Question

If $A=\left[\begin{array}{cc}\lambda & 1\\ -1& -\lambda \end{array}\right]$ , 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:
$A=\left[\begin{array}{cc}\lambda & 1\\ -1& -\lambda \end{array}\right]$
And
$A^2=O$
Substituting the values of $A$,
$$\left[\begin{array}{cc}\lambda & 1\\ -1& -\lambda \end{array}\right]\times \left[\begin{array}{cc}\lambda & 1\\ -1& -\lambda \end{array}\right]=O$$
Performing Matrix Multiplication,
$\left[\begin{array}{cc}\lambda \times \lambda +1\times \left(-1\right)& \lambda \times 1+1\times \left(-\lambda \right)\\ \left(-1\right)\times \lambda +\left(-\lambda \right)\times \left(-1\right)& \left(-1\right)\times 1+\left(-\lambda \right)\times \left(-\lambda \right)\end{array}\right]=O$
Simplifying further,
$\left[\begin{array}{cc}{\lambda }^2-1& \lambda -\lambda \\ -\lambda +\lambda & -1+{\lambda }^2\end{array}\right]=O$
We know that the Identity Matrix of order $2$ is $I=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ .
Therefore, substituting this,
$\left[\begin{array}{cc}{\lambda }^2-1& 0\\ 0& {\lambda }^2-1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
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 $B$.