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Find the inverse of nxn matrix by using minors, cofactors, and adjugate

The Mathematician — Wed, 29 Jul 2020 12:18:12 +0000

This article will explain how to find the inverse of a $n \times n$ matrix by using minors, cofactors, and adjugate. First, we will start with the general form, how to apply the above tools. After that, we will see an example of how to apply that on the $2 \times 2$ and $3 \times 3$ matrix. If you are unfamiliar with the invertible matrix, check this article.

How to find the inverse of a nxn matrix

Let $A$ be a $n \times n$ matrix. Specifically:

A = \begin{bmatrix}\begin{array}{cccc}a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\\vdots & \vdots & \ddots & \vdots \\a_{n,1} & a_{n,2} & \cdots & a_{n,n}\end{array}\end{bmatrix}

First step: finding the minors of a nxn matrix

We already have written an article about finding minors, so we give here the resulting matrix of minors $M$ for the matrix $A$ :

M = \begin{bmatrix}\begin{array}{cccc}M_{1,1} & M_{1,2} & \cdots & M_{1,n} \\M_{2,1} & M_{2,2} & \cdots & M_{2,n} \\\vdots & \vdots & \ddots & \vdots \\M_{n,1} & M_{n,2} & \cdots & M_{n,n}\end{array}\end{bmatrix}

Second step: finding the cofactor matrix

If we look at the matrix of minors $M$ above, then each minor needs to be rewritten as

\begin{equation*}M_{i,j} = (-1)^{i + j} M_{i,j}\end{equation*}

to get the cofactor matrix. The resulting cofactor matrix $C$ will be

C = \begin{bmatrix}\begin{array}{cccc}(-1)^{1 + 1}M_{1,1} & (-1)^{1 + 2}M_{1,2} & \cdots & (-1)^{1 + n}M_{1,n} \\(-1)^{2 + 1} M_{2,1} & (-1)^{2 + 2}M_{2,2} & \cdots & (-1)^{2 + n}M_{2,n} \\\vdots & \vdots & \ddots & \vdots \\(-1)^{n + 1}M_{n,1} & (-1)^{n + 2}M_{n,2} & \cdots & (-1)^{n + n}M_{n,n}\end{array}\end{bmatrix}

If $i + j$ is odd, then the minor will be multiplied by -1. If $i + j$ is even, then it will be multiplied by 1 (so nothing will happen). It is also evident that the minors on the diagonal remain the same since adding two equal integers will result in an even integer.

Third step: adjucate the matrix

Now we applied the first two steps, and so far, we have found the cofactor $C$ of the matrix $A$ . Our third step is to find the adjugate of a matrix $A$ . That is finding the transpose of the cofactor matrix $C$ of $A$ :

\text{adj}(A) = C^{T} = \begin{bmatrix}\begin{array}{cccc}(-1)^{1 + 1}M_{1,1} & (-1)^{2 + 1} M_{2,1} & \cdots & (-1)^{n + 1}M_{n,1} \\(-1)^{1 + 2}M_{1,2} & (-1)^{2 + 2}M_{2,2} & \cdots & (-1)^{n + 2}M_{n,2} \\\vdots & \vdots & \ddots & \vdots \\(-1)^{1 + n}M_{1,n} & (-1)^{2 + n}M_{2,n} & \cdots & (-1)^{n + n}M_{n,n}\end{array} \end{bmatrix}

Fourth step: finding the determinant of the original matrix

Our last step is to find the determinant of the matrix $A$ . If we have find that, we can plug everything together, and we will get the resulting inverse of matrix $A$ :

\begin{equation*}A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)\end{equation*}

Example of an inverse of a 2×2 matrix

We will give an example of how to find the inverse of a 2×2 matrix.

A = \begin{bmatrix}\begin{array}{cc}5 & -2 \\7 & 3\end{array}\end{bmatrix}

First step: finding the minors of matrix A

We have done an example of finding the matrix of minors of the same $2\times 2$ matrix $A$ in this article.

M = \begin{bmatrix}\begin{array}{cc}3 & 7 \\-2 & 5\end{array}\end{bmatrix}

Second step: finding the cofactor matrix of A

The minors on diagonal will stay the same, but the minors $M_{1,2}$ and $M_{2,1}$ have an odd integer (1 + 2 = 2 + 1 = 3). Therefore, those two minors will be multiplied with -1:

C = \begin{bmatrix}\begin{array}{cc}3 & -7 \\2 & 5\end{array}\end{bmatrix}

Third step: adjucate the matrix A

\text{adj}(A) = C^{T} = \begin{bmatrix}\begin{array}{cc}3 & 2 \\-7 & 5\end{array}\end{bmatrix}

Fourth step: finding the determinant of the original matrix A

First we need to calculate the determinant: $\text{det}(A) = 5 \cdot 3 - (-2 \cdot 7) = 29$ . Now we plug everything together, and we will get the resulting inverse matrix of $A$ :

A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{29} \begin{bmatrix}\begin{array}{cc}3 & 2 \\-7 & 5\end{array}\end{bmatrix}

Example of an inverse of a 3×3 matrix

We will give an example of how to find the inverse of a 3×3 matrix.

