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.

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

1. Put the A in an n \times 2n block matrix [A | I].
2. 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.

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

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

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

\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}

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