Inverse of a 3 by 3 Matrix

Inverse of a 3 by 3 Matrix: A 3 x 3 matrix has 3 rows and 3 columns. Elements of the matrix are the numbers which make up the matrix. A singular matrix is the one in which the determinant is not equal to zero. For every m×m square matrix there exist an inverse of it. It is represented by M-1. The inverse of a matrix cannot be evaluated by calculators and using shortcuts will be inappropriate. We should practice problems to understand the concept. It has a property as follows:

MM-1 = M-1 M = I2

In the above property, I2 represents the m x m matrix. Suppose, take an example of a 2 x 2 matrix.

inverse of a 3 × 3 matrix

Any m x m square matrix M, which has zero determinant always has an inverse M-1. It is mostly true for all the square matrix and is given by MM-1 = M-1M =Im

How to find the inverse of 3 x 3 matrix?

Let’s see how 3 x 3 matrix looks :

M = \(\begin{bmatrix} a & b &c \\ d& e &f \\ g& h &i \end{bmatrix}\)

Let’s see what are the steps to find Inverse

First, find the determinant of 3 × 3 × 3 Matrix and then find it’s minor, cofactors and adjoint and insert the results in the Inverse Matrix formula given below:

M-1 = (1 / det M ) adj M

Inverse of 3x3 matrix

    1. First let’s check whether the matrix is invertible:

This can be proved if its determinant is non zero.

det M = m det \(\begin{bmatrix} e & f\\ h & i \end{bmatrix} – b det\begin{bmatrix} d & f\\ g & i \end{bmatrix} + c det \begin{bmatrix} d & e\\ g & h \end{bmatrix}\)

    1. Co-factors of the matrix

For all the (i, j)the entries of the matrix M, let’s denote the co-factor Mij by (-1) i+j Mij, where Mij is a 2 x 2 matrix obtained by A by removing the ith row and jth column. For example:

M23 = det \(\begin{bmatrix} a & b &- \\ -& – &- \\ g& h & – \end{bmatrix} = Det \begin{bmatrix} a & b\\ g & h \end{bmatrix}\) = ah – bg

So, C23 = ( -1) )2+3 (ah -bg)

= bg – ah.

    1. Adjugate matrix

First, find all the adjugate cofactors and then Adjugate of M is

Adj M = \(\begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23}\\ C_{31} & C_{32} & C_{33} \end{pmatrix}^{T}\) \(\begin{pmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32}\\ C_{13} & C_{23} & C_{33} \end{pmatrix}\)

This is a matrix made out of Cofactors and then it’s transposed.

  1. And finally M-1 = (1 / det M ) adj M

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