Inverse of a matrix becomes important when we multiply by a matrix \( A \) and then want to perform the reverse operation. After all, multiplying by a matrix changes a vector (or another matrix), so a natural question comes up: how can we recover the original data? That’s exactly why we introduce the inverse matrix.
However, there’s a key detail: an inverse does not exist for every matrix. So before doing any calculations, it’s important to clearly understand when a matrix has an inverse and when it does not. Only after that does it make sense to move on to the adjoint (cofactor) method, because it provides a concrete formula and a clear sequence of steps.
Inverse of a Matrix and the Existence Condition: When Is It Defined?
Let \( A \) be a square matrix of order \( n \). A square matrix \( A^{-1} \) of the same order is called the inverse of \( A \) if the following equality holds:
\[ A\cdot A^{-1}=A^{-1}\cdot A=I, \]
where \( I \) is the identity matrix:
\[ I= \begin{pmatrix} 1&0&\cdots&0\\ 0&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&1 \end{pmatrix}. \]
In other words, multiplying by \( A \) and multiplying by \( A^{-1} \) ultimately give the identity matrix—and this works both on the left and on the right. It sounds logical, but can we always find such a matrix \( A^{-1} \)?
Let’s recall the main fact: for any square matrix, the inverse exists and is unique only when the matrix is non-singular, that is,
\[ \det(A)\neq 0. \]
If \( \det(A)=0 \), then it is impossible to construct an inverse matrix, because after multiplying by \( A \), the same result may correspond to different original data. So restoring the initial values uniquely is no longer possible.
That’s why this rule is very practical: the first thing we check is the determinant. If \( \det(A)=0 \), the method stops immediately. If \( \det(A)\neq 0 \), we move on to constructing \( A^{-1} \).
Inverse of a Matrix by the Adjoint Method: What Does the Algorithm Look Like?
To find an inverse of a matrix using the adjoint (cofactor) method, we follow a clear step-by-step procedure. First, we compute the determinant of the matrix \( A \) (for example, using Gaussian elimination or determinant expansion) and immediately check the existence condition:
\[ \det(A)\neq 0. \]
If \( \det(A)=0 \), we stop right there—an inverse matrix does not exist. If \( \det(A)\neq 0 \), we continue and, for every entry \( a_{ij} \) of the matrix \( A \), we find the minor \( M_{ij} \). The minor \( M_{ij} \) is the determinant of the \( (n-1)\times(n-1) \) submatrix obtained from \( A \) by deleting the \( i \)-th row and the \( j \)-th column (the row and column where the element \( a_{ij} \) is located).
Next, we compute the cofactors (also called algebraic complements):
\[ C_{ij}=(-1)^{i+j}\cdot M_{ij}. \]
So we take the corresponding minor and apply the sign rule \( (-1)^{i+j} \). It’s easy to make a sign mistake here, so it helps to remember: if \( i+j \) is even, the sign is “(+)”; if \( i+j \) is odd, the sign is “(-)”. For example, for a \( 4\times4 \) matrix, the sign pattern looks like this:
\[ \begin{pmatrix} +&-&+&-\\ -&+&-&+\\ +&-&+&-\\ -&+&-&+ \end{pmatrix}. \]
When all \( C_{ij} \) values are found, we build the cofactor matrix:
\[ C= \begin{pmatrix} C_{11}&C_{12}&\cdots&C_{1n}\\ C_{21}&C_{22}&\cdots&C_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ C_{n1}&C_{n2}&\cdots&C_{nn} \end{pmatrix}. \]
Now we transpose this matrix:
\[ C^{T}= \begin{pmatrix} C_{11}&C_{21}&\cdots&C_{n1}\\ C_{12}&C_{22}&\cdots&C_{n2}\\ \vdots&\vdots&\ddots&\vdots\\ C_{1n}&C_{2n}&\cdots&C_{nn} \end{pmatrix}. \]
The matrix \( C^{T} \) is called the adjoint of \( A \) and is denoted by \( \operatorname{adj}(A) \):
\[ \operatorname{adj}(A)=C^{T}. \]
Note. \( \operatorname{adj}(A) \) is not the matrix \( C \), but specifically its transpose \( C^{T} \).
Finally, the inverse matrix is computed by the formula:
\[ A^{-1}=\frac{1}{\det(A)}\cdot \operatorname{adj}(A) =\frac{1}{\det(A)}\cdot C^{T}. \]
This method is especially convenient for \( 2\times2 \)—\( 4\times4 \) matrices, where the amount of computation is still manageable. For larger sizes, the formulas stay the same, but the number of minors and determinants grows very quickly. That’s why, in practice, people often switch to algorithms such as Gauss–Jordan elimination.
One last tip—a quick self-check that often saves you from small sign mistakes: after finding \( A^{-1} \), you can multiply \( A\cdot A^{-1} \) and make sure the result is the identity matrix \( I \). If you don’t get \( I \), the issue is most often either in the signs \( (-1)^{i+j} \) or in computing the minors \( M_{ij} \).