A = \begin{bmatrix}\begin{array}{ccc}3 & -1 & 6 \\2 & 5 & 7 \\1 & -3 & 1\end{array}\end{bmatrix}

First step: finding the minors of matrix A

We have done an example of finding the matrix of minors of the same $3\times 3$ matrix $A$ in this article.

M = \begin{bmatrix}\begin{array}{ccc}26 & -5 & -11 \\17 & -3 & -8 \\-37 & 9 & 17\end{array}\end{bmatrix}

Second step: finding the cofactor matrix of A

The minors $M_{1,2}$ , $M_{2,1}$ , $M_{2,3}$ , and $M_{3,2}$ will be multiplied by -1, since the sum of the row and column position is odd.

C = \begin{bmatrix}\begin{array}{ccc}26 & 5 & -11 \\-17 & -3 & 8 \\-37 & -9 & 17\end{array}\end{bmatrix}

Third step: adjucate the matrix A

\text{adj}(A) = C^{T} = \begin{bmatrix}\begin{array}{ccc}26 & -17 & -37 \\5 & -3 & -9 \\-11 & 8 & 17\end{array}\end{bmatrix}

Fourth step: finding the determinant of the original matrix A

Finding the determinant of this example will cost more calculations than the previous one:

\text{det}(A) = 3 \cdot \begin{vmatrix}5 & 7 \\-3 & 1\end{vmatrix} - 2 \cdot \begin{vmatrix}-1 & 6 \\-3 & 1\end{vmatrix} + \cdot \begin{vmatrix}-1 & 6 \\5 & 7\end{vmatrix}

= 3(5\cdot 1 - 7\cdot (-3)) - 2((-1)\cdot 1 - 6\cdot (-3)) + (-1)\cdot 7 - 6 \cdot 5

= 3\cdot26 - 2\cdot 17 - 7 - 30 = 7

Now we plug everything together, and we will get the resulting inverse matrix of $A$ :

A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{7} \begin{bmatrix}\begin{array}{ccc}26 & -17 & -37 \\5 & -3 & -9 \\-11 & 8 & 17\end{array}\end{bmatrix}

Conclusion

This is probably one of the easiest ways to find the inverse of a matrix. There is also another way to find the inverse, that is, by using the Gauss-Jordan Method.

The post Find the inverse of nxn matrix by using minors, cofactors, and adjugate appeared first on Epsilonify.

How to find the minors of a nxn matrix?

The Mathematician — Sat, 25 Jul 2020 10:58:49 +0000

Finding a minor of a $n \times n$ matrix is easy, and the steps are almost the same like finding a determinant. It is the first step to find a cofactor matrix. We will start in this article with the general form of finding a minor, how to find a minor of a $2 \times 2$ , $3 \times 3$ , and $4\times 4$ matrix, where each section ends with an example.

What are the minors of a matrix

Let $A$ be a $n \times n$ matrix. Specifically:

A = \begin{bmatrix}\begin{array}{cccc} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n} \end{array}\end{bmatrix}

The minor $M_{i,j}$ of $A$ is the determinant of $n-1 \times n-1$ submatrix $A$ , where the $i$ th row and $j$ th column are deleted. In mathematical notation, we will get

M_{i,j} = \begin{vmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,j-1} & a_{1,j+1} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,j-1} & a_{2,j+1} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \cdots & \vdots \\ a_{i-1,1} & a_{i-1,2} & \cdots & a_{i-1,j-1} & a_{i-1,j+1} & \cdots & a_{i-1,n} \\ a_{i+1,1} & a_{i+1,2} & \cdots & a_{i+1,j-1} & a_{i+1,j+1} & \cdots & a_{i+1,n} \\ \vdots & \vdots & \cdots & \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,j-1} & a_{n,j+1} & \cdots & a_{n,n} \end{vmatrix}

Finding all minors of a matrix $A$ , and plugging them together in a new matrix is called the matrix of minors, which we will denote as $M$ :

M = \begin{bmatrix}\begin{array}{cccc} M_{1,1} & M_{1,2} & \cdots & M_{1,n} \\ M_{2,1} & M_{2,2} & \cdots & M_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ M_{n,1} & M_{n,2} & \cdots & M_{n,n} \end{array}\end{bmatrix}

In the end, it is almost the same as calculating the determinant. The only difference is that we need to delete one row and one column. We will do some examples in the next sections for more clearness.

