Question

If \[A = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&{ - 4}\\{ - 1}&3&4\\1&{ - 3}&{ - 4}\end{array}} \right]\], then find nilpotent index of the matrix.
A. 2
B. 3
C. 4
D. 6

Answer 1

Hint: We will calculate the value of n such that \[{A^n}\] becomes a zero matrix. The value of n is a nilpotent index.

Formula used:
A matrix is known as nilpotent, if \[{A^n}\] is zero matrices.

Complete step by step solution:
Given matrix is \[A = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&{ - 4}\\{ - 1}&3&4\\1&{ - 3}&{ - 4}\end{array}} \right]\].
Now we will calculate \[{A^2}\].
\[{A^2} = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&{ - 4}\\{ - 1}&3&4\\1&{ - 3}&{ - 4}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 3}&{ - 4}\\{ - 1}&3&4\\1&{ - 3}&{ - 4}\end{array}} \right]\]
\[ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}{1 \cdot 1 + \left( { - 3} \right) \cdot \left( { - 1} \right) + \left( { - 4} \right) \cdot 1}&{1 \cdot \left( { - 3} \right) + \left( { - 3} \right) \cdot 3 + \left( { - 4} \right) \cdot \left( { - 3} \right)}&{1 \cdot \left( { - 4} \right) + \left( { - 3} \right) \cdot 4 + \left( { - 4} \right) \cdot \left( { - 4} \right)}\\{ - 1 \cdot 1 + 3 \cdot \left( { - 1} \right) + 4 \cdot 1}&{ - 1 \cdot \left( { - 3} \right) + 3 \cdot 3 + 4 \cdot \left( { - 3} \right)}&{ - 1 \cdot \left( { - 4} \right) + 3 \cdot 4 + 4 \cdot \left( { - 4} \right)}\\{1 \cdot 1 + \left( { - 3} \right) \cdot \left( { - 1} \right) + \left( { - 4} \right) \cdot 1}&{1 \cdot \left( { - 3} \right) + \left( { - 3} \right) \cdot 3 + \left( { - 4} \right) \cdot \left( { - 3} \right)}&{1 \cdot \left( { - 4} \right) + \left( { - 3} \right) \cdot 4 + \left( { - 4} \right) \cdot \left( { - 4} \right)}\end{array}} \right]\]
\[ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right]\]
Since \[{A^2}\] is zero and the exponent is 2, the nilpotent index of the matrix is 2.
Hence option A is the correct option.

Additional information:
The determinate value of the nilpotent matrix is always zero. Since the determinant of a nilpotent matrix is zero, the matrix is a noninvertible matrix. The eigenvalues of a nilpotent matrix are also zero. Conversely, we can say that if all eigenvalue of a matrix is zero, then the matrix is a nilpotent matrix. The sum of two nilpotent is also zero.

Note: Students often confuse with singular matrix and nilpotent matrix. The determinate value of all nilpotent matrices is zero. But all singular matrices are not nilpotent. For example: \[\left[ {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right]\]