Instructions: Use this calculator to find the adjoint of a matrix you provide showing all the steps. First, click on one of the buttons below to specify the dimension of the matrix.

Then, click on the first cell and type the value, and move around the matrix by pressing "TAB" or by clicking on the corresponding cells, to define ALL the matrix values.

\begin{bmatrix} & \\ & \end{bmatrix}

Same way as cofactors, the adjoint matrix is tightly associated with the inverse of a matrix. Indeed, the inverse matrix and the adjoint matrix are close lookalikes.

In all fairness, the concept of adjoint of a matrix plays a very important role in advanced math (where instead of matrices we deal with linear operators). But in college math, the only times you will likely stumble upon the adjoint is when you compute the inverse of a matrix using the adjoint formula.

### How do you find the adjoint of a matrix?

First, in terms of how adjoint of a matrix is calculated, let us recall the minors matrix which is computed by calculating the determinant of the sub-matrices formed by removing the i-th row and j-th column of the given matrix $$A$$.

So then, the minors were defined as:

$M_{ij} = \det A^{i,j}$ ### How to get to the cofactor matrix?

The cofactor matrix, $$C$$ is obtained from the minors by adding certain "signs", and defined as:

$C_{ij} = (-1)^{i+j} M_{ij}$

### Finally, how do you get to the adjoint matrix? What is the adjoint formula?

Simple! Once you have the cofactor matrix calculated already, the you need to transpose the matrix in order to get the adjoint. Concretely:

$adj(A) = C^T$

So, in order to make it easier to remember we have broken-down the adjoint formula into 3 steps: First, you compute the matrix of minors, then you compute the cofactors, and then, you transpose the cofactors to get the adjoint.

### Are the adjoint and transpose same?

Although the adjoint involves transposing a matrix, in general the adjoint and the transpose matrices are different from each other.

### How do you find the adjoint of a 4x4 matrix or bigger?

The process of finding the adjoint can numerically extensive, considering that you need to compute $$n^2$$ sub-determinants, which can grow fast with $$n \ge 4$$. ### Example of adjoint Matrix calculation

Question: Consider the following matrix

$\begin{bmatrix} \displaystyle 2&\displaystyle 3&\displaystyle 1\\[0.6em]\displaystyle 2&\displaystyle 4&\displaystyle 1\\[0.6em]\displaystyle 1&\displaystyle 1&\displaystyle 1 \end{bmatrix}$

Compute the associated adjoint matrix $$adj A$$.

Solution:

We need to compute the adjoint matrix of the $$3 \times 3$$ matrix that has been provided:

### Step 1: Compute the Cofactor Matrix

First we compute the minors matrix. We have that, by definition, the minors matrix $$M$$ is defined by the formula

$M_{ij} = \det A^{i,j}$

where in this case $$A^{i,j}$$ is the matrix $$A$$ after deleting row $$i$$ and column $$j$$.

Therefore, and based on the matrix $$A$$ provided we get the following coefficients of the minors matrix:

For $$A^{ 1, 1}$$:

$M_{ 1 1} = \det A^{ 1 1} = \begin{vmatrix} \displaystyle 4&\displaystyle 1\\[0.6em]\displaystyle 1&\displaystyle 1 \end{vmatrix} = 4 \cdot \left( 1 \right) - 1 \cdot \left(1 \right) = 3$

For $$A^{ 1, 2}$$:

$M_{ 1 2} = \det A^{ 1 2} = \begin{vmatrix} \displaystyle 2&\displaystyle 1\\[0.6em]\displaystyle 1&\displaystyle 1 \end{vmatrix} = 2 \cdot \left( 1 \right) - 1 \cdot \left(1 \right) = 1$

For $$A^{ 1, 3}$$:

$M_{ 1 3} = \det A^{ 1 3} = \begin{vmatrix} \displaystyle 2&\displaystyle 4\\[0.6em]\displaystyle 1&\displaystyle 1 \end{vmatrix} = 2 \cdot \left( 1 \right) - 1 \cdot \left(4 \right) = -2$

For $$A^{ 2, 1}$$:

$M_{ 2 1} = \det A^{ 2 1} = \begin{vmatrix} \displaystyle 3&\displaystyle 1\\[0.6em]\displaystyle 1&\displaystyle 1 \end{vmatrix} = 3 \cdot \left( 1 \right) - 1 \cdot \left(1 \right) = 2$

For $$A^{ 2, 2}$$:

$M_{ 2 2} = \det A^{ 2 2} = \begin{vmatrix} \displaystyle 2&\displaystyle 1\\[0.6em]\displaystyle 1&\displaystyle 1 \end{vmatrix} = 2 \cdot \left( 1 \right) - 1 \cdot \left(1 \right) = 1$

For $$A^{ 2, 3}$$:

$M_{ 2 3} = \det A^{ 2 3} = \begin{vmatrix} \displaystyle 2&\displaystyle 3\\[0.6em]\displaystyle 1&\displaystyle 1 \end{vmatrix} = 2 \cdot \left( 1 \right) - 1 \cdot \left(3 \right) = -1$

For $$A^{ 3, 1}$$:

$M_{ 3 1} = \det A^{ 3 1} = \begin{vmatrix} \displaystyle 3&\displaystyle 1\\[0.6em]\displaystyle 4&\displaystyle 1 \end{vmatrix} = 3 \cdot \left( 1 \right) - 4 \cdot \left(1 \right) = -1$

For $$A^{ 3, 2}$$:

$M_{ 3 2} = \det A^{ 3 2} = \begin{vmatrix} \displaystyle 2&\displaystyle 1\\[0.6em]\displaystyle 2&\displaystyle 1 \end{vmatrix} = 2 \cdot \left( 1 \right) - 2 \cdot \left(1 \right) = 0$

For $$A^{ 3, 3}$$:

$M_{ 3 3} = \det A^{ 3 3} = \begin{vmatrix} \displaystyle 2&\displaystyle 3\\[0.6em]\displaystyle 2&\displaystyle 4 \end{vmatrix} = 2 \cdot \left( 4 \right) - 2 \cdot \left(3 \right) = 2$

Summarizing, the minors matrix is:

$M = \begin{bmatrix} \displaystyle 3&\displaystyle 1&\displaystyle -2\\[0.6em]\displaystyle 2&\displaystyle 1&\displaystyle -1\\[0.6em]\displaystyle -1&\displaystyle 0&\displaystyle 2 \end{bmatrix}$

Now, we can compute the elements of the cofactor matrix $$C$$ using the formula

$C_{ij} = (-1)^{i+j}M_{ij}$

The above formula can be used directly because the minors are known already. We get

$C_{ 1 1} = (-1)^{ 1+1} \cdot 3 = (-1)^{ 2} \cdot 3 = 3$ $C_{ 1 2} = (-1)^{ 1+2} \cdot 1 = (-1)^{ 3} \cdot 1 = -1$ $C_{ 1 3} = (-1)^{ 1+3} \left(-2\right)= (-1)^{ 4} \left(-2\right) = -2$ $C_{ 2 1} = (-1)^{ 2+1} \cdot 2 = (-1)^{ 3} \cdot 2 = -2$ $C_{ 2 2} = (-1)^{ 2+2} \cdot 1 = (-1)^{ 4} \cdot 1 = -1$ $C_{ 2 3} = (-1)^{ 2+3} \left(-1\right)= (-1)^{ 5} \left(-1\right) = 1$ $C_{ 3 1} = (-1)^{ 3+1} \left(-1\right)= (-1)^{ 4} \left(-1\right) = 1$ $C_{ 3 2} = (-1)^{ 3+2} \cdot 0 = (-1)^{ 5} \cdot 0 = 0$ $C_{ 3 3} = (-1)^{ 3+3} \cdot 2 = (-1)^{ 6} \cdot 2 = -2$

Summarizing, the cofactor matrix is:

$C = \begin{bmatrix} \displaystyle 3&\displaystyle -1&\displaystyle -2\\[0.6em]\displaystyle -2&\displaystyle -1&\displaystyle 1\\[0.6em]\displaystyle 1&\displaystyle 0&\displaystyle -2 \end{bmatrix}$

### Step 2: Compute the Adjoint Matrix from the cofactor Matrix

Now, we need just need to transpose the cofactor matrix we have found to compute the adjoint matrix. We get:

$adj(A) = C^T = \begin{bmatrix} \displaystyle 3&\displaystyle -1&\displaystyle -2\\[0.6em]\displaystyle -2&\displaystyle -1&\displaystyle 1\\[0.6em]\displaystyle 1&\displaystyle 0&\displaystyle -2 \end{bmatrix} ^T = \begin{bmatrix} \displaystyle 3&\displaystyle -2&\displaystyle 1\\[0.6em]\displaystyle -1&\displaystyle -1&\displaystyle 0\\[0.6em]\displaystyle -2&\displaystyle 1&\displaystyle -2 \end{bmatrix}$

which concludes the calculation of the adjoint matrix.