Inverse Matrix
by
Dr. Shorouk Ossama
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.
• Example:
𝟐 −𝟓
𝟑 𝟓
Let A =
,𝐁 =
−𝟏 𝟑
𝟏 𝟐
Find AB and BA.
THEOREM
𝑎
• The matrix A =
𝑐
𝑏
is invertible if and only
𝑑
if ad – bc ≠ 0, in which case the inverse is
given by the formula:
Where the quantity ad – bc in the theorem is called the
determinant of the 2x2 matrix A and is denoted by
𝑎
det (A) = ad – bc or by
𝑐
𝑏
= 𝑎𝑑 − 𝑏𝑐
𝑑
* 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
No Inverse
Singular
det A ≠ o
Has Inverse
Non-Singular
The matrix
a
A= c
b
d
is invertible if ad-bc≠0, in witch case the inverse is
given by the formula
• Example:
Calculating the Inverse of a 2 × 2 Matrix
The determinant of A is: det(A) = (6)(2) – (1)(5)
= 7, which is nonzero. Thus, it is invertible and
its inverse is:
(b) The matrix is not invertible since:
det(A) = (-1)(-6) – (2)(3) = 0.
•We leave it for you to confirm that:
A A-1 = I and A-1 A = I
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
0
0
-3
0
0
0
2
(AT )ˉ¹ = (A ˉ¹) T
Example: Consider the matrices
=
a11 x1 + a12 x2 = d1
a21 x1 + a22 x2 = d2
These are linear equations in the unknowns x1, x2.
Where:
Thus:
AX=d
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.
If we have any system:
2x + 3y = 1
x – 5y = 4
So:
2
AX=B
Multiply from left by A-1
A-1 A X = A-1 B
1. Get A-1
2. Multiply A-1 by B
3. These are x and y
X = A-1 B
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:
By Using Row Operations to Find A−1
𝟏 𝟐 𝟑
• Example: If A = 𝟐 𝟓 𝟑 ,
𝟏 𝟎 𝟖
- We want to reduce A to the identity matrix by
row operations and simultaneously apply these
operations to I to produce A-1
-Thereby producing a partitioned matrix of the
form 𝑨 𝑰 .
-so the final matrix will have the form 𝑰 𝑨−𝟏 .
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-R2+R1
By Using Cofactors
• We can find the inverse of the matrix 3x3 By
Using Cofactors. The transposed matrix of
cofactors given by:
𝐶11 𝐶21 𝐶31
•adj A = 𝐶12 𝐶22 𝐶32 is known as the
𝐶13 𝐶23 𝐶33
adjoint of A. Hence the inverse matrix A given
by:
+
-
+
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 (-1)] = 2 x 3 - 1 x 7 = -1
Then get :
-1
A
= (1/det A) * [adj matrix]
a = + ( -1 x 2 ) – ( -1 x 5 ) = 3
e = + ( 2 x 2 ) – ( -1 x 0 ) = 4
h = - [( 2 x 5 ) – ( 1 x 0 )] = -10
Thus:
Example: Verify by direct multiplication that
has the inverse matrix
And then check the matrix product BA = I
Hence B = A-1
Thanks