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

Asked by mishrk261 | 28th Aug, 2019, 09:07: PM

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 Renu Varma | 30th Aug, 2019, 10:14: AM

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