The One Matrix That Is Its Own Undo Button
In arithmetic, the number 1 is the value that changes nothing when you multiply by it, and it is its own reciprocal, since $1 \div 1 = 1$. Matrices have their own version of the number 1: the identity matrix. Ask for the inverse of the identity matrix - the matrix that "undoes" it - and the honest answer is that there is nothing to undo, so $I$ is its own inverse: $I^{-1} = I$. That single fact quietly sits under every equation you solve with matrices.
The reason is not a trick of notation. It falls straight out of what "identity" and "inverse" each mean, and you can prove it in two short lines.
What Is An Identity Matrix?
An identity matrix, written $I_n$ (or just $I$), is a square matrix of order $n$ with 1s all along the main diagonal and 0s in every other position. "Main diagonal" means the entries running from the top-left corner to the bottom-right corner. The 2×2 and 3×3 identity matrices look like this:
$$I_2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}, \qquad I_3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}.$$
Its defining behaviour is that it leaves any compatible matrix unchanged under multiplication:
$$A I = I A = A.$$
That is why $I$ is called the multiplicative identity for matrices - it plays the exact role the number 1 plays for ordinary numbers. You can read more on the base object in the identity matrix article.
What Is The Inverse Of A Matrix?
The inverse of a matrix $A$, written $A^{-1}$, is the matrix that multiplies with $A$ to give the identity:
$$A A^{-1} = A^{-1} A = I.$$
Only square matrices can have an inverse, and even then only when their determinant is not zero. A matrix with a non-zero determinant is called invertible (or non-singular). The standard formula for the inverse of a matrix is
$$A^{-1} = \frac{1}{\det A},\text{adj}(A),$$
where $\text{adj}(A)$ is the adjoint (adjugate) - the transpose of the cofactor matrix. Keep both defining ideas in view: the inverse is whatever matrix returns $I$ under multiplication, and the formula is one way to find it.
What Is The Inverse Of The Identity Matrix?
The inverse of the identity matrix of order $n$ is the identity matrix itself:
$$I_n^{-1} = I_n.$$
The definition of an inverse asks: which matrix $B$ satisfies $I , B = I$? Because the identity leaves every matrix unchanged, $I , I = I$ already holds, so $B = I$ does the job. The identity matrix is the one square matrix whose inverse you never have to compute - you already have it.
This is the matrix version of a fact you have used since primary school. The reciprocal of 1 is 1, since $1 \times 1 = 1$; the inverse of $I$ is $I$, since $I , I = I$.
How Do You Prove The Inverse Of The Identity Matrix Is Itself?
There are two clean proofs, and each reinforces a different definition.
Proof 1: straight from the definition of inverse.
The inverse of $I$ is any matrix $B$ with $I , B = B , I = I$. Test $B = I$:
$$I , I = I.$$
Both required equations hold at once, and a matrix inverse is unique, so:
$$I^{-1} = I.$$
Proof 2: from the adjoint-over-determinant formula.
Apply $A^{-1} = \frac{1}{\det A},\text{adj}(A)$ to $A = I$. Two facts feed in. First, the determinant of the identity matrix is 1:
$$\det(I) = 1.$$
Second, the adjoint of the identity matrix is the identity matrix:
$$\text{adj}(I) = I.$$
Substitute both into the formula:
$$I^{-1} = \frac{1}{\det(I)},\text{adj}(I)$$ $$I^{-1} = \frac{1}{1}, I$$ $$I^{-1} = I.$$
Both routes land in the same place, which is the point: the result does not depend on which definition of inverse you start from.
What Are The Properties Of The Inverse Of The Identity Matrix?
Self-inverse (involutory). Because $I , I = I$, the identity is its own inverse. Any matrix $A$ with $A A = I$ is called an involutory matrix, and $I$ is the simplest example.
Determinant is 1. Since $\det(I) = 1 \neq 0$, the identity is always invertible, for every order $n$.
Order is preserved. The inverse of $I_n$ is again $I_n$, with the same size and entries.
Equal to its own transpose. The identity is symmetric, so $I^{T} = I$, and therefore $I^{-1} = I^{T}$ as well.
Powers stay identity. For any whole number $k$, $I^{k} = I$, so raising the identity to a power (including the power $-1$) never changes it.
Where Is The Inverse Of The Identity Matrix Used?
"Every undo needs a do-nothing to measure against."
The identity being self-inverse is not a curiosity — it is the anchor that makes matrix algebra behave:
Solving matrix equations. To solve $A X = B$ you multiply by $A^{-1}$, and the whole method works because $A^{-1} A = I$ and $I X = X$. The identity is the resting state the algebra collapses to.
Checking an inverse is correct. You confirm a computed inverse by multiplying it against the original and looking for $I$. If the product is the identity, the inverse is right.
Computer graphics and transformations. The identity matrix represents "no transformation." Composing a transformation with its inverse must return the identity, which is how a rotation and its reverse cancel exactly.
Iterative and numerical methods. Algorithms that refine an approximate inverse test their progress by how close the product sits to $I$.
The deeper idea is that invertible matrices form a group under multiplication, and every group needs an identity element that is its own inverse. For matrices, that element is $I$.
Examples Of Inverse Of Identity Matrix
The set runs from a direct 2×2 check, through the most common misconception, to the 3×3 case, an equation, the formula route, and the involutory idea.
Example 1
Verify that the inverse of $I_2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ is itself by multiplication.
Multiply $I_2$ by $I_2$ entry by entry:
$$\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}.$$
The product is $I_2$, which is exactly the condition $I_2 , I_2 = I_2$.
Final answer: $I_2^{-1} = I_2$.
Example 2
Find the inverse of $I_3$. A student tries to invert it entry by entry — follow their reasoning.
Wrong attempt. The student remembers that the inverse of a number is its reciprocal, so they take the reciprocal of every entry of $I_3$. The diagonal 1s become $\tfrac{1}{1} = 1$, but each off-diagonal 0 becomes $\tfrac{1}{0}$.
Where it breaks. $\tfrac{1}{0}$ is undefined, so "reciprocal of each entry" cannot even be written down. That is the signal the method is wrong: a matrix inverse is not the entry-wise reciprocal.
Correct. The matrix inverse is defined by multiplication, not by reciprocating entries. Since $I_3 , I_3 = I_3$, the inverse is the identity itself:
$$I_3^{-1} = I_3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}.$$
Final answer: $I_3^{-1} = I_3$. The entry-wise reciprocal is never how a matrix inverse is found.
Example 3
Confirm $I_3 , I_3 = I_3$ by direct multiplication.
Each entry of the product is a row of the first identity dotted with a column of the second. Row 1 with column 1 gives $(1)(1) + (0)(0) + (0)(0) = 1$; row 1 with column 2 gives $(1)(0)+(0)(1)+(0)(0)=0$, and so on:
$$\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}.$$
Final answer: the product is $I_3$, confirming $I_3$ is its own inverse.
Example 4
Solve the matrix equation $I , X = A$, where $A = \begin{bmatrix} 5 & 2 \ 3 & 4 \end{bmatrix}$.
Multiply both sides on the left by $I^{-1}$. Since $I^{-1} = I$:
$$I^{-1} (I , X) = I^{-1} A$$ $$(I^{-1} I), X = I , A$$ $$I , X = A$$ $$X = A.$$
Final answer: $X = \begin{bmatrix} 5 & 2 \ 3 & 4 \end{bmatrix}$. Multiplying by the identity, or its inverse, changes nothing.
Example 5
Find $I_2^{-1}$ using the adjoint-over-determinant formula.
The determinant of $I_2$ is $\det(I_2) = (1)(1) - (0)(0) = 1$. The adjoint of $I_2$ is $I_2$ itself. Apply the formula:
$$I_2^{-1} = \frac{1}{\det(I_2)},\text{adj}(I_2)$$ $$I_2^{-1} = \frac{1}{1}\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$ $$I_2^{-1} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}.$$
Final answer: $I_2^{-1} = I_2$, matching the direct multiplication check.
Example 6
Show that the identity matrix is involutory, and contrast it with a non-trivial involutory matrix.
A matrix $A$ is involutory when $A A = I$, which makes it its own inverse. For the identity, $I , I = I$, so it is involutory. A different involutory matrix is
$$A = \begin{bmatrix} 1 & 0 \ 0 & -1 \end{bmatrix}, \qquad A A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I.$$
Final answer: $I$ is the simplest involutory matrix; the reflection matrix above is a non-trivial one, and both equal their own inverse.
Common Mistakes
Mistake 1: Inverting the identity entry by entry
Where it slips in: Carrying the number rule "inverse means reciprocal" over to matrices, and reciprocating each entry.
Don't do this: Turn the 0s into $\tfrac{1}{0}$; the operation is undefined and the idea is wrong.
The correct way: A matrix inverse is defined by the product rule $A A^{-1} = I$, not by reciprocating entries. The learner who first meets matrices right after fractions is the one who reaches for entry-wise reciprocals, and the fix is to always return to "what do I multiply by to get $I$."
Mistake 2: Confusing the identity matrix with the zero matrix
Where it slips in: Treating $I$ as the matrix version of 0 because both look sparse.
Don't do this: Assume $I$ has no inverse, the way the zero matrix has no inverse.
The correct way: The identity is the matrix version of 1, not 0. Its determinant is 1, so it is always invertible, while the zero matrix has determinant 0 and no inverse. The habit that catches this is checking the determinant before claiming a matrix is non-invertible.
Mistake 3: Thinking the inverse of the identity changes with its size
Where it slips in: Expecting $I_3^{-1}$ to differ in form from $I_2^{-1}$.
Don't do this: Look for a "bigger" or different inverse just because the order grew.
The correct way: For every order $n$, $I_n^{-1} = I_n$. The size of the identity changes, but the rule that it is its own inverse does not.
The Mathematicians Behind The Identity Matrix
The identity matrix and the algebra of matrix inverses were shaped by Arthur Cayley (1821–1895, England), who set out matrix multiplication and the identity element in his 1858 memoir on matrices. Cayley showed that matrices form an algebra with a unit element that behaves like the number 1, which is exactly the property that makes the identity its own inverse.
Conclusion
The inverse of the identity matrix is the identity matrix itself: $I^{-1} = I$, for every order $n$.
It follows from the definition of an inverse ($I I = I$) and from the adjoint formula ($\det I = 1$, $\text{adj},I = I$).
The identity is involutory, symmetric, always invertible, and unchanged by any power.
The most common error is reciprocating entries; a matrix inverse is defined by multiplication, not entry-wise reciprocals.
To build fluency with matrix inverses alongside a teacher, explore Bhanzu's algebra tutor sessions or a high school math tutor for the full matrices chapter.
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