We state and prove some theorems on non-singular matrices.

**Properties of inverses of matrices**

We state and prove some theorems on non-singular matrices.

If *A *is
non-singular, then

, where Î» is a
non-zero scalar.

Let A be non-singular. Then |A| â‰ 0 and A^{âˆ’1} exists. By
definition,

Let *A*, *B*, and *C *be square matrices of order
*n*. If *A *is non-singular and *AB *= *AC*, then *B *= *C*.

**Proof**

Since *A *is non-singular, *A*^{âˆ’}^{1} exists and *AA*^{âˆ’}^{1} = *A*^{âˆ’}^{1} *A *= *I _{n} *.
Taking

Let *A*, *B*, and *C *be square matrices of order
*n*. If *A *is non-singular and *BA *= *CA*, then *B *= *C*.

**Proof**

Since *A *is non-singular, *A*^{âˆ’}^{1} exists and *AA*^{âˆ’}^{1} = *A*^{âˆ’}^{1} *A *= *I _{n} *.
Taking

**Note**

If *A *is singular and *AB *= *AC *or *BA *= *CA*, then *B *and
*C *need not be equal. For instance, consider the following matrices:

We note that |A| = 0 and AB = AC ; . but B â‰ C

**Theorem
1.7 (Reversal Law for Inverses)**

If *A *and *B *are
non-singular matrices of the same order, then the product *AB *is also
non-singular and ( *AB*)^{âˆ’}^{1} = *B*^{âˆ’}^{1} *A*^{âˆ’}^{1}.

**Proof**

Assume that *A *and
*B *are non-singular matrices of same order *n*. Then,| A | â‰ 0, | B | â‰ 0, both *A*^{âˆ’}^{1} and *B*^{âˆ’}^{1} exist and they are of
order *n*. The products *AB *and *B*^{âˆ’}^{1} *A*^{âˆ’}^{1} can be found and they
are also of order *n*. Using the product rule for determinants, we get |*AB|
*=| *A *|| *B *|â‰ 0. So, *AB *is
non-singular and

*( AB)(B ^{-1} A^{-1} ) = ( A(BB^{-1} )) A^{-1}
= ( AI_{n} ) A^{-1} = AA^{-1} = I_{n} ;*

*(B ^{-1} A^{-1} )( AB) = (B^{-1} ( A^{-1}
A))B = (B^{-1}I_{n} )B = B^{-1}B = I_{n} .*

Hence ( *AB*)^{âˆ’}^{1} = *B*^{âˆ’}^{1} *A*^{âˆ’}^{1}.

**Theorem
1.8 (Law of Double Inverse)**

If *A *is
non-singular, then *A*^{âˆ’}^{1} is also non-singular and ( *A*^{âˆ’}^{1} )^{âˆ’}^{1} = *A*.

**Proof**

Assume that *A *is non-singular. Then |*A | *â‰ 0, and *A*^{âˆ’}^{1} exists.

Now, |A^{-1}| = 1/|A| â‰ 0

â‡’ A^{-1} is also non-singular, and AA^{-1}
= A^{-1}A = I

Now, A^{-1}A = I â‡’ (AA^{-1})^{-1} = I â‡’ (A^{-1})^{-1}
A^{-1} = I.

Post-multiplying
by A on both sides of equation (1), we get (A^{-1})^{-1}
=A.

If *A *is a non-singular square matrix of order *n *,
then

**Proof**

Since *A *is a non-singular square matrix, we have |*A|* â‰ 0 and so, we get

**Note**

If
*A* is a non-singular matrix of order
3, then, |A | â‰ 0 . By theorem 1.9 (ii), we get |adjA| = | A|^{2} and
so, | adj A | is positive. Then, we get |A| = Â± âˆš|adjA| .

So,
we get

Further,
by property (iii), we get

Hence,
if A is a non-singular matrix of order 3, then we get

**Example 1.4**

If *A *is a non-singular matrix of odd order, prove that |adj
A| is positive.

**Solution**

Let A be a non-singular matrix of order 2*m* + 1 , where *m* = 0,1, 2,
.. . . Then, we get |A| â‰ 0 and, by theorem 1.9 (ii), we have |adj A| = |A|^{(2m+1)-1}
= |A|^{2m}

Since |A|^{2m} is always positive, we get that |adj A| is
positive.

**Example 1.5**

Find a matrix *A *if adj( *A*) =

**Solution**

**Example 1.6**

If adj *A *= find *A*^{âˆ’}^{1}.

**Solution**

**Example1.7**

If *A *is symmetric, prove that adj *A *is also
symmetric.

**Solution**

Suppose A is symmetric. Then, A^{T} = A and so, by theorem
1.9 (vi), we get

adj (A^{T}) = (adj A)^{ T}_{ } â‡’ adj A =
(adj A)^{T}_{ }â‡’ adj A is symmetric_{}

**Theorem
1.10**

If *A *and *B *are
any two non-singular square matrices of order *n *, then

adj( *AB*) = (adj *B*)(adj *A*).

**Proof**

Replacing A by *AB* in
adj(A) = |A|A^{âˆ’1} , we get

adj(AB) = |AB| (AB)^{-1} = (| B | B^{-1}) (| A | A^{-1})
= adj(B) adj(A)

Verify the property ( *A ^{T} *)

For the given A, we get |A |= (2) (7) - (9)(1) = 14 âˆ’ 9 = 5 .

From
(1) and (2), we get (A^{-1}) = (A^{T})^{-1}. Thus, we
have verified the given property.

**Example 1.9**

Verify ( *AB*)^{âˆ’}^{1} = *B*^{âˆ’}^{1} *A*^{âˆ’}^{1} with

Solution

As
the matrices in (1) and (2) are same, (AB) ^{âˆ’1} = B^{-1} A^{-1}
is verified.

If *A = **, *find *x *and *y
*such that *A*^{2} + *xA *+ *yI _{2}=O_{2,}* Hence, find

Solution

So,
we get 22 + 4x + y =0, 31+5x+y=0, 27+3x=0 and 18+2x=0

Hence
x = âˆ’9 and y =14.Then, we get A^{2} - 9A + 14I_{2} = O_{2}

Post-multiplying
this equation by A^{âˆ’1} , we get A â€“ 9I_{2} + 14A^{-1}
= O_{2}. Hence, we get

Tags : Definition, Theorem, Formulas, Solved Example Problems | Inverse of a Non-Singular Square Matrix , 12th Mathematics : UNIT 1 : Applications of Matrices and Determinants

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

12th Mathematics : UNIT 1 : Applications of Matrices and Determinants : Properties of inverses of matrices | Definition, Theorem, Formulas, Solved Example Problems | Inverse of a Non-Singular Square Matrix

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright Â© 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.