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The inverse of matrix acts similarly in matrix algebra as the reciprocal of number takes in the division in general Mathematics. Just as we can solve a simple mathematical equation 3x = 6 for x by multiplying both sides by the reciprocal.

$3x = 6Â 3^{-1} 3x = 3^{-1}6Â x= \dfrac{6}{3}= 2$

Similarly, we can solve matrix equations like P x = q for the vector x by multiplying both the sides by the inverse of matrix P.

$Px = qÂ P^{-1} Px = P^{-1}q Â x= P^{-1} q$

Here, we will look at the matrix inverse using minors, cofactors, and adjugate.

The inverse of a matrix is calculated by determining the determinant and adjoint of a given matrix. Adjugate or adjoint of the matrix is given by the transpose of the cofactors of a given matrix. The formula to find the inverse of matrix is given by:

$A^{-1} = \dfrac{adj(A)}{|A|}; |A| \neq 0$

Let us consider A = [a]_{n} as the square matrix of order n. Let A_{pq} represent the cofactor of the elements whose indices are (p,q).

The matrix of cofactors A is the square matrix for order N.

\[C = \begin{bmatrix} A_{11}& A_{12} & . & . & . & A_{1n}\\Â A_{21}& A_{22 }& . & . & . & A_{2n} \\Â . &Â . &Â . & . & . & .\\Â . &Â . &Â . & . & . & .\\Â . &Â . & . & . & . & .\\Â A_{n1} & A_{n2} & . & . & . & A_{nn} \end{bmatrix}\]

The matrix of cofactors is also known as the cofactor matrix or comatrix.

In Mathematics, a cofactor is a number used to find the inverse of a matrix, adjoined. The cofactor is defined as the number that is obtained when the rows or columns of selected elements in the given matrix are removed, which is just a numerical grid in the form of a square or a rectangle.

The cofactors are always introduced by a positive (+) or negative (-) symbol. Let us consider X as n x n matrix and let M_{ij} as (n - 1) x (n - 1) matrix obtained by removing the i^{th} row and j^{th} column. Then det (M_{ij} is known as the minor of a_{ij}. The cofactor A_{ij}Â of a_{ij}Â is defined in terms of the minor is: A_{ij} = (-1)^{i + j} det (M_{ij}}).

Let $AÂ = \begin{bmatrix} 2 & 6 & -1 \\Â 0 & 1 & 4 \\ 3 & -2 & -6 \end{bmatrix}$

Let us consider $M_{ij}$ as the minor of elements of $i^{th}$ row and $j^{th}$ column. Accordingly,

$M_{32}Â = \begin{bmatrix}2 & -1 \\ 0 & 4 \\ \end{bmatrix}$

Therefore, the minor of $A_{32}$ is the determinant of the above 2 by 2 matrix.

As the given matrix is triangular, the determinant is the product of diagonals.

Accordingly $(2 \times 4)(-1 \times 0)Â = (8 - 0) = 8$

Applying the Matrix of Cofactors formula, we get:

$A_{32} = (-1)^{3+2} (8) = -8$

Let X = [a_{ij}] be a square matrix of order n. The adjugate of matrix X (also known as adjoint of Matrix X) is defined as the transpose of the cofactor matrix X. It is represented by adj X. An adjugate matrix is also known as an adjoint matrix.

To determine the adjugate of a matrix, first, find the cofactor of the given matrix. Then find the transpose of the cofactors of the matrix.Â

