calculus – ii inverse matrix by dr. eman saad & dr. shorouk ossama
TRANSCRIPT
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CALCULUS – IIInverse Matrix
by
Dr. Eman Saad & Dr. Shorouk Ossama
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References
Robert Wrede and Murrary R. Spiegel, Theory
and Problems of Advanced Calculas, 2nd Edition,
2002.
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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.
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a11 x1 + a12 x2 = d1
a21 x1 + a22 x2 = d2
These are linear equations in the unknowns x1, x2.
Where: A X = d
Thus:
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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
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It follows that:
Where: ( a11 a22 - a21 a12 ) not equal to Zero
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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.
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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
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det A = o
det A ≠ o
No Inverse Singular
Has Inverse Non-Singular
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The matrixA=
is invertible if ad-bc≠0, in witch case the inverse is given by the formula
a bc d
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Example: Consider the matrices
we obtain:
Also
Therefore, (AB) -1 = B -1 A-1
?
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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
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Example: Let A and Aˉ¹ :
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Example: If
A=
Then
1 0 00 -3 00 0 2
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(AT )ˉ¹ = (A ˉ¹) T
Example: Consider the matrices
=
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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
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Where:
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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]
+ - +
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a = + ( -1 x 2 ) – ( -1 x 5 )e = + ( 2 x 2 ) – ( -1 x 0 )h = - [( 2 x 5 ) – ( 1 x 0 )]
Thus:
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Example: Find the inverse of
Solution [A ¦ I ]
[ I ¦ Aˉ¹ ]
The computations are as follows:
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We added -2 times the first row to the second -2R1+R2
and -1 times the first row to the third -R1+R3
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We added 2 times the second row to the third 2R2+R3
We multiplied the third row by -1 -R3
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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
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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
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Example: Verify by direct multiplication that
has the inverse matrix
And then check the matrix product BA = I Hence B = A-1
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Thanks