CALCULUS – II Inverse Matrix by Dr. Eman Saad & Dr. Shorouk Ossama

References  Robert Wrede and Murrary R. Spiegel, Theory and Problems of Advanced Calculas, 2 nd Edition, 2002.

Inverse Matrix : If A is a square matrix, and if a matrix B of the same size can be found such that AB = BA = I, then A is said to be invertible and B is called an inverse of A.

a 11 x 1 + a 12 x 2 = d 1 a 21 x 1 + a 22 x 2 = d 2 These are linear equations in the unknowns x 1, x 2. Where: A X = d Thus:

To get the value for X 1 & X 2 Eliminate x 2 by multiplying the first equation by a 22 and the second equation by a 12 then subtracting the two equations so that: (a 11 a 22 - a 21 a 12 ) x 1 = a 22 d 1 - a 12 d 2 Similarly, elimination of x 1 leads to: - (a 11 a 22 - a 21 a 12 ) x 2 = a 21 d 1 - a 11 d 2

It follows that: Where: ( a 11 a 22 - a 21 a 12 ) not equal to Zero

If AX = d is multiplied on the left by the inverse A -1, then: A -1 A X = A -1 d І X = A -1 d X = A -1 d Hence A -1 can be identified with the matrix C.

To get the value of A -1 where A = Where: det A (is the determinate of the matrix A), if det A = 0, then the matrix has no inverse Said to be Singular If det A ≠ 0, then the matrix has inverse Said to be non- Singular

det A = o det A ≠ o No InverseSingular Has InverseNon-Singular

The matrix A= is invertible if ad-bc≠ 0, in witch case the inverse is given by the formula abcdabcd

Example: Consider the matrices we obtain: Also Therefore, (AB) -1 = B -1 A -1 ?

If A is an invertible matrix, then: (a) Aˉ¹ is invertible and ( Aˉ¹ ) ˉ¹ = A (b) A ⁿ is invertible and ( A ⁿ ) ˉ¹ = ( Aˉ¹ ) ⁿ for n=0,1,2,… (c) For any nonzero scalar k, the matrix kA is invertible and ( kA ) ˉ¹ = (1/ k) Aˉ¹ (d) (AB) -1 = B -1 A -1

Example: Let A and Aˉ¹ :

Example: If A= Then

(A T )ˉ¹ = (A ˉ¹) T Example: Consider the matrices =

For the inverse of a 3x3 matrix we can adopt the same approach as for the 2x2 case by eliminating x 1, x 2,….. Successively between the set of equations AX = d or : a 11 x 1 + a 12 x 2 + a 13 x 3 = d 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = d 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = d 3

Where:

Example: Find A -1 where: We first find det A: det A = 2 x [(-1) x 2 – (-1) x 5 ] -1 x [1 x 2 – (-1) x 5] + 0 x [1x (-1) – (-1) x (-10)] = 2 x x 7 = -1 Then get : A -1 = (1/det A) * [adj matrix] +-+

a = + ( -1 x 2 ) – ( -1 x 5 ) e = + ( 2 x 2 ) – ( -1 x 0 ) h = - [( 2 x 5 ) – ( 1 x 0 )] Thus:

Example: Find the inverse of Solution [ A ¦ I ] [ I ¦ Aˉ¹ ] The computations are as follows:

We added -2 times the first row to the second -2R 1 +R 2 and -1 times the first row to the third -R 1 +R 3

We added 2 times the second row to the third 2R 2 +R 3 We multiplied the third row by -1 -R 3

We added 3 times the third row to the second and - 3 times the third row to the first. We added -2 times the second row to the first -2R 2 +R 1

If we have any system: x + 3y = 1 x – 5y = 4 So: Multiply from left by A -1 A -1 A X = A -1 B X = A -1 B 1.Get A -1 2.Multiply A -1 by B 3.These are x and y A X = B

Example: Verify by direct multiplication that has the inverse matrix And then check the matrix product BA = I Hence B = A -1

Thanks