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Question 23 : A 2 × 2 matrix is filled with four distinct integers randomly chosen from the set {1,2,3,4,5,6} Then the probability that the matrix generated in such a way is singular is

- \\frac{2}{45})
- \\frac{1}{45})
- \\frac{4}{15})
- \\frac{1}{15})

\\frac{2}{45})

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A matrix is singular if and only if the determinant of the matrix is 0.

In a 2 × 2 matrix,

\\left|\begin{array}{cc}a & c \\d & b\end{array}\right| \\) = ab - cd = 0

Or ab = cd

If one of a,b,c or d is 5, then ab ≠ cd, because in the set {1,2,3,4,5,6} there is only one multiple of 5.

So, we just have the set {1,2,3,4,6} to deal with and out of these we'll have select 4 integers.

We can't pick either 3 or 6. We'll have to pick them both.

That is 3 and 6 have to stay on either side of the equation ab = cd

So, the equation now becomes, 3b = 6d

Clearly, b > d

{b, d} ⊂ {1, 2, 4}

If b = 4, d = 2

If b = 2, d = 1

So, only two such combinations exist.

{a, b, c, d} = {1, 2, 3, 6} or {a, b, c, d} = {4, 2, 3, 6}

We can select four numbers from the six in ^{6}C_{4} = 15 ways

After selecting the 4 integers, we can arrange them in order in 4! ways.

So, the total number of matrices that can be constructed = 15 × 4!

Once we select {a, b, c, d} = {1, 2, 3, 6} or {a, b, c, d} = {4, 2, 3, 6},

To construct a singular matrix, for example, 2 & 3 should be on the same diagonal. Similarly, 4 & 3 too in case of {a, b, c, d} = {4, 2, 3, 6}.

Only one-third of the arrangments yeild a singular matrix and the others dont.

But the probabilty of getting a singular matrix = \\frac{2 × 4! ÷ 3}{15 × 4!}) = \\frac{2}{45})

The question is **" A 2 × 2 matrix is filled with four distinct integers randomly chosen from the set {1,2,3,4,5,6} Then the probability that the matrix generated in such a way is singular is " **

Choicve A is the correct answer.

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