Inverse of a Matrix

Inverse of a Matrix

Inverse of a Matrix

If A is a non-singular square matrix i.e. |A| ≠ 0, and there exists another square matrix B such that AB = BA = I where I is an identity matrix of order same as that of A or B, then the matrix B is said to be the inverse of matrix A and vice-versa. The inverse of A is denoted by A-1. B = A-1.

Note: For the existence of the inverse of a matrix, the following requirements are necessary:

i)     The matrix must be a square matrix. This requirement is essential because, let A be a matrix of order 2×3 and B be another matrix of order 3×2, then AB and BA both exist but have different orders, namely 2×2 and 3×3 and hence cannot be equal.

ii)   The equation AB = BA = I (Identity Matrix) must be satisfied. For example, if

A = (■(3&0@0&3)) and B = (■(1&3@5&1)), then AB = (■(3&9@15&3)) = BA

But, AB = BA ≠ I. So, they are not inverse matrix to each other.

iii)  The matrix must be non-singular. i.e |A| ≠ 0.

 

Method of Finding an Inverse Matrix

Let A = (■(a&b@c&d)) be a square matrix, then its determinant |A| = ad – bc ≠ 0 (suppose) and A-1 exists.  Let A-1 = (■(p&q@r&s)), then AA-1 = I i.e. (■(a&b@c&d))(■(p&q@r&s)) = (■(1&0@0&1)) or,  (■(ap+br&aq+bs@cp+dr&cq+ds)) = (■(1&0@0&1)) Equating corresponding elements, we have ap + br = 1 ………………………… (i) aq + bs = 0 ……………………….. (ii) cp + dr = 0 ……………………….. (iii) cq + ds = 1 ……………………….. (iv) Multiplying equation (i) by d and (iii) by b, and subtracting (iii) from (i), we get adp + bdr – (bcp + bdr ) = d - 0 or, adp – bcp = d or, p(ad – bc) = d or, p = d/(ad-bc)  And similarly, multiplying (i) by c and (iii) by a, and then subtracting (iii) from (i), we get acp + bcr – (acp + adr) = c - 0 or, bcr – adr = c or, -r(ad – bc) = c or, r = - c/(ad-bc)  Similarly, solving equations (ii) and (iv), we get, q = - b/(ad-bc)   and   s = a/(ad-bc) ∴ A-1 = (■(p&q@r&s)) = (■(d/(ad-bc)&- b/(ad-bc)@- c/(ad-bc)&a/(ad-bc))) Or,  A-1 = 1/(ad-bc) (■(d&-b@-c&a)) Or,  A-1 = 1/(|A|) (■(d&-b@-c&a)) ∴ If A = (■(a&b@c&d)) and |A| ≠ 0, then inverse of A is given by A-1 = 1/(|A|) (■(d&-b@-c&a))


Some Properties of Inverses

a.   The inverse of the product of two non-singular matrices is equal to the product of their inverses taken in reverse order. i.e. If A and B are two non-singular matrices of same order then (AB)-1 = B-1A-1.

Let A = (■(2&5@1&3)) and B = (■(3&1@7&2)) then, 	|A| = |■(2&5@1&3)| = 6 – 5 = 1 	|B| = |■(3&1@7&2)| = 6 – 7 = -1 	So, 	A-1 = 1/(|A|) (■(d&-b@-c&a)) = 1/1 (■(3&-5@-1&2)) = (■(3&-5@-1&2)) 	B-1 = 1/(|B|) (■(d&-b@-c&a)) = 1/(-1) (■(2&-1@-7&3)) = (■(-2&1@7&-3)) 	Now, 	AB = (■(2&5@1&3))(■(3&1@7&2)) = (■(41&12@24&7)) 	And, 	|AB| = |■(41&12@24&7)| = 287 – 288 = -1 	So, 	(AB)-1 = 1/(|AB|) (■(d&-b@-c&a)) = 1/(-1) (■(7&-12@-24&41)) = (■(-7&12@24&-41)) 	Again, 	B-1A-1 = (■(-2&1@7&-3))(■(3&-5@-1&2)) = (■(-7&12@24&-41)) 	Hence, (AB)-1 = B-1A-1

b.   The transpose and inverse of a non-singular matrix is commutative. i.e. (A-1)t = (At)-1.

Let A = (■(2&-3@7&4)) then At = (■(2&7@-3&4)) and 	|A| = |■(2&-3@7&4)| = 8 + 21 = 29 	|At| = |■(2&7@-3&4)| = 8 + 21 = 29 	So, A-1 = 1/(|A|) (■(d&-b@-c&a)) = 1/29 (■(4&3@-7&2)) = (■(4/29&3/29@-7/29&2/29)) 	∴ (A-1)t = (■(4/29&-7/29@3/29&2/29)) 	And, 	(At)-1 = 1/|A^t |  (■(d&-b@-c&a)) = 1/29 (■(4&-7@3&2)) = (■(4/29&-7/29@3/29&2/29)) 	Hence, (A-1)t = (At)-1.


Worked Out Examples

Example 1: Find the inverse of the matrix A = (■(7&6@6&5)). Solution: Here, A = (■(7&6@6&5)), ∴ |A| = |■(7&6@6&5)| = 35 – 36 = -1 Now, A-1 = 1/(|A|) (■(d&-b@-c&a))               = 1/(-1) (■(5&-6@-6&7))               = (■(-5&6@6&-7))

Example 2: Find the inverse of the matrix B = (■(1&4@7&2)). If B-1 is the inverse of B, prove that BB-1 = I. Solution: Here, B = (■(1&4@7&2))  ∴ |B| = |■(1&4@7&2)| = 2 – 28 = -26 Now, B-1 = 1/(|B|) (■(d&-b@-c&a)) = 1/(-26) (■(2&-4@-7&1)) = (■((-2)/26&4/26@7/26&(-1)/26)) = (■((-1)/13&2/13@7/26&(-1)/26)) Now, BB-1 = (■(1&4@7&2))(■((-1)/13&2/13@7/26&(-1)/26)) = (■((-1)/13+14/13&2/13+(-2)/13@(-7)/26+7/13&14/13+(-1)/13)) = (■(1&0@0&1)). Proved.

Example 3: If the inverse of the matrix (■(x&2x-9@-y&3)) is the matrix (■(3&5@y&x)). Find the values of x and y. Solution:  Let A = (■(x&2x-9@-y&3)) and A-1 = (■(3&5@y&x)). We know that A.A-1 = I i.e. (■(x&2x-9@-y&3))(■(3&5@y&x)) = (■(1&0@0&1)) or,  (■(3x+(2x-9)y&5x+x(2x-9)@-3y+3y&-5y+3x)) = (■(1&0@0&1)) or,  (■(3x+2xy-9y&5x+2x^2-9x)@0&3x-5y)) = (■(1&0@0&1)) or,  (■(3x+2xy-9y&2x^2-4x)@0&3x-5y)) = (■(1&0@0&1)) Equating the corresponding elements,  2x2 – 4x = 0 Or, 2x(x – 2) = 0 ∴ Either, 2x = 0 or x = 0 Or, x – 2 = 0 or x = 2 ∴ x = 0 or 2 Similarly, 3x – 5y = 1 When x = 0, y = -1/5 When x = 2, y = 1 Hence, the required value of x = 0 or 2 and y = -1/5 or 1.

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