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the nos of 3X3 non singular matrices ,with4 entries as 1 and all other enterices as 0,is                        

Asked by mishrk261 28th August 2019, 9:07 PM
Answered by Expert
Answer:
FIrst consider all the non-singular matrices with 3 entries as 1 and other 6 entries as 0
Now, for a 3 x 3 matrix to be non-singular, each row and column must have a non-zero value and that is 1
So, the possible matrices are:
open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets space space open square brackets table row 0 1 0 row 1 0 0 row 0 0 1 end table close square brackets space space open square brackets table row 0 1 0 row 0 0 1 row 1 0 0 end table close square brackets space space
open square brackets table row 0 0 1 row 0 1 0 row 1 0 0 end table close square brackets space space open square brackets table row 1 0 0 row 0 0 1 row 0 1 0 end table close square brackets space space open square brackets table row 0 0 1 row 1 0 0 row 0 1 0 end table close square brackets
Now, in each of these possibilities, we can make any of the 6 0's into 1.
So, for each matrix there are 6 possibilities
Hence, the total number of possibilities = 6 x 6 =36
Answered by Expert 30th August 2019, 10:14 AM
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