Inverse of a Matrix in Practice: Examples with Solutions
To understand the method better, let’s move to practical examples. Step by step, we’ll go through a few typical problems where we apply what we’ve learned. Try to pay attention not only to the final answers, but also to the logic of each stage—this will help you feel more confident in similar exercises later.
Example 1. Find the inverse of the matrix
\[ A= \begin{pmatrix} 2 & -1 & 4 \\ 4 & -2 & 8 \\ 6 & -3 & 12 \end{pmatrix}. \]
The first step is to check whether an inverse of a matrix can exist. Since the second row is twice the first row, and the third row is three times the first row, the rows are linearly dependent. This means the determinant of the matrix is zero:
\[ \det(A)=0. \]
So the inverse matrix does not exist.
Example 2. Find the inverse of the matrix
\[ A= \begin{pmatrix} 5 & -6 & 7 & 1\\ 7 & 10 & -9 & 8\\ 3 & 3 & 5 & 1\\ -10 & 2 & 3 & 4 \end{pmatrix}. \]
First, we check the condition for the inverse matrix to exist, meaning we need to confirm that the determinant of \( A \) is nonzero:
\[ \det(A)= \begin{vmatrix} 5 & -6 & 7 & 1\\ 7 & 10 & -9 & 8\\ 3 & 3 & 5 & 1\\ -10 & 2 & 3 & 4 \end{vmatrix}. \]
We expand the determinant along the first row:
\[ \det(A)= 5 \cdot \begin{vmatrix} 10 & -9 & 8\\ 3 & 5 & 1\\ 2 & 3 & 4 \end{vmatrix} +6 \cdot \begin{vmatrix} 7 & -9 & 8\\ 3 & 5 & 1\\ -10 & 3 & 4 \end{vmatrix} +7 \cdot \begin{vmatrix} 7 & 10 & 8\\ 3 & 3 & 1\\ -10 & 2 & 4 \end{vmatrix} -1 \cdot \begin{vmatrix} 7 & 10 & -9\\ 3 & 3 & 5\\ -10 & 2 & 3 \end{vmatrix}. \]
After computing, we get:
\[ \det(A) = 5\cdot 252 + 6\cdot 789 + 7\cdot 138 – 1\cdot(-921) = 7881 \neq 0. \]
So the matrix \( A \) is square and non-singular, which means we can find its inverse. Now we can proceed with the calculations.
We construct the minors \( M_{ij} \) for each element \( a_{ij} \) of the matrix \( A \). To do this, we delete the \( i \)-th row and the \( j \)-th column, and then compute the determinant of the remaining submatrix:
\[ \begin{array}{llll} M_{11}=\begin{vmatrix}10&-9&8\\3&5&1\\2&3&4\end{vmatrix}=252, & M_{12}=\begin{vmatrix}7&-9&8\\3&5&1\\-10&3&4\end{vmatrix}=789, & M_{13}=\begin{vmatrix}7&10&8\\3&3&1\\-10&2&4\end{vmatrix}=138, \\[6pt] M_{14}=\begin{vmatrix}7&10&-9\\3&3&5\\-10&2&3\end{vmatrix}=-921, & M_{21}=\begin{vmatrix}-6&7&1\\3&5&1\\2&3&4\end{vmatrix}=-173, & M_{22}=\begin{vmatrix}5&7&1\\3&5&1\\-10&3&4\end{vmatrix}=-10, \\[6pt] M_{23}=\begin{vmatrix}5&-6&1\\3&3&1\\-10&2&4\end{vmatrix}=218, & M_{24}=\begin{vmatrix}5&-6&7\\3&3&5\\-10&2&3\end{vmatrix}=601, & M_{31}=\begin{vmatrix}-6&7&1\\10&-9&8\\2&3&4\end{vmatrix}=240, \\[6pt] M_{32}=\begin{vmatrix}5&7&1\\7&-9&8\\-10&3&4\end{vmatrix}=-1125, & M_{33}=\begin{vmatrix}5&-6&1\\7&10&8\\-10&2&4\end{vmatrix}=882, & M_{34}=\begin{vmatrix}5&-6&7\\7&10&-9\\-10&2&3\end{vmatrix}=624, \\[6pt] M_{41}=\begin{vmatrix}-6&7&1\\10&-9&8\\3&5&1\end{vmatrix}=469, & M_{42}=\begin{vmatrix}5&7&1\\7&-9&8\\3&5&1\end{vmatrix}=-64, & M_{43}=\begin{vmatrix}5&-6&1\\7&10&8\\3&3&1\end{vmatrix}=-181, \\[6pt] M_{44}=\begin{vmatrix}5&-6&7\\7&10&-9\\3&3&5\end{vmatrix}=694. \end{array} \]
Next, using the cofactor formula, we compute the cofactors:
\[ \begin{aligned} C_{11} &= (-1)^{1+1} \cdot M_{11} = +252, & C_{12} &= (-1)^{1+2} \cdot M_{12} = -789, & C_{13} &= (-1)^{1+3} \cdot M_{13} = +138, \\ C_{14} &= (-1)^{1+4} \cdot M_{14} = -921, & C_{21} &= (-1)^{2+1} \cdot M_{21} = +173, & C_{22} &= (-1)^{2+2} \cdot M_{22} = -10, \\ C_{23} &= (-1)^{2+3} \cdot M_{23} = -218, & C_{24} &= (-1)^{2+4} \cdot M_{24} = +601, & C_{31} &= (-1)^{3+1} \cdot M_{31} = +240, \\ C_{32} &= (-1)^{3+2} \cdot M_{32} = +1125, & C_{33} &= (-1)^{3+3} \cdot M_{33} = +882, & C_{34} &= (-1)^{3+4} \cdot M_{34} = -624, \\ C_{41} &= (-1)^{4+1} \cdot M_{41} = -469, & C_{42} &= (-1)^{4+2} \cdot M_{42} = -64, & C_{43} &= (-1)^{4+3} \cdot M_{43} = +181, \\ C_{44} &= (-1)^{4+4} \cdot M_{44} = +694. \end{aligned} \]
