Do you still need help with this? You may have figured it out by now since it follows the same method with the question 23.

>>Just in case, for any vector u and v, for the two to be parallel means v can be written as u=k x v.(for k any complex number, furthermore for any k in any field F) (I have used x for the multiplication sign and in matrix ; means the row ended that is if A=[1 2; 3 4] , first row of A from left to right is 1 and 2 and the second row is 3 and 4)

>>In your question for v = (8t, -2), k must be 1/2 since (k x (-2) = -1). Since k is 1/2=> 4 = 1/2 x 8t => t = 1. Hence in (a) they are parallel for t = 1.

>>(b) v = (8t, 2t). 4 = k x 8t => k x t = 1/2 and -1 = k x 2t => k x t = -1/2. But 1/2 is not equal to -1/2 so they are never parallel.

>>© v = (1, t^{2}) => 4 = k x 1 = k, Since k is 4, -1 = 4 x t^{2} Hence they would be parallel for the complex numbers i/2 and -i/2 (Usually in algebra courses you would be allowed to use complex numbers but you should still check if you are allowed to use them. If you are not allowed to use them, they are never parallel, again)

Also for determinant, (**pass this part if you are not interested in abstract linear algebra**) determinant is defined as for a matrix A, (there should be permutations under Σ, since you want to sum all of the permutations up to isomorphism possible) detA=Σ(sgn(j_{1}....j_{n})a_{1j1} a_{1j2} .... a_{njn}) where a_{ijn }denotes the element in the ith row and j_{n}th column (j_{n} is a permutation so it's not the same as the initial column number j) and the sgn(j_{1}....j_{n}) denotes the -1 to the power of number of permutations of elements 1,2,...,n to j_{1}....j_{n }(if you haven't taken any algebra course before, this is simply just saying the number of ways you can correctly and uniquely permute (1,2,...,n),,,, up to isomorphism of course) and defined as (-1)^{number of inversions in j1....jn} (>>further (-1)^{row num+col num} in the determinant comes from this sign of permutation)

**Moving on/further from the abstract definition for a 2x2 matrix A=[a b; c d] , det(A) = ad-bc ** (coming from definition sgn(1,2)a_{11}a_{12} + sgn(2,1)a_{12}a_{21}) (However, I still used the definition)

We can use the definition to calculate the determinant of any matrix using the determinant of the 2x2 matrix which is above and would usually be given to you as a lemma (but it's a simple application of the definition)**, **

**for B=[a b c; d e f; g h i] **I will use the first row to calculate the determinant however you are free to choose any row you like

**det(B)=(-1)**^{1+1}a det([e f; h i]) + (-1)^{1+2}b det([d f; g i]) **+ (-1)**^{1+3}c det([d e; g h]). (However, I still used the definition)

*That is, you are first choosing a row then by starting from the first column, you will add the column number and the row number (this is the number of permutations) and you will calculate-1 to the power of sum you have just calculated then you will multiply it with the rest: and then you will multiply by the element which is in the intersection of the row and column you are currently in then you will disregard all the elements in the row and column you are currently in, then you will multiply by the determinant of the smaller matrix created by disregarding that row and column. Then you will move on to the next column and continue the process until the end of the row. For the new determinants you have to calculate, you have to continue this process until you get to the matrix in the form 2x2.*

But since the abstract definition is practically useless** there are some correlations and shortcuts which can help you calculate the determinants more easily**, I suggest you to study them. Your teacher can probably help you. It's really easy actually, it just takes lots of works to actually use the determinant practically. If you have more questions about how to calculate them or how to prove their properties, I would love to help ^^

NOTE: Furthermore for question 22, __CLAIM:__ for vectors u=(x_{1}, y_{1}) and v=(x_{2}, y_{2}), (assuming u and v are not the zero vector) they are said to be parallel if and only if the determinant of the matrix A = [x_{1}, y_{1; }x_{2}, y_{2}] is zero.

>>*proof: if u and v are parallel then u = k x v for some k in a field. then u can be written as (kx*_{2, }ky_{2})

*then det(A) = x*_{1}y_{2 }- y_{1}x_{2 }= kx_{2}y_{2 }-_{ }ky_{2}x_{2} = 0

*if det(A) = 0 then x*_{1}y_{2 }=_{ }y_{1}x_{2 }=>_{ }x_{1}/x_{2}_{ }=_{ }y_{1}/y_{2 }= k for some k in a field. Hence the vectors are parallel.

Applying the claim to your question:

(a) det([4 -1; 8t -2]) = -8 - (-8t) = 8t - 8 Since we want u and (a) to be parallel determinant must be zero and 8t-8 is zero when t = 1.

(b)det([4 -1; 8t 2t]) = 8t - (-8t) = 16t Since we want u and (b) to be parallel determinant must be zero and 16t = 0 when t = 0 but if one of the vectors is zero then the definition fails and the vectors are not parallel.

©det([4 -1; 1 t^{2}) = 4t^{2} - (-1) = 4t^{2} + 1 Since we want u and © to be parallel determinant must be zero. 4t^{2} + 1 = 0 then 4t^{2} = -1 then t^{2 }= -1/4 if you take the square root you will get two roots in complex numbers that are i/2 and -i/2. So the vectors are parallel for t = i/2 or t = -i/2

!!When I first posted this, I have made many mistakes but currently everything seems fine.!!