Adjugate Matrix Example

Find the adjoint of Matrix,

$AÂ = \begin{bmatrix}3 & 1 & -1 \\ 2 & -2 & 0 \\ 1 & 2 & -1\end{bmatrix}$

Cofactor of 3 $= A_{11} = \begin{vmatrix}-2 & 0\\ 2 & -1\end{vmatrix} = 2$

Cofactor of 1 $= A_{12} = - \begin{vmatrix}2 & 0\\ 1 & -1\end{vmatrix} = 2$

Cofactor of -1 $= A_{13} = \begin{vmatrix}2 & -2\\ 1 & 2\end{vmatrix} = 6$

Cofactor of 2 $= A_{21} = - \begin{vmatrix}1 & -1\\ 2 & -1\end{vmatrix} = -1$

Cofactor of -2 $= A_{22} = \begin{vmatrix}3 & -1\\ 1 & -1\end{vmatrix} = -2$

Cofactor of 0 $= A_{23} = - \begin{vmatrix}3 & 1\\ 1 & 2\end{vmatrix} = -5$

Cofactor of 1 $= A_{31} = \begin{vmatrix}1 & -1\\ -2 & 0\end{vmatrix} = -2$

Cofactor of 2 $= A_{32} = - \begin{vmatrix}3 & -1\\ 2 & 0\end{vmatrix} = -2$

Cofactor of -1 $= A_{33} = \begin{vmatrix}3 & 1\\ 2 & -2\end{vmatrix} = -8$

$\therefore$ The cofactor of matrix $A$ is $A_{ij} = \begin{vmatrix}2 & 2 & 6\\ -1 & -2 & -5\\-2 & -2 & -8\end{vmatrix}$

Let us now find the transpose of $A_{ij}$

$Adj A = (A_{ij})^T$

$ \begin{vmatrix}2 & -1 & -2\\ 2 & -2 & -2\\6 & -5 & -8\end{vmatrix}$

Hence, the adjoint of matrix $A = \begin{vmatrix}2 & -1 & -2\\ 2 & -2 & -2\\6 & -5 & -8\end{vmatrix}$

In a square matrix, each element has its minor. The minor is defined as the value that is received from the determinants of a square matrix by deleting out a column and a row corresponding to the element of a matrix.

Given a square matrix X, by minor of an element we mean the value of the determinant is obtained by deleting the i^{th} row and j^{th} column of a matrix. It is represented by M^{ij}.

To find the minor of a square matrix, we have to remove a row and column one by one at a time and calculate their determinant, until all the minors are calculated. Following are the steps to calculate the minor from a matrix.

Hide i

^{th}row and j^{th}column individually from a given matrix, where i refers to m and j refers to n that is the total number of rows and columns in matrices.Calculate the value of the determinant of the matrix after hiding the rows and columns from Step 1.

Let us learn to find minors of 3 x 3 matrix below:

Find the minors of matrix $A = \begin{bmatrix}a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}$

$M_{11} = \begin{bmatrix}e & f\\ h & i \end{bmatrix} = (ei - hf)$

$M_{12} = \begin{bmatrix}d & f\\ g & i \end{bmatrix} = (di - ff)$

$M_{13} = \begin{bmatrix}d & e\\ g & h \end{bmatrix} = (dh - eg)$

$M_{21} = \begin{bmatrix}b & c\\ h & i \end{bmatrix} = (bi - ch)$

$M_{22} = \begin{bmatrix}a & c\\ g & i \end{bmatrix} = (ai - cg)$

$M_{23} = \begin{bmatrix}a & b\\ g & h \end{bmatrix} = (ah - bg)$

$M_{31} = \begin{bmatrix}b & c\\ e & f \end{bmatrix} = (bf - ce)$

$M_{32} = \begin{bmatrix}a & c\\ d & f \end{bmatrix} = (af - cd)$

$M_{33} = \begin{bmatrix}a & b\\ d & e \end{bmatrix} = (ae - bd)$

Following are the steps to finding the inverse of the matrix using minors, cofactors, and adjugate.

The first step to finding the inverse of the matrix is to determine the matrix of minors.

The second step is to transform the given matrix into a matrix of cofactors.

The third step is to find the adjoint of the matrix.

At the end, multiply by 1/Determinant.