After that, the cofactor matrix is:
\[ C= \begin{pmatrix} C_{11}&C_{12}&C_{13}&C_{14}\\ C_{21}&C_{22}&C_{23}&C_{24}\\ C_{31}&C_{32}&C_{33}&C_{34}\\ C_{41}&C_{42}&C_{43}&C_{44} \end{pmatrix}= \begin{pmatrix} 252&-789&138&-921\\ 173&-10&-218&601\\ 240&1125&882&-624\\ -469&-64&181&694 \end{pmatrix}. \]
Now we transpose this matrix to obtain the adjoint:
\[ \operatorname{adj}(A)= \begin{pmatrix} 252&173&240&-469\\ -789&-10&1125&-64\\ 138&-218&882&181\\ -921&601&-624&694 \end{pmatrix}. \]
And using the inverse matrix formula:
\[ A^{-1}=\frac{1}{\det A} \cdot \operatorname{adj}(A), \]
we get:
\[ A^{-1}=\frac{1}{7881} \begin{pmatrix} 252&173&240&-469\\ -789&-10&1125&-64\\ 138&-218&882&181\\ -921&601&-624&694 \end{pmatrix} \approx \begin{pmatrix} 0.032&0.022&0.030&-0.060 \\ -0.100&-0.001&0.143&-0.008 \\ 0.018&-0.028&0.112&0.023 \\ 0.117&0.076&-0.079&0.088 \end{pmatrix}. \]
Example 3. Find the inverse of the matrix
\[ A= \begin{pmatrix} 0 & -1 & 3 & 0 & 2 \\ 2 & -4 & 1 & 5 & 3 \\ -4 & 5 & 7 & -10 & 0 \\ -2 & 1 & 8 & -5 & 3 \end{pmatrix}. \]
This matrix is not square, because it has \( 4 \) rows and \( 5 \) columns. And an inverse of a matrix (in the classical sense) is defined only for square matrices. Therefore, it is impossible to find an inverse matrix for this one.
Practical takeaway. Before starting any calculations, it’s always worth asking yourself two questions: “Is the matrix square?” and “Is the determinant nonzero?” This saves time and helps you immediately see whether looking for an inverse matrix even makes sense.
Where to Go Next: Topics That Strengthen Your Matrix Skills
If you already feel confident working with an inverse of a matrix using the adjoint (cofactor) method, a natural question comes up: what should you study next to feel even more comfortable with more advanced problems? Below are topics that continue this path in a logical way and add new approaches to a familiar idea.
- Inverse of a Matrix and the Characteristic Polynomial: An Unusual but Powerful Approach — We will explain the connection with the characteristic polynomial and show how it can be used in practice.
- Inverse of a Matrix Using the Bordering Method: How Adding Rows and Columns Leads to the Result — We will present the main idea of the bordering method and explain how an inverse matrix can be obtained through step-by-step expansion.
- Inverse of a Matrix Using Block Partitioning: When a Matrix Becomes Convenient Blocks — We will look at the block approach and explain how to invert a matrix by working with its parts.
Inverse of a Matrix in Code: Try Implementing the Algorithm in Your Own Program
Theory and worked examples give you a strong foundation, but real confidence often appears when you try to implement the idea yourself. That’s exactly where the flowchart in this section becomes useful: it shows the full calculation route step by step. Then the most interesting part is on you—pick your favorite programming language and turn this logic into a small program that takes a square matrix, checks whether an inverse exists, and returns the result in the form \( A^{-1} \).
Want to test yourself on real input, catch sign mistakes, and confirm that everything works the way it should? Then use the flowchart below as your guide and try writing your own implementation. It’s a short practice task, but it helps the topic stick really well.
