If A = `[(lambda,1),(-1,-lambda)]`, then for what value of `lambda,A^(2)=0`?

Options

A. 0
B. `+-1`
C. `-1`
D. 1

Answer

To find the value of \( \lambda \) such that \( A^2 = 0 \) for the matrix \( A = \begin{pmatrix} \lambda & 1 \\ -1 & -\lambda \end{pmatrix} \), we will perform the following steps:

### Step 1: Calculate \( A^2 \)

We need to compute \( A^2 \) by multiplying matrix \( A \) with itself:

\[ A^2 = A \cdot A = \begin{pmatrix} \lambda & 1 \\ -1 & -\lambda \end{pmatrix} \cdot \begin{pmatrix} \lambda & 1 \\ -1 & -\lambda \end{pmatrix} \]

### Step 2: Perform the Matrix Multiplication

Using the formula for matrix multiplication, we calculate each element of the resulting matrix:

- The element at position (1,1):
\[ \lambda \cdot \lambda + 1 \cdot (-1) = \lambda^2 - 1 \]

- The element at position (1,2):
\[ \lambda \cdot 1 + 1 \cdot (-\lambda) = \lambda - \lambda = 0 \]

- The element at position (2,1):
\[ -1 \cdot \lambda + (-\lambda) \cdot (-1) = -\lambda + \lambda = 0 \]

- The element at position (2,2):
\[ -1 \cdot 1 + (-\lambda) \cdot (-\lambda) = -1 + \lambda^2 \]

Putting it all together, we have:

\[ A^2 = \begin{pmatrix} \lambda^2 - 1 & 0 \\ 0 & \lambda^2 - 1 \end{pmatrix} \]

### Step 3: Set \( A^2 \) Equal to the Zero Matrix

To find the value of \( \lambda \) such that \( A^2 = 0 \), we set the resulting matrix equal to the zero matrix:

\[ \begin{pmatrix} \lambda^2 - 1 & 0 \\ 0 & \lambda^2 - 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

This gives us the equation:

\[ \lambda^2 - 1 = 0 \]

### Step 4: Solve for \( \lambda \)

Now we solve the equation:

\[ \lambda^2 = 1 \]

Taking the square root of both sides, we find:

\[ \lambda = \pm 1 \]

### Conclusion

Thus, the values of \( \lambda \) for which \( A^2 = 0 \) are:

\[ \lambda = 1 \quad \text{or} \quad \lambda = -1 \]