00:01
In this problem, we have to find three two by two matrices that are their own inverses other than identity matrix and negative of the identity matrix.
00:17
So let us consider matrix a is equal to 01 .10.
00:32
Now we find the product of matrix a with itself that is a times a is equal to 0110 times 0110.
01:01
Since the two matrices are comfortable for multiplication because the number of columns in the first matrix are equal to the number of columns in the first matrix are equal to the number.
01:13
Of rows in the second matrix.
01:18
So the product will be multiplying the entries of r1 with the corresponding entries of c1, we will get 0 times 0 plus 1 times 1 is 1.
01:43
Now, the product of r1 with c2 is 0 plus 0.
01:56
Product of r2c1 is 1 times 0 and 0 times 1 is 0.
02:04
Now, the product of r2c2 is 1 and 0 % 1 is 1 and 0 times 1 % 0 is 0 so the product will be 1 0 0 that is the identity matrix of order 2 by 2 this implies that a square is equal to identity matrix thus the given matrix it's is its own inverse now consider the second matrix b is equal to square root 3 by 2, 1 by 2 half and minus square root 3 by 2.
03:25
Now b squared is equal to b times b is equal to b times b is equal to...