00:02
In problem 14, we have to check whether the given statements are true or false.
00:11
So first consider the statement in part a of this problem.
00:18
Now, to check the truth of the statement, let us consider two matrices satisfying the given condition.
00:29
That is matrix a is equal to, 2 to 2 and matrix b is equal to 4 .2.
00:53
Here we see that the entry, here is the entry a11 and entry a1 of the first matrix and entry b11 of the s3b1 of the 4 is the entry b11 of the second matrix and 2 is the entry a11 of the first matrix.
01:24
Clearly, a11 is equal to 2 times a11.
01:33
Now we are going to find determinants of the matrix a and b.
01:40
Clearly, we see that determinant of a is equal to 2 times 2 is 4 minus 2 times 2 is 4 is equal to 0, and determinant of b is equal to 4 times 2 is 8 minus 2 times 2 is 4.
02:02
That is determinant of a is equal to 0, and determinant of the matrix b, is equal to four.
02:14
That is determinant of the matrix b is not equal to two times determinant of matrix a.
02:35
Thus, the given statement is false.
02:44
Now we are going to check the statement b.
02:48
For this, the determinant of the permutation matrix that is determinant of a is equal to determinant of 0 -0 -0 -1 -1 -0 -0 -0 -0 -1 -0 -1 -0 -1 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0.
03:24
Now, we are going to evaluate this determinant using row operations.
03:37
So interchanging row 1 and row 4, we get this determinant is equal to 1 ,000, 0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -1.
03:59
And 1 and 0 1 0 by interchanging row 1 and row 4.
04:19
From the properties of determinant, we know that if we interchange any two rows, then the sign of determinant will be negative.
04:33
So this is equal to minus 1 power 2, that is we have to multiply it with by another negative sign.
04:50
When we interchange row 2 and row 4 of this determinant above determinant that is 1 000 0100 0100 0100 0 0 0 0 0 1 1 1 and 0 0 0 0 0 0 0 0 0 0 0.
05:14
Here we interchange row 2 and row 4 4.
05:21
Now, if we interchange row 2 and row 3, we have to multiply with another minus 1, so we have minus 1 cube, and the determinant will be 1 -0 -0 -1 -0 -0 -0 -0 -0 -0 -0 -0 -1 -0 -0 -0 -1 -0 and 0 -0 -0 -0 -0 -0 -1, and 0 -0 -0 -0 -0 -0 -1.
05:52
Row 1 by interchanging row 2 and row 3.
06:02
So as we know that determinant of identity matrix is equal to 1 because the product of the entries of pivots is equal to 1.
06:15
So the determinant will be minus 1 cube times 1 is equal to minus 1.
06:25
One cube is minus 1.
06:27
So minus 1 times 1 is equal to minus 1...