How to find the minors of a 2×2 matrix?

We will be starting by finding minors of a $2\times 2$ matrix. So let $A$ be $2\times 2$ matrix. Specifically:

A = \begin{bmatrix}\begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{array}\end{bmatrix}

For each minor $M_{1,1}$ , $M_{1,2}$ , $M_{2,1}$ and $M_{2,2}$ , we need to find it’s determinant of the $1\times 1$ submatrix of $A$ . But $1\times 1$ matrix is one element. Therefore, we can rewrite the minors as:

\begin{equation*} M_{1,1} = a_{2,2}, \quad M_{1,2} = a_{2,1}, \quad M_{2,1} = a_{1,2}, \quad \text{and} \quad M_{2,2} = a_{1,1} \end{equation*}

and this results in the next matrix of minors

M

M = \begin{bmatrix}\begin{array}{cc} M_{1,1} & M_{1,2} \\ M_{2,1} & M_{2,2} \end{array}\end{bmatrix} = \begin{bmatrix}\begin{array}{cc} a_{2,2} & a_{2,1} \\ a_{1,2} & a_{1,1} \end{array}\end{bmatrix}

Example.

A = \begin{bmatrix}\begin{array}{cc} 5 & -2 \\ 7 & 3 \end{array}\end{bmatrix} \ \Rightarrow \ M = \begin{bmatrix}\begin{array}{cc} 3 & 7 \\ -2 & 5 \end{array}\end{bmatrix}

How to find the minors of a 3×3 matrix?

Now we need to perform more calculations since we have 9 determinants. Concretely, let $A$ be the matrix

A = \begin{bmatrix}\begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \end{array}\end{bmatrix}

Then the matrix of minors of $A$ is

M = \begin{bmatrix}\begin{array}{ccc} M_{1,1} & M_{1,2} & M_{1,3} \\ M_{2,1} & M_{2,2} & M_{2,3} \\ M_{3,1} & M_{3,2} & M_{3,3} \end{array} \end{bmatrix}

\quad = \begin{bmatrix}\begin{array}{ccc} \begin{vmatrix} a_{2,2} & a_{2,3}\\ a_{3,2} & a_{3,3}\\ \end{vmatrix} & \begin{vmatrix} a_{2,1} & a_{2,3}\\ a_{3,1} & a_{3,3}\\ \end{vmatrix} & \begin{vmatrix} a_{2,1} & a_{2,2}\\ a_{3,1} & a_{3,2}\\ \end{vmatrix} \\\\ \begin{vmatrix} a_{1,2} & a_{1,3}\\ a_{3,2}& a_{3,3}\\ \end{vmatrix} & \begin{vmatrix} a_{1,1} & a_{1,3}\\ a_{3,1}& a_{3,3}\\ \end{vmatrix} & \begin{vmatrix} a_{1,1} & a_{1,2}\\ a_{3,1}& a_{3,2}\\ \end{vmatrix} \\\\ \begin{vmatrix} a_{1,2} & a_{1,3}\\ a_{2,2}& a_{2,3}\\ \end{vmatrix} & \begin{vmatrix} a_{1,1} & a_{1,3}\\ a_{2,1}& a_{2,3}\\ \end{vmatrix} & \begin{vmatrix} a_{1,1} & a_{1,2}\\ a_{2,1} & a_{2,2}\\ \end{vmatrix} \end{array} \end{bmatrix}

Example.

A = \begin{bmatrix}\begin{array}{ccc} 3 & -1 & 6 \\ 2 & 5 & 7 \\ 1 & -3 & 1 \end{array} \end{bmatrix} \ \Rightarrow \ M = \begin{bmatrix}\begin{array}{ccc} 26 & -5 & -11 \\ 17 & -3 & -8 \\ -37 & 9 & 17 \end{array}\end{bmatrix}

How to find the minors of a 4×4 matrix?

We saw in the previous sections that finding a minor is quite an easy and straightforward calculation. To tackle the $4 \times 4$ matrix, we need to calculate 16 determinants of $3\times 3$ submatrices. That is a lot of calculations! Normally, that would be done by computers, but here, we will give a few calculations how to find minors of a $4 \times 4$ matrix.

