin{array}{cccc|cccc} 1&2&1&2&1&0&0&0\\ 0&1&0&0&-1&1&0&0\\ 0&0&1&\alpha&0&0&1&0\\ 0&0&0&1&-1&0&0&1 \end{array}\right].

The four pivots in the left block are already $1$ . Clear the remaining entries in this order:

R_3\leftarrow R_3-\alpha R_4,\qquad R_1\leftarrow R_1-2R_2,\qquad R_1\leftarrow R_1-R_3,\qquad R_1\leftarrow R_1-2R_4.

After the first operation, the right half of row 3 is $(\alpha,0,1,-\alpha)$ . Rows 2 and 4 on the right are $(-1,1,0,0)$ and $(-1,0,0,1)$ ; the three operations on row 1 therefore give $(5-\alpha,-2,-1,\alpha-2)$ . Every step adds a multiple of another row, so no division by a parameter expression occurs. Each operation is reversible for every real $\alpha$ , and the final augmented matrix is

\left[\begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & 5-\alpha & -2 & -1 & \alpha-2 \\ 0 & 1 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & \alpha & 0 & 1 & -\alpha \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right].

Since the left block is $I_4$ for every real $\alpha$ , the matrix $A_\alpha$ is invertible for every real $\alpha$ , and

A_\alpha^{-1} = \begin{bmatrix} 5-\alpha & -2 & -1 & \alpha-2 \\ -1 & 1 & 0 & 0 \\ \alpha & 0 & 1 & -\alpha \\ -1 & 0 & 0 & 1 \end{bmatrix}.

Notice what this conclusion does and does not say. The parameter remains in the inverse; it is not a value to be solved for. The row reduction has shown that no value of $\alpha$ creates a missing pivot in the left block.

Common mistake

(Do not treat every parameter as a restriction)

Sometimes a parameter appears because the answer is a family of inverses, not because the matrix fails for special values. The restriction comes from a missing pivot or an illegal division by zero, not from the mere presence of a symbol.

Reading supplied row-reduction tables

Worked example 5.1.17

(An inverse pair and a transposed system)

Suppose a reduction table shows that $[A \mid I_4]$ is row-equivalent to $[I_4 \mid D]$ , where

D= \begin{bmatrix} 3 & -2 & 3 & -2 \\ 0 & 1 & -2 & 1 \\ -3 & 1 & -2 & 2 \\ 1 & 0 & 1 & -1 \end{bmatrix}.

Then the row-reduction result itself already says

A^{-1}=D.

The corresponding coefficient matrix is

A= \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 2 & 3 \\ 2 & 1 & 3 & 3 \\ 3 & 2 & 4 & 3 \end{bmatrix}.

Direct multiplication gives $AD=I_4$ , which verifies this inverse pair. Now let

g= \begin{bmatrix} 1 \\ 0 \\ 3 \\ 0 \end{bmatrix}.

To solve $A^t x=g$ , do not row-reduce again from the beginning. Since $A$ is invertible, $A^t$ is also invertible and

(A^t)^{-1}=(A^{-1})^t=D^t.

Therefore

x=(A^t)^{-1}g=D^t g = \begin{bmatrix} -6 \\ 1 \\ -3 \\ 4 \end{bmatrix}.

The computation is short because the reduction table has already done the hard work.

Worked example 5.1.18

(Four inverse computations read by pivot columns)

For each matrix below, the method is the same: row-reduce $[A \mid I_4]$ , look only at the pivot pattern of the left block first, and read the inverse from the right block only if the left block becomes $I_4$ .

Case (a). For

A_1= \begin{bmatrix} 1 & 0 & 3 & 3 \\ 2 & 1 & 2 & 3 \\ -1 & 3 & 0 & 3 \\ 4 & 2 & 2 & 4 \end{bmatrix},

one echelon stage has left block

A_1^\sharp= \begin{bmatrix} 1 & 0 & 3 & 3 \\ 0 & 1 & -4 & -3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}.

