The Inverse of A 2x2 Matrix

✎☞5.4 ✌✍IntroductionThe inverse of a 2 × 2 matrixOnce you know how to multiply matrices it is natural to ask whether they can be divided. Theanswer is no. However, by defining another matrix called the inverse matrix it is possible towork with an operation which plays a similar role to division. In this leaflet we explain what ismeant by an inverse matrix and how the inverse of a 2 × 2 matrix is calculated.1. The inverse of a 2 × 2 matrixThe inverse ofa2× 2 matrix A, is another 2 × 2 matrix denoted by A −1 with the propertythatAA −1 = A −1 A = I( )1 0where I is the 2 × 2 identity matrix . That is, multiplying a matrix by its inverse0 1produces an identity matrix. Note that in this context A −1 does not mean 1 . ANot all 2 × 2 matrices have an inverse matrix. If the determinant of the matrix is zero, thenit will not have an inverse, and the matrix is said to be singular. Only non-singular matriceshave inverses.2. A simple formula for the inverse( )a bIn the case of a 2 × 2 matrix a simple formula exists to find its inverse:c d(a bif A =c d)then A −1 =1ad − bc(d −b−c a)ExampleFind the inverse of the matrix A =(3 14 2).SolutionUsing the formulaA −1 1=(3)(2) − (1)(4)= 1 ( )2 −12 −4 3(2 −1−4 3)5.4.1 copyright c○ Pearson Education Limited, 2000

This could be written as (1 −12−232)You should check that this answer is correct( by performing ) the matrix multiplication AA −1 .1 0The result should be the identity matrix I = .0 1ExampleFind the inverse of the matrix A =SolutionUsing the formula(2 4−3 1).A −1 1=(2)(1) − (4)(−3)= 1 ( )1 −414 3 2(1 −43 2)This can be writtenA −1 =(1/14 −4/143/14 2/14although it is quite permissible to leave the factor 114Exercises1. Find the inverse of A =(1 53 2).)=(1/14 −2/73/14 1/7)at the front of the matrix.( )6 42. Explain why the inverse of the matrix cannot be calculated.3 2( )( )3 43 −43. Show that is the inverse of.2 3−2 3Answers1. A −1 = 1−13(2 −5−3 1) (−2=13513313− 1132. The determinant of the matrix is zero, that is, it is singular and so has no inverse.).5.4.2 copyright c○ Pearson Education Limited, 2000