CALCULUS – IIInverse Matrix

by

Dr. Eman Saad & Dr. Shorouk Ossama

References

Robert Wrede and Murrary R. Spiegel, Theory

and Problems of Advanced Calculas, 2nd 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.

a11 x1 + a12 x2 = d1

a21 x1 + a22 x2 = d2

These are linear equations in the unknowns x1, x2.

Where: A X = d

Thus:

To get the value for X1 & X2

Eliminate x2 by multiplying the first equation by

a22 and the second equation by a12 then

subtracting the two equations so that:

(a11 a22 - a21 a12 ) x1 = a22 d1 - a12 d2

Similarly, elimination of x1 leads to:

- (a11 a22 - a21 a12 ) x2 = a21 d1 - a11 d2

It follows that:

Where: ( a11 a22 - a21 a12 ) 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 Inverse Singular

Has Inverse Non-Singular

The matrixA=

is invertible if ad-bc≠0, in witch case the inverse is given by the formula

a bc d

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) Aⁿ is invertible and (Aⁿ) 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

1 0 00 -3 00 0 2

(AT )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 x1, x2,….. Successively between the

set of equations AX = d

or : a11 x1 + a12 x2 + a13 x3 = d1

a21 x1 + a22 x2 + a23 x3 = d2

a31 x1 + a32 x2 + a33 x3 = d3

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 3 - 1 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 -2R1+R2

and -1 times the first row to the third -R1+R3

We added 2 times the second row to the third 2R2+R3

We multiplied the third row by -1 -R3

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-2R2+R1

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 B1. Get A-1 2. Multiply A-1 by B3. 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