Problem
Let R =\( \left\{ \begin{pmatrix} a & 3 & b\\ c & 2 & d\\ 0 & 5 & 0 \end{pmatrix}: a, b, c, d \in \left\{0, 3, 5, 7, 11, 13, 17, 19\right\} \right\} \). Then find the number of invertible matrices in R.
Solution
To be invertible, the determinant of the given matrix must be non-zero. To find the determinant, we’ll expand along the third row to get |R| = -5(ad – bc). For this to be non-zero, (ad – bc) ≠ 0.
To find the number of matrices for which the above holds, we’ll find the total number of matrices and subtract those cases where (ad – bc) = 0, since that’ll be easier to count.
The total number of matrices will be 8⁴, since each of a, b, c, and d can be indepently chosen in 8 different ways (8 being the number of elements in the set {0, 3, 5, 7, 11, 13, 17, 19}).
Now, let’s come to the cases where (ad – bc) = 0 or ad = bc.
There are a number of ways in which ad can be equal to bc. Some are obvious – when atleast one of the two numbers of each side is 0. Let’s count these first.
So, ad and bc will both be 0 (and hence equal), when alteast one number on each side equals 0. We could count these by making cases for each side, i.e., {(a = 0, d ≠ 0), (a ≠ 0, d = 0), (a = 0, d = 0)}.
But a better way to count is to subtract the number of cases when each side is non-zero, from the total.
The number of cases where, say, LHS is non-zero is 7², since each of a and d can be indepently chosen in 7 different ways (leaving out the 0). And the total number of cases will be 8².
So, the number of ways in which LHS can be 0 is 8² – 7² or 15. Same is the number for the RHS. So, the total number of ways in which ad = bc, both sides being 0, equals 15 x 15 or 225.
When none of the four numbers is 0, when can ad be equal to bc? Let’s take a closer look at the set, after getting rid of the 0: {3, 5, 7, 11, 13, 17, 19}. Prime numbers!
Given the fundamental theorem of arithmetic, the only way ad can equal to bc is when either a = b and c = d, or a = c and b = d. Let’s count each case separately.
Case 1: When a = b and c = d
Here, a can be selected in 7 ways, and b gets selected automatically. Similarly, c can be selected in 7 ways, and d gets selected with the same value. Since a and c can be selected independently, the total number of ways will be 7 x 7 or 49.
Case 2: When a = c and b = d
Nothing’s different here – 49 cases again.
But there’s a catch! In the two cases above we’ve double counted the cases where a = b = c = d. Because, a could be equal to c in the first case, making all numbers equal. Similarly a could be equal to b in the second case.
So, we’ll have to remove the cases where a = b = c = d once from the above 49 + 49 or 98 cases.
Counting that is simple – 7 ways to select a, and the other three get selected automatically. So, the number of cases where ad = bc, both sides being non-zero, will be 98 – 7 or 91.
Finally, the total number of cases where ad = bc will be 225 + 91 or 316.
And, we’re done! The number of cases where (ad – bc) ≠ 0, i.e. the number of invertible matrices in R will be 4096 – 316 = 3780.
Comments
The problem was a combinatorics problem at heart, but was unnecessarily complicated by introducing the matrix – probably to make it 10% more difficult? Or, for a wider coverage across topics for the exam.
Here’s an alternate method to complicate: “Let ax + by + 1 = 0 and cx + dy + 2 = 0 be two lines such that a, b, c, d ∈ {0, 3, 5, 7, 11, 13, 17, 19}. Find the number of sets {a, b, c, d} such that the lines intersect.”
But yeah, don’t think that’s really needed. Apart from this complication, the problem seemed to be of a moderate level – one needs to list down all the cases carefully. Post that, the counting part is easy enough.
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