Let us learn to find inverse of matrix using minors, cofactors and adjugate with an example:

1. Find the inverse of matrix

$X = \begin{bmatrix}3 & 1 & 2\\ 2 & 1 & -2\\0 & 1 & 1 \end{bmatrix}$

Solution:

Step 1: To find the inverse of the matrix X, we will first find the matrix of minors.

Matrix of Minors $=\begin{bmatrix}3 & 2 & 2\\ -1 & 3 & 3\\-4 & -10 & 1 \end{bmatrix}$

Step 2:Â In this step, we will find the cofactors of the above matrix of minor

Cofactors of Matrix of Minor $ - \begin{bmatrix}3 & 2 & 2\\ -1 & 3 & 3\\-4 & -10 & 1 \end{bmatrix} \times \begin{bmatrix}+ & - & +\\ - & + & -\\+ & - & + \end{bmatrix} = \begin{bmatrix}3 & -2 & 2\\ 1 & 3 & -3\\-4 & 10 & 1 \end{bmatrix}$

Step 3: Now, we will find the adjugate or adjoint of the above matrix by swapping the position of elements diagonally such that:

Adjoint of Cofactor $= \begin{bmatrix}3 & 1 & 4\\ -2 & 3 & 10\\2 & -3 & 1 \end{bmatrix}$

Step 4: Now, we will find the determinants of original matrix X using the following determinants formula:

$\text{det} \begin{bmatrix}a & b & c\\ d & e & f\\g & h & i \end{bmatrix} = a \cdot \text{det} \begin{bmatrix}e & f\\ h & i\end{bmatrix} - b \cdot \text{det} \begin{bmatrix}d & f\\ g & i\end{bmatrix} + c \cdot \text{det} \begin{bmatrix}d & e\\ g & h \end{bmatrix}$

$\text{det} \begin{bmatrix}3 & 1 & 2\\ 2 & 1 & -2\\0 & 1 & 1 \end{bmatrix} = 3 \cdot \text{det} \begin{bmatrix}1 & -2\\ 1 & 1\end{bmatrix} - (1) \cdot \text{det} \begin{bmatrix}2 & -2\\ 0 & 1\end{bmatrix} + 2 \cdot \text{det} \begin{bmatrix}2 & 1\\ 0 & 1 \end{bmatrix}$

$\Rightarrow 3 [1 - (- 2)] -1[2 - (-0)] + 2 [2 -0]$

$\Rightarrow 3(1 + 2) -1(2+ 0) + 2 (1)$

$\Rightarrow 3(3) -1(2) + 2$

$\Rightarrow 9 -2 + 2$

$\Rightarrow 11$

Determinant $= 11$

Step 5: Now, we will multiply the adjoint by 1/Determinant to get the inverse of the original matrix X.

$X^{-1} = \begin{bmatrix}3 & 1 & -4\\ -2 & 3 & 10\\2 & -3 & 1 \end{bmatrix}$

$X^{-1} = \begin{bmatrix}\dfrac{3}{11} &\dfrac{1}{11} & \dfrac{-4}{11}\\ \dfrac{-2}{11} & \dfrac{3}{11} & \dfrac{10}{11}\\\dfrac{2}{11} & \dfrac{-3}{11} & \dfrac{1}{11} \end{bmatrix}$

19th Century English Mathematician James Sylvester introduced the term Matrix.

Algebraic aspects of Matrix were developed in two papers in the 1850s by Arthur Cayley.

You can easily find the inverse of Matrix using the formula $A^{-1} = \dfrac{adj(A)}{|A|}; |A| \neq 0$.

FAQ (Frequently Asked Questions)

Q1. What is a Matrix?

Ans. A matrix is a rectangular arrangement of numbers, placed in rows and columns.

Q2. How to Determine the Inverse of the 3 by 3 Matrix?

Ans. For determining the inverse of a 3 by 3 matrix, first, we will calculate the determinant of the matrix, and if the determinant of the matrix is 0, then it implies that there is no matrix. Further, we will arrange the matrix by rewriting the first row as the first column, the second row as the second column, and the third row as the third column.