There is no pivot in the fourth column. Hence $A_1$ is not invertible.

Case (b). Take $A_2=A$ from the preceding inverse-pair example. Its reduction reaches $[I_4 \mid D]$ , so all four columns have pivots and $A_2^{-1}=D$ . The same matrix used there to solve the transposed system provides an invertible case here.

Case (c). For

A_3= \begin{bmatrix} 1 & 2 & -1 & 30 \\ 3 & 2 & 0 & 15 \\ 0 & 1 & 3 & 15 \\ 1 & 1 & 1 & 10 \end{bmatrix},

one echelon stage has left block

A_3^\sharp= \begin{bmatrix} 1 & 2 & -1 & 30 \\ 0 & 1 & 3 & 15 \\ 0 & 0 & 5 & -5 \\ 0 & 0 & 0 & 0 \end{bmatrix}.

Again a pivot column is missing, so $A_3$ is not invertible.

Case (d). For

A_4= \begin{bmatrix} 1 & 1 & 2 & 1 \\ 2 & 3 & 4 & 1 \\ 3 & 3 & 3 & 1 \\ 1 & 2 & 3 & 1 \end{bmatrix},

the reduction reaches $[I_4 \mid A_4^{-1}]$ , with

A_4^{-1}= \begin{bmatrix} 1 & 1 & 0 & -2 \\ -2 & -2 & 1 & 3 \\ 1 & 2 & -1 & -2 \\ 0 & -3 & 1 & 3 \end{bmatrix}.

The identity left block certifies invertibility in cases (b) and (d). The missing pivots in cases (a) and (c) rule it out, regardless of the entries in the right block.

Worked example 5.1.19

(A final identity block controls the whole chain)

Suppose a row-operation chain carries $[A \mid I_5]$ through an intermediate matrix $[B \mid C]$ and finally to $[D \mid E]$ . After the last few operations, the final augmented matrix is

[D \mid E] = \left[ \begin{array}{ccccc|ccccc} 1 & 0 & 0 & 0 & 0 & 1 & 0 & -2 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 9 & 1 & -18 & -3 & -2 \\ 0 & 0 & 1 & 0 & 0 & -6 & 2 & 11 & 2 & 0 \\ 0 & 0 & 0 & 1 & 0 & -3 & 0 & 6 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -6 & 0 & 12 & 2 & 1 \end{array} \right].

The left block is $I_5$ , so $D=I_5$ and

A^{-1}=E= \begin{bmatrix} 1 & 0 & -2 & 0 & 0 \\ 9 & 1 & -18 & -3 & -2 \\ -6 & 2 & 11 & 2 & 0 \\ -3 & 0 & 6 & 1 & 0 \\ -6 & 0 & 12 & 2 & 1 \end{bmatrix}.

It follows immediately that $A^t$ is invertible and

(A^t)^{-1}=E^t = \begin{bmatrix} 1 & 9 & -6 & -3 & -6 \\ 0 & 1 & 2 & 0 & 0 \\ -2 & -18 & 11 & 6 & 12 \\ 0 & -3 & 2 & 1 & 2 \\ 0 & -2 & 0 & 0 & 1 \end{bmatrix}.

The same chain can also be read as one left multiplication. Let $H$ be the product of all row-operation matrices in the chain. Since $HA=I_5$ and $HI_5=E$ , we have $H=A^{-1}$ . Therefore, if the same row operations carry $F+I_5$ to $A^3+3A+I_5$ , then

A^3+3A+I_5=A^{-1}(F+I_5).

Multiplying by $A$ on the left gives

F+I_5=A(A^3+3A+I_5)=A^4+3A^2+A,

F=A^4+3A^2+A-I_5.

The same row-operation chain acts on every matrix by the same left multiplier $H=A^{-1}$ . This is why the numerical reduction also determines the equation for $F$ .

