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find all values of x for which A is invertible



To find the values of \( x \) for which the matrix \( A \) is invertible, we need to determine when the determinant of \( A \) is non-zero. 

The matrix \( A \) is given as:

\[
A = \begin{bmatrix}
x - \frac{1}{2} & 0 & 0 \\
x & x - \frac{1}{3} & 0 \\
x^2 & x^3 & x + \frac{1}{4}
\end{bmatrix}
\]

### Step 1: Compute the determinant of \( A \)

The determinant of a 3x3 matrix:

\[
\text{det}(A) = \begin{vmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{vmatrix}
\]

can be calculated using cofactor expansion along the first row:

\[
\text{det}(A) = a_{11} \cdot \text{det}(A_{11}) - a_{12} \cdot \text{det}(A_{12}) + a_{13} \cdot \text{det}(A_{13})
\]

where \( A_{ij} \) are the minors (2x2 determinants) of matrix \( A \).

For the matrix \( A \):

\[
\text{det}(A) = (x - \frac{1}{2}) \cdot \text{det}\begin{bmatrix} x - \frac{1}{3} & 0 \\ x^3 & x + \frac{1}{4} \end{bmatrix}
\]

### Step 2: Compute the 2x2 determinant

The determinant of the 2x2 matrix is:

\[
\text{det}\begin{bmatrix} x - \frac{1}{3} & 0 \\ x^3 & x + \frac{1}{4} \end{bmatrix} = (x - \frac{1}{3})(x + \frac{1}{4}) - 0 \cdot x^3
\]
\[
= (x - \frac{1}{3})(x + \frac{1}{4}) = x^2 + \frac{1}{4}x - \frac{1}{3}x - \frac{1}{12}
\]
\[
= x^2 + \frac{1}{12}x - \frac{1}{12}
\]

### Step 3: Final expression for determinant

Now substitute this result back into the determinant formula for \( A \):

\[
\text{det}(A) = (x - \frac{1}{2}) \cdot \left(x^2 + \frac{1}{12}x - \frac{1}{12}\right)
\]

### Step 4: Set the determinant to zero and solve

To find when \( A \) is invertible, we set:

\[
(x - \frac{1}{2}) \cdot \left(x^2 + \frac{1}{12}x - \frac{1}{12}\right) = 0
\]

This gives two cases:

1. \( x - \frac{1}{2} = 0 \Rightarrow x = \frac{1}{2} \)
2. \( x^2 + \frac{1}{12}x - \frac{1}{12} = 0 \)

To solve the second equation, apply the quadratic formula:

\[
x = \frac{-\frac{1}{12} \pm \sqrt{\left(\frac{1}{12}\right)^2 - 4(1)(-\frac{1}{12})}}{2(1)}
\]
\[
x = \frac{-\frac{1}{12} \pm \sqrt{\frac{1}{144} + \frac{4}{48}}}{2}
\]
\[
x = \frac{-\frac{1}{12} \pm \sqrt{\frac{1 + 12}{144}}}{2}
\]
\[
x = \frac{-\frac{1}{12} \pm \sqrt{\frac{13}{144}}}{2}
\]
\[
x = \frac{-\frac{1}{12} \pm \frac{\sqrt{13}}{12}}{2}
\]
\[
x = \frac{-1 \pm \sqrt{13}}{24}
\]

Thus, the roots of the quadratic equation are:

\[
x = \frac{-1 + \sqrt{13}}{24}, \quad x = \frac{-1 - \sqrt{13}}{24}
\]

### Final Answer:

The matrix \( A \) is invertible for all values of \( x \) except:
1. \( x = \frac{1}{2} \)
2. \( x = \frac{-1 + \sqrt{13}}{24} \)
3. \( x = \frac{-1 - \sqrt{13}}{24} \)

Would you like more details, or have any questions?

Here are five related questions you might consider:
1. How do you calculate the inverse of a 3x3 matrix?
2. What are the properties of a matrix when it is not invertible?
3. How does the determinant of a matrix relate to its eigenvalues?
4. What are some efficient methods for computing determinants of larger matrices?
5. How can we apply Cramer's rule to solve systems of equations using matrices?

**Tip:** For any matrix, if the determinant is zero, the matrix is singular (non-invertible). Always check the determinant first when dealing with matrix inversion!

Find all values of x for which A is invertible.

This problem involves finding the values of x for which matrix A is invertible. It covers key concepts such as the determinant of a 3x3 matrix, cofactor expansion, and solving a quadratic equation. The solution walks through calculating the determinant and solving for x.

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