A = \begin{bmatrix}\begin{array}{cccc} 3 & -1 & 6 & 2 \\ 2 & 5 & 7 & 4 \\ 1 & -3 & 1 & 9 \\ 4 & 1 & 3 & 7 \end{array}\end{bmatrix}

We will give an example for calculating the minors

M_{1,1}

and

M_{4,2}

M_{1,1} = \begin{vmatrix} 5 & 7 & 4\\ -3 & 1 & 9\\ 1 & 3 & 7 \end{vmatrix} = 5 \cdot \begin{vmatrix} 1 & 9\\ 3 & 7 \end{vmatrix} + 3 \cdot \begin{vmatrix} 7 & 4\\ 3 & 7 \end{vmatrix} + 1 \cdot \begin{vmatrix} 7 & 4\\ 1 & 9 \end{vmatrix}

= 5 \cdot (-20) + 3 \cdot 37 + 1 \cdot 59 = 70

M_{4,2} = \begin{vmatrix} 3 & 6 & 2 \\ 2 & 7 & 4 \\ 1 & 1 & 9 \\ \end{vmatrix} = 3 \cdot \begin{vmatrix} 7 & 4 \\ 1 & 9 \\ \end{vmatrix} - 2 \cdot \begin{vmatrix} 6 & 2 \\ 1 & 9 \\ \end{vmatrix} + 1 \cdot \begin{vmatrix} 6 & 2 \\ 7 & 4 \\ \end{vmatrix}

= 3\cdot 59 - 2 \cdot 52 + 1 \cdot 14 = 83

Conclusion

Finding minors of a matrix can be done easily and quickly for matrices of $2 \times 2$ and $3 \times 3$ , but bigger than that will increase the complexity of calculations. In the end, minors can be quite handy to find the inverse matrix.

The post How to find the minors of a nxn matrix? appeared first on Epsilonify.

How to find the inverse of a matrix

The Mathematician — Wed, 22 Jul 2020 13:16:55 +0000

What is the inverse of a matrix?

Let $A$ be a $n \times n$ matrix. We define the inverse matrix of $A$ if there exists a $n \times n$ matrix $B$ such that $AB = I = BA$ . Then we write the inverse $B$ of $A$ as $A^{-1}$ . We also say that $A$ is invertible if an inverse matrix of $A$ exists. Note that not every matrix has an inverse.

Theorem 1. Let $A$ be $n \times n$ matrix. The matrix $A$ has an inverse if and only if it has $n$ pivots.

This theorem is possible the easiest one. It is also quite clear why the matrix needs $n$ pivots. Take the assumption that this is not true. Then we have at least one zero row, without loss of generality, and take the last row as the zero row. Then it implies that multiplying two matrices will also contain at least one zero row. That is in contradiction with the definition of an inverse because we need $AA^{-1} = A^{-1}A = I$ . Another way to check if the matrix $A$ has an inverse is by checking its determinant.

Theorem 2. Let $A$ be $n \times n$ matrix. $A$ has an inverse if and only if $\det(A) \neq 0$ . Few notes to end this part. Let $A$ be an invertible matrix. $A$ has only one inverse (so it is unique). Let $\vec{x},\vec{b} \in V$ , where $V$ is a vector space, and $A\vec{x} = \vec{b}$ . Then $A\vec{x} = \vec{b}$ if and only if $\vec{x} = A^{-1}\vec{b}$ .

The Gauss-Jordan Method

The Gauss-Jordan Method is the same as Gaus elimination, but then finding the inverse of that specific matrix. To perform the Gaus-Jordan, we need to follow the next steps: Algorithm. Algorithm to find the inverse:

Put the $A$ in an $n \times 2n$ block matrix $[A | I]$ .
Now perform the Gaus elimination till you get $[I | A^{-1}]$ . If that fails, then $A$ has no inverse as it doesn’t have $n$ pivots.

We will end this by giving an example of performing the Gauss-Jordan Method.

[A | I] =

\begin{bmatrix}\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 2 & 1 & 0 & 1 \end{array} \end{bmatrix}

\sim

\begin{bmatrix}\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 3 & 2 & -1 \end{array} \end{bmatrix}

\sim

\begin{bmatrix}\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{2}{3} & -\frac{1}{3} \end{array} \end{bmatrix}

\sim

\begin{bmatrix}\begin{array}{cc|cc} 1 & 0 & -\frac{1}{3} & \frac{2}{3} \\ 0 & 1 & \frac{2}{3} & -\frac{1}{3} \end{array} \end{bmatrix}

= [I | A^{-1}]

When matrices are getting bigger, mistakes can be easily made. But after the calculation with the Gauss-Jordan method, the reader can check if the answer is correct by applying

AA^{-1} = A^{-1}A = I

. There is a more simple way to find the inverse matrix by using minors, cofactors, and adjugate.

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