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Question 5 : If A is a 3 X 3 non-zero matrix such that A^{2} = 0 then determinant of [(1 + A)^{2} - 50A] is equal to

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A non-zero Square matrix A is said to be a nilpotent matrix of order n if A^{n} = 0. Determinant of[(I+ A)^{50} - 50A] should be the same for any nilpotent matrix of order 2.

So, Let us consider a simple 3 x 3 nilpotent matrix. A = \\left[\begin{array}{ccc}0 & 1 & 1\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]\\)

Check that A^{2} = 0.

I + A = \\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]\\) + \\left[\begin{array}{ccc}0 & 1 & 1\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]\\) = \\left[\begin{array}{ccc}1 & 1 & 1\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]\\).

Observe that I + A is an Upper triangular matrix with all the principal diagonal elements being 1.

(Upper triangular matrix, means that the elements below the diagonal, that is the elements on the South – East are all 0.)

Any number of times a triangular matrix is multiplied to itself, the result is always a triangular matrix.

Something more special about I + A can be known by keen observation, Since the principal diagonal elements are unity(1), and it is a upper triangular matrix, any number of times (I + A) is multiplied to itself, it will remain an Upper triangular matrix with all the principal diagonal elements maintained at unity(1).

Check (I + A)^{2}.

Therefore, (I + A)^{50}, will also be an Upper triangular matrix with all the principal diagonal elements being 1.

So, (I + A)^{50} may look something like this

(I + A)^{50} = \\left[\begin{array}{ccc}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{array}\right]\\)

50 A = 50 \\left[\begin{array}{ccc}0 & 1 & 1\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]\\) = \\left[\begin{array}{ccc}0 & 50 & 50\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]\\)

Observe that both (I + A)^{50} and 50 A are Upper triangular matrices. So will their sum or difference be.

(I + A)^{50} - 50 A = \\left[\begin{array}{ccc}1 & x & y\\0 & 1 & z\\0 & 0 & 1\end{array}\right]\\) - \\left[\begin{array}{ccc}0 & 50 & 50\\0 & 0 & 0\\0 & 0 & 0\end{array}\right]\\) = \\left[\begin{array}{ccc}1 & x-50 & y-50\\0 & 1 & z\\0 & 0 & 1\end{array}\right]\\)

Clearly, (I + A)^{50} - 50 A is also an Upper triangular matrix with principal diagonal elements being 1.

We are asked to find the determinant of (I + A)^{50} - 50 A ,

Determinant of any triangular matrix is the sum of its principal diagonal elements.

So, determinant of (I + A)^{50} - 50 A = 1 + 1 + 1 = 3

Determinant of (I + A)^{50} - 50 A = 3.

**Tip:**You should choose your nilpotent matrix very carefully, For this problem, It has to be a triangular matrix with Principal Diagonal Elements being 0 too.

The question is **"If A is a 3 X 3 non-zero matrix such that A ^{2} = 0 then determinant of [(1 + A)^{2} - 50A] is equal to" **

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