Question

If $A = \left[ {\begin{array}{*{20}{c}}
  \lambda &1 \\
  { - 1}&{ - \lambda }
\end{array}} \right]$ , then for what value of $\lambda $ is ${A^2} = O$ ?
A. $0$
B. $ \pm 1$
C. $ - 1$
D. $1$

Answer 1

Hint: In the above question, we are provided with a matrix $A$ which has some elements in terms of $\lambda $ . We are also given that ${A^2} = O$ . Perform Matrix Multiplication and evaluate the value of ${A^2}$. Now equate each of its elements to the corresponding elements of the zero matrices. Then, evaluate the value of $\lambda $.

Complete step by step Solution:
We are given a matrix $A$ such that:
$A = \left[ {\begin{array}{*{20}{c}}
  \lambda &1 \\
  { - 1}&{ - \lambda }
\end{array}} \right]$
And
${A^2} = O$
Substituting the values of $A$,
\[\left[ {\begin{array}{*{20}{c}}
  \lambda &1 \\
  { - 1}&{ - \lambda }
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  \lambda &1 \\
  { - 1}&{ - \lambda }
\end{array}} \right] = O\]
Performing Matrix Multiplication,
$\left[ {\begin{array}{*{20}{c}}
  {\lambda \times \lambda + 1 \times \left( { - 1} \right)}&{\lambda \times 1 + 1 \times \left( { - \lambda } \right)} \\
  {\left( { - 1} \right) \times \lambda + \left( { - \lambda } \right) \times \left( { - 1} \right)}&{\left( { - 1} \right) \times 1 + \left( { - \lambda } \right) \times \left( { - \lambda } \right)}
\end{array}} \right] = O$
Simplifying further,
$\left[ {\begin{array}{*{20}{c}}
  {{\lambda ^2} - 1}&{\lambda - \lambda } \\
  { - \lambda + \lambda }&{ - 1 + {\lambda ^2}}
\end{array}} \right] = O$
We know that the Identity Matrix of order $2$ is $I = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right]$ .
Therefore, substituting this,
$\left[ {\begin{array}{*{20}{c}}
  {{\lambda ^2} - 1}&0 \\
  0&{{\lambda ^2} - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right]$
Comparing the elements of the matrix on the Left-Hand side with the one on the Right-Hand side,
${\lambda ^2} - 1 = 0$
This gives: $\lambda = \pm 1$
Hence, the value of $\lambda $ is $ \pm 1$ .

Hence, the correct option is (B).

Note: Two matrices $A$ and $B$ are said to be equal to each other when they both are of the same order and when every element of matrix $A$ is equal to the corresponding elements of matrix $B$.