Question

If \left[ {\begin{array}{*{20}{c}} 2&4 \\\ { - 1}&k; \end{array}} \right] is an nilpotent matrix of index $2$ then find the value of $k$ .
(A) $3$
(B) $- 3$
(C) $4$
(D) $- 2$

Answer 1

Hint : In this problem, first we will find ${A^2}$ . Then, we will equate the corresponding elements of the matrix ${A^2}$ and the zero matrix. After that, we will get a linear equation in $k$ . On solving the linear equation, we will get the required value.

Complete step-by-step answer :
In this problem, the given matrix \left[ {\begin{array}{*{20}{c}} 2&4 \\\ { - 1}&k; \end{array}} \right] is an nilpotent matrix of index $2$ . Let us say A = \left[ {\begin{array}{*{20}{c}} 2&4 \\\ { - 1}&k; \end{array}} \right] . Let us find ${A^2}$ . So, we can write
{A^2} = A \cdot A \\\ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} 2&4 \\\ { - 1}&k; \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&4 \\\ { - 1}&k; \end{array}} \right] \;
Multiply each element of the first row of the first matrix with corresponding elements of the first column of the second matrix and add all results. Then, multiply each element of the first row of the first matrix with the corresponding elements of the second column of the second matrix and add all results. Then, multiply each element of the second row of the first matrix with corresponding elements of the first column of the second matrix and add all results. Then, multiply each element of the second row of the first matrix with corresponding elements of the second column of the second matrix and add all results. Hence, we get
{A^2} = \left[ {\begin{array}{*{20}{c}} {2\left( 2 \right) + 4\left( { - 1} \right)}&{2\left( 4 \right) + 4\left( k \right)} \\\ { - 1\left( 2 \right) + k\left( { - 1} \right)}&{ - 1\left( 4 \right) + k\left( k \right)} \end{array}} \right] \\\ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} 0&{8 + 4k} \\\ { - 2 - k}&{{k^2} - 4} \end{array}} \right] \;
Since $A$ is a nilpotent matrix of index $2$ , we can say that ${A^2}$ is zero matrix. Note that zero matrix can be written as \left[ {\begin{array}{*{20}{c}} 0&0 \\\ 0&0 \end{array}} \right] . So, we can write \left[ {\begin{array}{*{20}{c}} 0&{8 + 4k} \\\ { - 2 - k}&{{k^2} - 4} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\\ 0&0 \end{array}} \right] . Let us equate corresponding elements. So, we get
$8 + 4k = 0 \Rightarrow k = - 2 \\\ \- 2 - k = 0 \Rightarrow k = - 2 \\\ {k^2} - 4 = 0 \Rightarrow \left( {k + 2} \right)\left( {k - 2} \right) = 0 \Rightarrow k = \pm 2 \\\$
Observe the values of $k$ which we obtained from three different equations. So, we can say that the common value of $k$ is $- 2$ . Hence, we can say that if \left[ {\begin{array}{*{20}{c}} 2&4 \\\ { - 1}&k; \end{array}} \right] is an nilpotent matrix of index $2$ then value of $k$ is $- 2$ . Hence, option D is correct.
So, the correct answer is “Option D”.

Note : A square matrix $A$ is called nilpotent matrix of index $n$ if ${A^n}$ is zero matrix and ${A^{n - 1}}$ is non-zero matrix where $n$ is a positive integer. To solve this type of problems, we must know the matrix multiplication.