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Question 34779: Find the inverse of the following matrix.
A = [2 4]
[1 -3]
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! A = [2 4]
[1 -3]
I AM NOT SURE WHAT METHOD YOU HAVE BEEN TAUGHT BUT LET ME GIVE YOU THE FUNDAMENTAL APPROACH METHOD FROM DEFINITION BY A 4 STEP PROCESS.FIRSTLY BEFORE START LABEL THE ROWS AND COLUMNS FOR UNDERSTANDING...MATRIX A IS
.............COLUMN 1..............COLUMN 2
ROW 1...........2.....................4
ROW 2...........1....................-3
NOW WE HAVE AN ADDRESS SYSTEM.E-R1C1 MEANS ELEMENT IN ROW 1 & COLUMN 1..THE ELEMENT THERE IS 2..ETC......
STEP 1.
1.REPLACE EACH ELEMENT WITH ITS MINOR...IT MEANS...SUPPOSE WE TAKE E-R1C1 THAT IS 2.WE SHOULD REPLACE IT WITH ITS MINOR..THAT IS THE ELEMENT REMAINING AFTER THE ROW AND COLUMN CONTAINING OUR ORIGINAL ELEMENT E-R1C1 ARE REMOVED..THAT IS REMOVE R1 AND C1..WHAT DO WE HAVE ..IT IS -3.SO WE PUT -3 INPLACE OF 1 IN R1C1.
A)TO REPEAT E-R1C1=1 IS REPLACED BY E-R2C2=3 WHICH IS THE ELEMENT REMAINING AFTER R1 AND C1 ARE REMOVED
SIMILARLY.......
B)NEXT E-R1C2=4 IS REPLACED BY E-R2C1=1 WHICH IS THE ELEMENT REMAINING AFTER R1 AND C2 ARE REMOVED
C)NEXT E-R2C1=1 IS REPLACED BY E-R1C2=4 WHICH IS THE ELEMENT REMAINING AFTER R2 AND C1 ARE REMOVED
D)NEXT E-R2C2=-3 IS REPLACED BY E-R1C1=1 WHICH IS THE ELEMENT REMAINING AFTER R2 AND C2 ARE REMOVED
SO AFTER STEP 1 WE HAVE
.............COLUMN 1..............COLUMN 2
ROW 1...........-3.....................1
ROW 2............4.....................2
STEP 2.
MULTIPLY EACH OF THE ELEMENTS WITH (-1)^(RN+CN)...THAT IS SUM OF ITS ROW AND COLUMN NUMBERS....FOR EXAMPLE FOR E-R1C1..WE MULTIPLY WITH(-1)^(1+1)=1
FOR EXAMPLE FOR E-R1C2..WE MULTIPLY WITH(-1)^(1+2)=-1
FOR EXAMPLE FOR E-R2C1..WE MULTIPLY WITH(-1)^(2+1)=-1
FOR EXAMPLE FOR E-R2C2..WE MULTIPLY WITH(-1)^(2+2)=1
SO AFTER STEP 2 WE HAVE
.............COLUMN 1..............COLUMN 2
ROW 1...........-3....................-1
ROW 2...........-4.....................2
STEP 3.
TRANSPOSE THE ABOVE MATRIX..THAT IS CHANGE ROWS INTO COLUMNS AND COLUMNS INTO ROWS...
SO WE GET AFTER STEP 3
.............COLUMN 1..............COLUMN 2
ROW 1...........-3....................-4
ROW 2...........-1.....................2
STEP 4.
FIND DETERMINANT OF ORIGINAL MATRIX AND DIVIDE THE MATRIX WE GOT IN STEP 3 WITH THAT .THAT IS DIVIDE EACH ELEMENT WITH THE DETERMINANT.THE RESULTANT MATRIX IS THE INVERSE OF A.
NOW DETERMINANT OF ORIGINAL MATRIX IS =2*(-3)-1*4=-6-4=-10
SO WE GET AFTER STEP 4 THE INVERSE MATRIX OF A AS A^-1 EQUAL TO
-3/-10,-4/-10
-1/-10,2/-10
THAT IS
3/10,4/10
1/10,-2/10
YOU CAN CHECK BY MULTIPLYING A WITH A^-1 AND SEE WHETHER YOU GET UNIT MATRIX OF 2 ND. ORDER NAMELY
1,0
0,1
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