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Is a Matrix Invertible?
Find the Inverse of a Matrix


A square matrix in Math can have what is called an inverse matrix.

For a matrix labelled  A,  the notation for the inverse matrix is  A-1.


We’ll explain how to establish if a matrix is invertible, and also how to find the inverse matrix if it indeed does exist.




Is a Matrix Invertible?
Inverse of a 2×2 Matrix

To find out if a matrix is invertible, you want to establish the determinant of the matrix.

Details of how to find the determinant of a matrix can be seen here.


The key thing to note is that a matrix only has an inverse if its determinant does not equal  0.

So if  A  is a matrix, the inverse matrix  A-1  exists as long as  det(A) ≠ 0.



For a  2 × 2  matrix  \begin{bmatrix} a & b \\ c & d \end{bmatrix},  the determinant is found by   ad − bc.


The role the determinant plays in finding the inverse of a 2 × 2 matrix is in the following calculation.

For a matrix   A = \begin{bmatrix} a & b \\ c & d \end{bmatrix},   the inverse matrix is:

A-1\bf{\frac{1}{det(A)}} × \begin{bmatrix} d & {\text{-}}b \\ {\text{-}}c & a \end{bmatrix}.



Example    


(1.1) 

Find the inverse of the matrix   B  =  \begin{bmatrix} {\text{-}}1 & 4 \\ {\text{-}}2 & 4 \end{bmatrix}.

Solution   

det( B )  =  (-1 × 4) − (4 × –2)  =  –4 − –8  =  4

B-1  =   {\frac{1}{4}} × \begin{bmatrix} 4 & {\text{-}}4 \\ 2 & {\text{-}}1 \end{bmatrix}   =   \begin{bmatrix} 1 & {\text{-}}1 \\ {\frac{1}{2}} & {\text{-}}{\frac{1}{4}} \end{bmatrix}






Multiplying a Matrix by its Inverse


One property of an inverse matrix, is that when multiplied with its original matrix, regardless of order, the result is the identity matrix.


A × A-1  =  I     ,     A-1 × A  =  I

As such, when working out the inverse of a matrix, a good way to check if you have the right answer is to multiply your inverse with the original and observe the result.


We can check our inverse matrix found in example (1.1).


B  =  \begin{bmatrix} {\text{-}}1 & 4 \\ {\text{-}}2 & 4 \end{bmatrix}     ,     B-1  =   \begin{bmatrix} 1 & {\text{-}}1 \\ {\frac{1}{2}} & {\text{-}}{\frac{1}{4}} \end{bmatrix}

B × B-1  =  \begin{bmatrix} {\text{-}}1 & 4 \\ {\text{-}}2 & 4 \end{bmatrix} × \begin{bmatrix} 1 & {\text{-}}1 \\ {\frac{1}{2}} & {\text{-}}{\frac{1}{4}} \end{bmatrix}

=   \begin{bmatrix} ({\text{-}}1\times1)+(4\times{\small{\frac{1}{2}}}) & ({\text{-}}1\times{\text{-}}1)+(4 \times {\text{-}}{\small{\frac{1}{4}}}) \\ ({\text{-}}2\times1)+(4\times{\small{\frac{1}{2}}}) & ({\text{-}}2\times{\text{-}}1)+(4 \times {\text{-}}{\small{\frac{1}{4}}}) \end{bmatrix}   =   \begin{bmatrix} {\text{-}}1+2 & 1-1 \\ {\text{-}}2+2 & 2-1 \end{bmatrix}  =  \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}


So the inverse matrix obtained in example (1.1) is indeed the correct one.





Following on from the case of a (2 × 2) matrix inverse, is the inverse of a (3 × 3) matrix inverse.





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