`A=[{:(1,0,k),(2, 1,3),(k,0,1):}]` is invertible for

Options

A. PQ
B. QAP
C. PAQ
D. `PA^(-1)Q`

Answer

To determine the values of \( k \) for which the matrix

\[ A = \begin{pmatrix} 1 & 0 & k \\ 2 & 1 & 3 \\ k & 0 & 1 \end{pmatrix} \]

is invertible, we need to find the determinant of \( A \) and set the condition that it should not equal zero.

### Step 1: Calculate the Determinant of Matrix A

The determinant of a \( 3 \times 3 \) matrix

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

is given by the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

- \( a = 1, b = 0, c = k \)
- \( d = 2, e = 1, f = 3 \)
- \( g = k, h = 0, i = 1 \)

Substituting these values into the determinant formula:

\[ \text{det}(A) = 1 \cdot (1 \cdot 1 - 3 \cdot 0) - 0 \cdot (2 \cdot 1 - 3 \cdot k) + k \cdot (2 \cdot 0 - 1 \cdot k) \]

This simplifies to:

\[ \text{det}(A) = 1 \cdot (1 - 0) + k \cdot (0 - k) \]

\[ \text{det}(A) = 1 - k^2 \]

### Step 2: Set the Determinant Not Equal to Zero

For the matrix \( A \) to be invertible, we need:

\[ 1 - k^2 \neq 0 \]

### Step 3: Solve the Inequality

Setting the equation to zero gives:

\[ 1 - k^2 = 0 \]

This can be factored as:

\[ (1 - k)(1 + k) = 0 \]

Thus, the solutions are:

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

### Step 4: Conclusion

The matrix \( A \) is invertible for all values of \( k \) except:

\[ k \neq 1 \quad \text{and} \quad k \neq -1 \]