Question

for what values of c will a = 1 1 c c4 be invertible? (enter your answers as a comma-separated list.)

Answer 1

The matrix A is invertible for all values of c except for c = 0.

Step-by-step explanation:

Determinant: The determinant of a 2x2 matrix can be calculated as:

| a b |

| c d | = ad - bc

For matrix A, the determinant is:

| 1 1 |

| c c4 | = 1 * c4 - 1 * c = c(c^3 - 1)

Invertibility: A matrix is invertible if its determinant is non-zero. Therefore, for A to be invertible, c(c^3 - 1) must not be equal to 0.

Solving for exceptions:

c = 0: If c = 0, the determinant becomes 0 * (0^3 - 1) = 0, making A non-invertible.

c ≠ 0: For c ≠ 0, the determinant becomes c(c^3 - 1) which can be factored as c(c - 1)(c^2 + c + 1). The quadratic factor (c^2 + c + 1) has no real roots, so it is always positive. Therefore, as long as c ≠ 0, the determinant is non-zero, and A is invertible.

Therefore, the only value of c for which A is not invertible is c = 0. The matrix is invertible for all other values of c.

Answer 2

The values of
`\(c\)` for which
`\(A\)` is invertible are all real numbers except
`\(c = 0\)` and
`\(c = 1\)`. So, the answer would be
`\(c \\eq 0, 1\)`.

For a matrix to be invertible, its determinant must be non-zero. The determinant of a 2x2 matrix
`\(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\)` is
`\(ad - bc\)`.

So, for the matrix
`\(A = \begin{bmatrix} 1 & 1 \\ c & c^8 \end{bmatrix}\)` to be invertible, its determinant must not be zero:

`\[\text{det}(A) = (1 * c^8) - (1 * c) = c^8 - c \\eq 0\]`

Factoring out
`\(c\)` gives
`\(c(c^7 - 1) \\eq 0\)`. This is true when
`\(c \\eq 0\)` and
`\(c^7 \\eq 1\)`.

However,
`\(c^7 = 1\) implies \(c = 1\) (since for any positive integer \(n\), \(a^n = 1\) implies \(a = 1\))`.

Therefore, The answer is
`\(c \\eq 0, 1\)`.

The complete question is here: