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Inverse Matrix Calculator

Created by Maciej Kowalski, PhD candidate
Reviewed by Bogna Szyk and Jack Bowater
Last updated: Sep 19, 2023


Welcome to the inverse matrix calculator, where you'll have the chance to learn all about inverting matrices. This operation is similar to searching for the fraction of a given number, except now we're multiplying matrices and want to obtain the identity matrix as a result.

But don't worry. Before we give, say, the inverse of a 4×44\times4 matrix, we'll look at some basic definitions, including a singular and nonsingular matrix. Then we'll move on to the general inverse matrix formula with a neat simplification for the inverse of a 2×22\times2 matrix and some useful matrix inverse properties. Last but not least, we give an example with thorough calculations of how to find the inverse of a 3×33\times3 matrix.

What is a matrix?

In primary school, they teach you the natural numbers, 11, 22, or 143143, and they make perfect sense – you have 11 toy car, 22 comic books, and terribly long 143143 days until Christmas. Then they tell you that there are also fractions (or rational numbers, as they call them), such as 1/21/2, or decimals, like 1.251.25, which still seems reasonable. After all, you gave 1/21/2 of your chocolate bar to your brother, and it cost $1.25\text{\textdollar}1.25. Next, you meet the negative numbers like 2-2 or 30-30, and they're a bit harder to grasp. But, once you think about it, one guy from your class got 2-2 points on a test for cheating, and there was a $30-\text{\textdollar}30 discount on jeans on Black Friday.

Lastly, the school introduces real numbers and some weird worm-like symbols that they keep calling square roots. What's even worse, while 4\sqrt{4} is a simple 22, 3\sqrt{3} is something like 1.73205...1.73205... and the digits go on forever. They convince you that such numbers describe, for example, the diagonal of a rectangle. And then there's π\pi, which somehow appeared out of nowhere when you talked about circles. Fair enough, maybe those numbers are real in some sense. But that's just about as far as it can go, right?

Wrong. Mathematicians are busy figuring out various interesting and, believe it or not, useful extensions of real numbers. The most important one is complex numbers, which are the starting point for any modern physicist. Fortunately, that's not the direction we're taking here. There is another.

A matrix is an array of elements (usually numbers) that has a set number of rows and columns. An example of a matrix would be:

A=(310211)\scriptsize A=\begin{pmatrix} 3&-1\\ 0&2\\ 1&-1 \end{pmatrix}

Moreover, we say that a matrix has cells, or boxes, in which we write the elements of our array. For example, matrix AA above has the value 22 in the cell that is in the second row and the second column. The starting point here is 1-cell matrices, which are basically the same thing as real numbers.

As you can see, matrices are a tool used to write a few numbers concisely and operate with the whole lot as a single object. As such, they are extremely useful when dealing with:

  • Systems of equations, especially when using Cramer's rule or as we've seen in our condition numbers calculator;
  • Vectors and vector spaces;
  • 3-dimensional geometry (e.g., the dot product and the cross product);
  • Eigenvalues and eigenvectors; and
  • graph theory and discrete mathematics.

Calculations with matrices are a great deal trickier than with numbers. For instance, if we want to add them, we first have to make sure that we can. But, since we're here on the inverse matrix calculator, we leave addition for later. First, however, let's familiarize ourselves with a few definitions.

Singular and nonsingular matrix, the identity matrix

Whether you want to find the inverse of a 2×22\times2 matrix or the inverse of a 4×44\times4 matrix, you have to understand one thing first: it doesn't always exist. Think of a fraction, say a/ba / b. Such a thing is perfectly fine as long as bb is non-zero. If it is, the expression doesn't make sense, and a similar thing happens for matrices.

A singular matrix is one that doesn't have an inverse. A nonsingular matrix is (surprise, surprise) one that does. Therefore, whenever you face an exercise with an inverse matrix, you should begin by checking if it's nonsingular. Otherwise, there's no point sweating over calculations. It just cannot be done.
You can still get pretty close to a singular matrix's inverse by instead calculating its Moore-Penrose pseudoinverse. If you don't know what the pseudoinverse is, wait no more and jump to the pseudoinverse calculator!

By definition, the inverse of a matrix AA is a matrix A1A^{-1} for which:

AA1=A1A=IA\cdot A^{-1} = A^{-1}\cdot A = \mathbb{I}

Where I\mathbb{I}denotes the identity matrix, i.e., a square matrix that has 11s on the main diagonal and 00s elsewhere. For example, the 3×33\times3 identity matrix is:

I=(100010001)\scriptsize\mathbb{I} = \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}

In other words, when given an arbitrary matrix AA, we want to find another one for which the product of the two (in whatever order) gives the identity matrix. Think of I\mathbb{I} as 11 (the identity element) in the world of matrices. After all, for a fraction a/ba / b, its inverse is b/ab / a but not just because we "flip it" (at least, not by definition). It's because of a similar multiplication property:

ab×ba=ba×ab=1\scriptsize\frac{a}{b}\times\frac{b}{a}=\frac{b}{a}\times\frac{a}{b}=1

That was enough time spent reading through definitions, don't you think? Let's finally see the inverse matrix formula and learn how to find the inverse of a 2×22\times2, 3×33\times3, and 4×44\times4 matrix.

How to find the inverse of a matrix: inverse matrix formula

Before we go into special cases, like the inverse of a 2×22\times2 matrix, let's take a look at the general definition.

Let AA be a square nonsingular matrix of size nn. Then the inverse A1A^{-1} (if it exists) is given by the formula:

1A×((1)1+1×A11(1)1+2×A12(1)1+n×A1n(1)2+1×A21(1)2+2×A22(1)2+n×A2n(1)n+1×An1(1)n+2×An2(1)n+n×Ann)\scriptsize\frac{1}{|A|}\times\begin{pmatrix}(-1)^{1+1}\times A_{11}&(-1)^{1+2}\times A_{12}&\cdots&(-1)^{1+n}\times A_{1n}\\ (-1)^{2+1}\times A_{21}&(-1)^{2+2}\times A_{22}&\cdots&(-1)^{2+n}\times A_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ (-1)^{n+1}\times A_{n1}&(-1)^{n+2}\times A_{n2}&\cdots&(-1)^{n+n}\times A_{nn}\end{pmatrix}

The A|A| is the determinant of AA (not to be confused with the absolute value of a number). The AijA_{ij} denotes the i,ji,j-minor of AA, i.e., the determinant of the matrix obtained from AA by forgetting about its ithi^{\mathrm{th}} row and jthj^{\mathrm{th}} column (it is a square matrix of size n1n-1). What we have obtained in called the cofactor matrix of AA. Lastly, the T^{\mathrm{T}} outside the array is the transposition. It means that once we know the cells inside, we have to "flip them" so that the ithi^{\mathrm{th}} row will become its ithi^{\mathrm{th}}h column and vice versa, as we taught you at the matrix transpose calculator. This leads to the adjoint matrix of AA. All these steps are detailed at Omni's adjoint matrix calculator, in case you need a more formal explanation.

Phew, that was a lot of symbols and a lot of technical mumbo-jumbo, but that's just the way mathematicians like it. Some of us wind down by watching romcoms, and others write down definitions that sound smart. Who are we to judge them?

In the next section, we point out a few important facts to take into account when looking for the inverse of a 4×44\times4 matrix, or whatever size it is. But before we see them, let's take some time to look at what the above matrix inverse formula becomes when it's the inverse of a 2×22\times2 matrix that we're looking for.

Let:

A=(abcd)\scriptsize A = \begin{pmatrix}a&b\\c&d\end{pmatrix}

Then the minors (the AijA_{ij}s above) come from crossing out one of the rows and one of the columns. But if we do that, we'll be left with a single cell! And the determinant of such a thing (a 1×11\times1 matrix) is just the number in that cell. For example, A12A_{12} comes from forgetting the first row and the second column, which means that only cc remains (or rather (c)\begin{pmatrix}c\end{pmatrix} since it's a matrix). Therefore,

A1=1A×((1(1+1×d(1)1+2×c(1)2+1×b(1)2+2×a)T\scriptsize A^{-1} = \frac{1}{|A|}\times\begin{pmatrix}(-1(^{1+1}\times d& (-1)^{1+2}\times c\\(-1)^{2+1}\times b&(-1)^{2+2}\times a\end{pmatrix}^{\mathrm{T}}

Also, in this special case, the determinant is simple enough: A=a×db×c|A| = a\times d - b\times c. So after taking the minuses and the transposition, we arrive at a nice and pretty formula for the inverse of a 2×22\times2 matrix:

A1=1a×db×c×(dbca)\scriptsize A^{-1} = \frac{1}{a\times d-b\times c}\times\begin{pmatrix}d&-b\\-c&a\end{pmatrix}

Arguably, the inverse of a 4×44\times4 matrix is not as easy to calculate as the 2×22\times2 case. There is an alternative way of calculating the inverse of a matrix; the method involves elementary row operations and the so-called Gaussian elimination (for more information, be sure to check out the (reduced) row echelon form calculator). As an example, we describe below how to find the inverse of a 3×33\times3 matrix using the alternative algorithm.

Say that you want to calculate the inverse of a matrix:

(a1a2a3b1b2b3c1c2c3)\scriptsize\begin{pmatrix} a_1&a_2&a_3 \\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{pmatrix}

We then construct a matrix with three rows and twice as many columns like the one below:

(a1a2a3100b1b2b3010c1c2c3001)\scriptsize\begin{pmatrix} a_1&a_2&a_3&\vdots&1&0&0 \\ b_1&b_2&b_3&\vdots&0&1&0\\ c_1&c_2&c_3&\vdots&0&0&1 \end{pmatrix}

and use Gaussian elimination on the 6-element rows of the matrix to transform it into something of the form:

(100x1x2x3010y1y2y3001z1z2z3)\scriptsize\begin{pmatrix} 1&0&0&\vdots &x_1&x_2&x_3\\ 0&1&0&\vdots&y_1&y_2&y_3\\ 0&0&1&\vdots&z_1&z_2&z_3 \end{pmatrix}

where the xx's, yy's, and zz's are obtained along the way from the transformations. Then:

A1=(x1x2x3y1y2y3z1z2z3)\scriptsize A^{-1} = \begin{pmatrix} x_1&x_2&x_3\\ y_1&y_2&y_3\\ z_1&z_2&z_3 \end{pmatrix}

Whichever method you prefer, it might be useful to check out a few matrix inverse properties to make our studies a little easier.

Matrix inverse properties

Below we list a few observations and matrix inverse properties.

  1. The inverse of a matrix doesn't always exist. Let's take a closer look at the inverse matrix formula in the section above. It contains the determinant of the matrix. This means that, first of all, we need to have a square matrix even to start thinking about its inverse. Secondly, the determinant appears in the denominator of a fraction in the inverse matrix formula. Therefore, if that determinant is equal to 00, then that expression doesn't make any sense, and the inverse doesn't exist.

  2. The inverse of an inverse is the initial matrix. In other words, if you invert a matrix twice, you'll obtain what you started with. Symbolically, we can write this property as (A1)1=A(A^{-1})^{-1} = A for an arbitrary nonsingular matrix AA.

  3. The inverse of a product is the product of the inverses in the reverse order. This means that if you have two square matrices AA and BB of the same size and want to calculate the inverse of their product, then, alternatively, you can find their individual inverses and multiply them but in the reverse order. In short, (AB)1=B1A1(A\cdot B)^{-1} = B^{-1}\cdot A^{-1}.

  4. The inverse of the transpose is the transpose of the inverse. In essence, it doesn't matter if you first transpose a matrix and then calculate its inverse or first find the inverse and only transpose it then. In symbolic notation, this translates to (AT)1=(A1)T(A^{\mathrm{T}})^{-1} = (A^{-1})^{\mathrm{T}}. In particular, observe that this relies on the fact that the determinant of a matrix stays the same after transposition.

We hope that you're sufficiently intrigued by the theory and can't wait to tell your friends about it over a cup of coffee. However, before you go spreading knowledge, let's go together through an example and see how to find the inverse of a 3×33\times3 matrix in practice.

Example: using the inverse matrix calculator

We'll now study step-by-step how to find the inverse of a 3×33\times3 matrix. Say that you're given an array:

A=(105216340)\scriptsize A = \begin{pmatrix} 1&0&5\\ 2&1&6\\ 3&4&0 \end{pmatrix}

Before we move on to the calculations, let's see how we can use the inverse matrix calculator to do it all for us.

First of all, we're dealing with a 3×33\times3 matrix, so we have to tell the calculator that by choosing the proper option under "Matrix size." This will show us a symbolic example of such an array with cells denoted a1a_1, a2a_2, and so on. We have to input the numbers given by our matrix under the correct symbols from the picture. For example, a3a_3 is in the first row in the third column, so we find the corresponding cell in our matrix and check that it has 55 in there. Therefore, we put a3=5a_3 = 5 into the inverse matrix calculator. Similarly, we get the other cells:

We define the other cells:

a1=1a2=0a3=5\scriptsize \begin{split} a_1&=1\\ a_2&=0\\ a_3&=5 \end{split}

Then:

b1=2b2=1b3=6\scriptsize \begin{split} b_1&=2\\ b_2&=1\\ b_3&=6 \end{split}

And:

c1=3c2=4c3=0\scriptsize \begin{split} c_1&=3\\ c_2&=4\\ c_3&=0 \end{split}

The moment we input the last number, the inverse matrix calculator will spit out the answer or tell us that the inverse doesn't exist. But, if you don't want any spoilers, we can also do the calculations by hand.

A priori, we don't even know if A1A^{-1} exists, maybe it's just a fairytale like vampires? To make sure, let's calculate its determinant:

A=1×1×0+0×6×3+5×2×45×1×30×2×01×6×4=0+0+4015024=1\scriptsize \begin{split} |A| = 1\times1\times0+0\times6\times3+5\times2\times4\\-5\times1\times3-0\times2\times0-1\times6\times4\\=0+0+40-15-0-24=1 \end{split}

Phew, no vampires today, just a nonsingular matrix and good ol' mathematics.

Recall the matrix inverse formula and observe that it's now time to calculate the AijA_{ij}s for ii and jj between 11 and 33. As an example, let's take, say, A11A_{11}, and A23A_{23}. The first of the two is the determinant of what we get by forgetting the first row and the first column of AA. This means that:

A11=1640\scriptsize A_{11} = \begin{vmatrix}1&6\\4&0\end{vmatrix}

Similarly, A23A_{23} comes from crossing out the second row and the third column:

A23=1034\scriptsize A_{23} = \begin{vmatrix}1&0\\3&4\end{vmatrix}

This gives:

A11=1×06×4=23\scriptsize A_{11} = 1\times0-6\times4=-23

And:

A23=1×40×3=4\scriptsize A_{23} = 1\times4-0\times3=4

The complete first row is:

A11=23A12=18A13=5\scriptsize \begin{split} A_{11} & = -23\\ A_{12} & = -18\\ A_{13} & = 5\\ \end{split}

For the second row, we find:

A21=20A22=15A23=4\scriptsize \begin{split} A_{21} & = -20\\ A_{22} & = -15\\ A_{23} & = 4\\ \end{split}

And the third row is:

A31=5A32=4A33=1\scriptsize \begin{split} A_{31} & = -5\\ A_{32} & = -4\\ A_{33} & = 1\\ \end{split}

It only remains to use the inverse matrix formula and plug in all the numbers we've calculated above:

A1=1A×((1)1+1×A11(1)1+2×A12(1)1+3×A13(1)2+1×A21(1)2+2×A22(1)2+3×A23(1)3+1×A31(1)3+2×A32(1)3+3×A33)T=1A×((1)1+1×(23)(1)1+2×(18)(1)1+3×5(1)2+1×(20)(1)2+2×(15)(1)2+3×4(1)3+1×(5)(1)3+2×(4)(1)3+3×1)T=1×(2318520154541)T=(2320518154541)\scriptsize \begin{split} A^{-1} &= \frac{1}{|A|}\!\times\!\begin{pmatrix}(-1)^{1+1}\times A_{11}&(-1)^{1+2}\times A_{12}&(-1)^{1+3}\times A_{13}\\ (-1)^{2+1}\times A_{21}&(-1)^{2+2}\times A_{22}&(-1)^{2+3}\times A_{23}\\ (-1)^{3+1}\times A_{31}&(-1)^{3+2}\times A_{32}&(-1)^{3+3}\times A_{33}\end{pmatrix}^{\mathrm{\!\!T}}\\[1.5em] &= \frac{1}{|A|}\!\times\!\begin{pmatrix}(-1)^{1+1}\times (-23)&(-1)^{1+2}\times (-18)&(-1)^{1+3}\times 5\\ (-1)^{2+1}\times (-20)&(-1)^{2+2}\times(-15)&(-1)^{2+3}\times 4\\ (-1)^{3+1}\times (-5)&(-1)^{3+2}\times(-4)&(-1)^{3+3}\times1\end{pmatrix}^{\mathrm{\!\!T}}\\[1.5em] &= 1\!\times\!\begin{pmatrix} -23&18&5\\ 20&-15&-4\\ -5&4&1 \end{pmatrix}^{\mathrm{\!\!T}}\\[1.5em] &= \begin{pmatrix} -23&20&-5\\ 18&-15&4\\ 5&-4&1 \end{pmatrix} \end{split}