Prove or disprove invertibility without determinants

Worked example 5.1.20

(A vector identity that proves invertibility)

Suppose $A$ is a $7 \times 7$ matrix and

A^2x = Ax + x

for every vector $x \in \mathbb R^7$ . Move the terms involving $A$ to the left:

(A^2-A)x = x \qquad\text{for every }x.

Since $A^2-A=(A-I_7)A$ , this says

((A-I_7)A)x = I_7x \qquad\text{for every }x.

Two matrices that agree on every vector are the same matrix, so

(A-I_7)A = I_7.

Thus $A-I_7$ is a left inverse of $A$ . Because $A$ is square, a one-sided identity is enough: $A$ is invertible, and its inverse is $A-I_7$ .

Worked example 5.1.21

(A contradiction proof for non-invertibility)

Suppose $A$ and $B$ are distinct $6 \times 6$ matrices, and

A^3=B^3, \qquad A^2B=B^2A.

We prove that $A^2+B^2$ is not invertible. Let

C=A^2+B^2.

First compare $CB$ and $CA$ :

CB=(A^2+B^2)B=A^2B+B^3,

while

CA=(A^2+B^2)A=A^3+B^2A.

Using the assumptions $A^2B=B^2A$ and $B^3=A^3$ , we obtain

CB=CA.

If $C$ were invertible, multiplying by $C^{-1}$ on the left would give $B=A$ , contradicting the assumption that $A$ and $B$ are distinct. Therefore $C=A^2+B^2$ is not invertible.

Common mistakes

Common mistake

(Do not confuse one-sided inverses in the rectangular case)

For a non-square matrix, a left inverse need not come with a right inverse, and a right inverse need not come with a left inverse. If both exist, they coincide. For a square matrix, the equivalence theorem proves that either one-sided inverse guarantees a unique two-sided inverse.

Common mistake

(Do not guess invertibility from appearance)

A matrix can look simple and still fail to be invertible. The correct test is to row-reduce it, or to use one of the equivalent conditions above.

Common mistake

(Do not permute factors without justification)

From $ABCD=I$ , it is legitimate to derive certain cyclic identities such as $DABC=I$ by using inverse theory. It is not legitimate to rearrange the factors as if $A,B,C,D$ commute.

Quick checks

Quick check

(If $A$ is invertible, what is $A^{-1}A$ ?)

Use the defining property of an inverse.

Solution · Answer

It is $I_p$ .

Quick check

(If $A$ is invertible, can the homogeneous system $Ax = 0$ have a nonzero solution?)

Use the unique-solution statement.

Solution · Answer

No. Since $Ax = b$ has the unique solution $x = A^{-1}b$ , the homogeneous case $b = 0$ has only the trivial solution $x = 0$ .

Quick check

(If $A$ is invertible, is $A^t$ invertible?)

Use the transpose rule above.

Solution · Answer

Yes. $A^t$ is invertible, and $(A^t)^{-1} = (A^{-1})^t$ .

Quick check

(If $B = GA$ with $G$ invertible and the columns of $A$ satisfy $a_3 = 2a_1 - a_2$ , what relation must hold among the columns of $B$ ?)

Keep the same coefficients and use $b_j = Ga_j$ .

Solution · Answer

The corresponding columns of $B$ satisfy

b_3 = 2b_1 - b_2.

Row-equivalence changes the columns themselves, but not the coefficient pattern in a linear relation.

Quick check

(Suppose $A,B,C,D$ are square matrices and $ABCD=I$ . Which cyclic identity is forced: $DABC=I$ or $DCBA=I$ ?)

Group the original product as $(ABC)D=I$ .

Solution · Answer

$DABC=I$ is forced. Since $(ABC)D=I$ , the square one-sided identity theorem implies that $D$ is the inverse of $ABC$ , so $D(ABC)=I$ . The equality $DCBA=I$ would reverse the order of the factors and is not forced in general.

Quick check

(In the parameterized inverse above, what is the $(1,4)$ entry of $A_\alpha^{-1}$ when $\alpha=3$ ?)

Substitute $\alpha=3$ into the first row, fourth column.

