The value of the determinant

Key Concepts

Determinant of a Matrix

The determinant is a scalar value computed from a square matrix that provides important information about the matrix, such as whether it is invertible and the scaling factor applied by the linear transformation it represents. In problems involving determinants, various methods like cofactor expansion, use of row operations, or recognition of patterns in matrix entries are used for efficient evaluation.

Binomial Coefficients

Binomial coefficients, typically expressed as nCk or 'n choose k', represent the number of ways to choose k elements from a set of n elements without regard to order. They obey various combinatorial identities, such as Pascal's identity, and appear in polynomial expansions and probability, often simplifying the relationships between the entries in a matrix when evaluating determinants.

Row Operations and Matrix Simplification

Row operations, including swapping rows, scaling rows, and adding multiples of one row to another, are instrumental in simplifying a matrix to a form in which the determinant can be computed more easily. These operations utilize the inherent properties of determinants, such as the effect on the determinant value when rows are manipulated, and help in revealing structural patterns or dependencies among the rows.

Transcript

00:01 So in the given question we have a determinant that is given as m m m c1 n plus 1 c1 m plus 2 c1 and in the third row we have nc2 n plus 1 c2 and in the third row we have nc 2 n plus 1 c2 and n plus 2 c2 and we are told to evaluate this determinant right so what we can do first is we can write we can take m as a common factor from this determinant and simplify this determinant by evaluating the combinations that are given right so to evaluate the combination we should know the formula for nc r that is n factorial divided by r factorial times n minus r factorial right so this relation this formula should be kept in mind in order to simplify this matrix this determinant so n c1 would then be equal to n factorial divided by 1 factorial times n minus 1 factorial which is equal to n factorial can be written as n fact to n n minus 1 factorial n times n minus 1 factorial divided by n minus 1 factorial which gives us n c1 as equal to n right similarly we have n plus 1 c1 which is n plus 1 factorial divided by 1 factorial times n plus 1 minus 1 factorial which is factorial so we can simplify this and we would get n plus 1 over here and for n plus 2 c1 we would get n plus 2 next we have n c2 right so for nc 2 it is the same way that is n factorial divided by 2 factorial times n minus 2 factorial n minus 2 factorial which we can write as n times n minus 1 times n minus 2 factorial divided by 2 factorial is 2 so 2 times n minus 2 factorial gives us n times n minus 1 divided by 2 as n c2 right so similarly for n plus 1 c2 we would have n plus 1 c2 we would have n plus 1 times n times n minus 1 right we would have n plus 1 factorial divided by 2 factorial times n plus 1 minus 2 which is n minus 1 factorial so what we would have is n plus 1 times n divided by 2 next we have n plus 1 sorry n plus 2 c2 which would be equal to n plus 2 factorial n plus 2 factorial times divided by 2 factorial times n plus 2 minus 2 which would be equal to n factorial so this would simplify to n plus 1 times n plus n plus 2 times n plus n plus 2 times n plus 2.

03:52 So these are the substitutions that we are going to make in the above determinant first so what we would have over here is we can take this as a m as a common factor and for n c1 we just found it is equal to n n plus 1 and n plus 2 right that is what we found right and for n c2 we have n times n minus 1 n times n minus 1 by 2 n plus 1 n plus 1 times n n divided by 2 is for n plus 1 c2 and for n plus 2 c2 we have n plus 2 times n plus 1 divided by 2 so this is what we have as the simplified determinant right so now what we can do is we can simplify this further right so let's do the column operation c2 changes to c2 minus c1 c2 minus c1 and this would give us the determinant as 1 n times n minus 1 by 1 divided by 2 1 minus 0 n plus 1 minus n is equal to 1 and n plus 1 times n divided by 2 minus n minus n minus 1 we can just solve it over here n square plus n minus n square minus n would be equal to 2 times n divided by 2 which is equal to n right so over here it would be n and the rest we have the same form right so this is what we have so next what we can do is before doing this, we should have done the column 3, right? so first let's simplify column 3, then we can simplify column 2, right? so if we do that, we can simplify the determinant even better, right? so let's simplify c3 changes to c3 minus c2, then we would have 1 n plus 1, n plus 1 times n divided by 2, here and for column 3 we would have 1 minus 1 which is 0 n plus 2 minus n plus 1 is 1 and n plus 2 times n plus 1 is n square plus n square plus 2 n which is 3 n plus 2 minus we have n square minus n this is divided by 2 so n square minus n square is 0 3 n minus n is 2 n so we have 2 n plus 2 divided by 2 which is n plus 2 so this is what we would have as the determinant after doing the column operation c3 changes to c3 minus c2 so next up what we are doing is we are going to change c2 changes to c2 minus c1 and what we would have is 1 n times n minus 1 by 2 0 1 and we would have n over here and over here we have 1 0 1 n plus 2 right so this is what we have so let's check this again right so n plus 1 to times n is n square n square plus n minus n square plus n which gives us 2 times n divided by 2 which is equal to n and for the third element over here we would have n square plus n plus n plus 2 n plus 2 minus n square plus n squared minus n so n square and n square would cancel off 2 n minus n minus n plus n is 0 and what we would have is 2 n plus 2 divided by 2 which would be equal to n plus 1 right it is not n plus 2 over here it is n plus 1 so this is is what we get after simplifying the matrix doing these two column operations and now we can easily find the value of this determinant by taking 1 times n plus 1 minus n and this would be equal to 1 times 1 which is equal to 1.

09:31 So in the question we have a few options and we should know that there is an m outside that we took as a common factor in the beginning of the solution right we have an m outside the determinant so the answer is actually m times one right we have an m outside the determinant so it is 1 times m which is equal to m so the required answer or the solution or the value of this determinant is equal to m and it is given by the option a...

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