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Example Determinant: Determine the determinant of each of the following matrices.
Solution:
For the 2 x 2 there isn't much to perform other than to plug this in the formula.
= ( -9) + 4 - (18) (2) = 0
For the 3 x 3 we could plug this in the formula, though not like the 2 x 2 case it is difficult formula to remember. There is an easier manner to get similar result. An earlier way of finding the same result is to perform the following. Firstly write down the matrix and then tack a copy of the first two columns on the end as given below.
Here, notice that there are three diagonals which run from left to right and three diagonals which run from right to left. What we perform is multiply the entries on all diagonal up and whether the diagonal runs from left to right we add them up and whether the diagonal runs from right to left we subtract them.
Now there is the work for this matrix.
= (2)(-6)(-1) + (3) (7)(4) + (1)(-1)(5)-(3)(-1)(-1) - (2)(7) (5)- (1)(-6)(4)
= 42
You can either utilize the formula or the shortcut to find the determinant of a 3 x 3.
If the determinant of a matrix is zero then we call which matrix singular and if the determinant of a matrix isn't zero so we call the matrix nonsingular. The 2 x 2 matrix into the above illustration was singular whereas the 3 x 3 matrix is nonsingular.
a) Let n = (abc) 7 . Prove that n ≡ a + b + c (mod 6). b) Use congruences to show that 4|3 2n - 1 for all integers n ≥ 0.
3/45
Multiply the given below and write the answer in standard form. (2 - √-100 )(1 + √-36 ) Solution If we have to multiply this out in its present form we would get, (2 -
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15(4*4*4*4*+5*5*5)+(13*13*13+3*3*3)
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