Solution · Answer

The $(1,4)$ entry is $\alpha-2$ , so when $\alpha=3$ it equals $1$ .

Quick check

(If $A^2x=Ax+x$ for every vector $x$ , which matrix acts as the inverse of $A$ ?)

Move all terms to one side and look for a factorization involving $A$ .

Solution · Answer

The inverse is $A-I$ . Indeed, the vector identity gives $(A-I)A=I$ ; for square matrices, this one-sided identity is enough to conclude that $A$ is invertible.

Quick check

(If row operations transform $[A \mid I_p]$ into $[I_p \mid D]$ , what is $D$ ?)

Read the right block only after confirming the left block is the identity.

Solution · Answer

$D=A^{-1}$ . The same row operations that turn $A$ into $I_p$ turn $I_p$ into the inverse of $A$ .

Quick check

(Row operations take $[A\mid I_p]$ to $[I_p\mid E]$ , where $A$ is square. Explain why $AE=I_p$ follows as well as $EA=I_p$ .)

Distinguish what the row-operation product gives directly from what the square one-sided inverse theorem adds.

Solution · Answer

If $H$ is the product of the elementary matrices, the two blocks give $HA=I_p$ and $HI_p=E$ . Hence $H=E$ and $EA=I_p$ directly. Because $A$ and $E$ are square of the same order, a left inverse is also a right inverse. Thus $E=A^{-1}$ and $AE=I_p$ . The square hypothesis is essential to this last implication.

Exercise

Quick check

(Suppose $A$ is invertible and $AB = I_p$ . Prove that $B = A^{-1}$ .)

Use the fact that the inverse of $A$ is unique.

Solution · Guided solution

Since $A$ is invertible, it has an inverse $A^{-1}$ . Multiply the equation $AB = I_p$ on the left by $A^{-1}$ :

A^{-1}(AB) = A^{-1}I_p.

By associativity,

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

Because $A^{-1}A = I_p$ , this becomes

B = A^{-1}.

So any right inverse of an invertible square matrix must be the inverse itself.

Quick check

(Suppose $(A-B)A=I$ for square matrices $A$ and $B$ . Prove that $A$ and $B$ commute.)

First use the square one-sided identity theorem, then expand $AB-BA$ .

Solution · Guided solution

Since $(A-B)A=I$ , the square one-sided identity theorem gives

A(A-B)=I.

Now compute the commutator:

AB-BA = AB-A^2+A^2-BA.

Regroup the two parts:

AB-A^2 = A(B-A) = -A(A-B),

and

A^2-BA = (A-B)A.

Therefore

AB-BA = -A(A-B)+(A-B)A = -I+I=0.

So $AB=BA$ , which means $A$ and $B$ commute.

This page depends especially on 2.3 Gaussian elimination and RREF, 3.2 Matrix multiplication, identity matrices, and linear systems, and 3.3 Transpose, symmetric, and skew-symmetric matrices.

The connection between a zero null space and uniqueness is developed in 4.1 Homogeneous systems and null space.

Website function storage (used when needed)

5.1 Invertible matrices

Before you start

Left and right inverses

What invertibility means

Equivalent formulations

Row reduction and the inverse

How to use the invertibility dictionary

Row-equivalence through invertible matrices

Products and inverse order

Why the transpose also matters

Products that equal the identity

Compute an inverse by row reduction

A family of inverse matrices

Reading supplied row-reduction tables

Prove or disprove invertibility without determinants

Common mistakes

Quick checks

Exercise

Prerequisites

Key terms in this section

Before you start

Left and right inverses

What invertibility means

Equivalent formulations

Row reduction and the inverse

How to use the invertibility dictionary

Row-equivalence through invertible matrices

Products and inverse order

Why the transpose also matters

Products that equal the identity

Compute an inverse by row reduction

A family of inverse matrices

Reading supplied row-reduction tables

Prove or disprove invertibility without determinants

Common mistakes

Quick checks

Exercise

Related notes

Prerequisites

Key terms in this section