If `A=[(3,-2),(4,-1)]`, then find all the possible values of `lambda` such that the matrix `(A-lambdaI)` is singular.

Explanation

To find all possible values of \( \lambda \) such that the matrix \( A - \lambda I \) is singular, we follow these steps:

### Step 1: Define the given matrix and identity matrix
Let \( A = \begin{pmatrix} 3 & -2 \\ 4 & -1 \end{pmatrix} \) and the identity matrix \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

### Step 2: Formulate \( A - \lambda I \)
We subtract \( \lambda I \) from \( A \):
\[ A - \lambda I = \begin{pmatrix} 3 & -2 \\ 4 & -1 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} = \begin{pmatrix} 3 - \lambda & -2 \\ 4 & -1 - \lambda \end{pmatrix} \]

### Step 3: Set the determinant to zero
For the matrix to be singular, its determinant must be zero:
\[ \text{det}(A - \lambda I) = 0 \]

### Step 4: Calculate the determinant
The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \). Thus, we compute:
\[ \text{det}(A - \lambda I) = (3 - \lambda)(-1 - \lambda) - (-2)(4) \]
Expanding this:
\[ = (3 - \lambda)(-1 - \lambda) + 8 \]
\[ = -3 - 3\lambda + \lambda + \lambda^2 + 8 \]
\[ = \lambda^2 - 2\lambda + 5 \]

### Step 5: Set the determinant equal to zero
Now we set the determinant equal to zero:
\[ \lambda^2 - 2\lambda + 5 = 0 \]

### Step 6: Use the quadratic formula
To solve for \( \lambda \), we apply the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \), and \( c = 5 \):
\[ \lambda = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \]
\[ = \frac{2 \pm \sqrt{4 - 20}}{2} \]
\[ = \frac{2 \pm \sqrt{-16}}{2} \]
\[ = \frac{2 \pm 4i}{2} \]
\[ = 1 \pm 2i \]

### Step 7: Conclusion
Thus, the possible values of \( \lambda \) such that the matrix \( A - \lambda I \) is singular are:
\[ \lambda = 1 + 2i \quad \text{and} \quad \lambda = 1 - 2i \]

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