Are the following matrices invertible ?
(i) `|{:(2,-3),(1,4):}|`
(ii) `|{:(7,0),(3,1):}|`
(iii) `|{:(1,-2,-3),(1,-3,-4),(1,-4,-5):}|`
(iv) `|{:(2,3,-1),(0,1,4),(-5,0,-2):}|`
(v) `|{:(0,1,2),(1,2,3),(3,1,1):}|`

Answer

To determine whether the given matrices are invertible, we need to calculate the determinant of each matrix. A matrix is invertible if its determinant is non-zero.

### Step-by-Step Solution:

**(i)** For the matrix \( A_1 = \begin{pmatrix} 2 & -3 \\ 1 & 4 \end{pmatrix} \):

1. Calculate the determinant:
\[ \text{det}(A_1) = (2)(4) - (-3)(1) = 8 + 3 = 11 \]
2. Since \( \text{det}(A_1) = 11 \neq 0 \), the matrix is **invertible**.

**(ii)** For the matrix \( A_2 = \begin{pmatrix} 7 & 0 \\ 3 & 1 \end{pmatrix} \):

1. Calculate the determinant:
\[ \text{det}(A_2) = (7)(1) - (0)(3) = 7 - 0 = 7 \]
2. Since \( \text{det}(A_2) = 7 \neq 0 \), the matrix is **invertible**.

**(iii)** For the matrix \( A_3 = \begin{pmatrix} 1 & -2 & -3 \\ 1 & -3 & -4 \\ 1 & -4 & -5 \end{pmatrix} \):

1. Calculate the determinant using cofactor expansion:
\[ \text{det}(A_3) = 1 \cdot \text{det}\begin{pmatrix} -3 & -4 \\ -4 & -5 \end{pmatrix} - (-2) \cdot \text{det}\begin{pmatrix} 1 & -4 \\ 1 & -5 \end{pmatrix} + (-3) \cdot \text{det}\begin{pmatrix} 1 & -3 \\ 1 & -4 \end{pmatrix} \]
2. Calculate the 2x2 determinants:
\[ \text{det}\begin{pmatrix} -3 & -4 \\ -4 & -5 \end{pmatrix} = (-3)(-5) - (-4)(-4) = 15 - 16 = -1 \]
\[ \text{det}\begin{pmatrix} 1 & -4 \\ 1 & -5 \end{pmatrix} = (1)(-5) - (-4)(1) = -5 + 4 = -1 \]
\[ \text{det}\begin{pmatrix} 1 & -3 \\ 1 & -4 \end{pmatrix} = (1)(-4) - (-3)(1) = -4 + 3 = -1 \]
3. Substitute back:
\[ \text{det}(A_3) = 1(-1) + 2(-1) - 3(-1) = -1 - 2 + 3 = 0 \]
4. Since \( \text{det}(A_3) = 0 \), the matrix is **not invertible**.

**(iv)** For the matrix \( A_4 = \begin{pmatrix} 2 & 3 & -1 \\ 0 & 1 & 4 \\ -5 & 0 & -2 \end{pmatrix} \):

1. Calculate the determinant using cofactor expansion:
\[ \text{det}(A_4) = 2 \cdot \text{det}\begin{pmatrix} 1 & 4 \\ 0 & -2 \end{pmatrix} - 3 \cdot \text{det}\begin{pmatrix} 0 & 4 \\ -5 & -2 \end{pmatrix} + (-1) \cdot \text{det}\begin{pmatrix} 0 & 1 \\ -5 & 0 \end{pmatrix} \]
2. Calculate the 2x2 determinants:
\[ \text{det}\begin{pmatrix} 1 & 4 \\ 0 & -2 \end{pmatrix} = (1)(-2) - (4)(0) = -2 \]
\[ \text{det}\begin{pmatrix} 0 & 4 \\ -5 & -2 \end{pmatrix} = (0)(-2) - (4)(-5) = 20 \]
\[ \text{det}\begin{pmatrix} 0 & 1 \\ -5 & 0 \end{pmatrix} = (0)(0) - (1)(-5) = 5 \]
3. Substitute back:
\[ \text{det}(A_4) = 2(-2) - 3(20) - 5 = -4 - 60 - 5 = -69 \]
4. Since \( \text{det}(A_4) = -69 \neq 0 \), the matrix is **invertible**.

**(v)** For the matrix \( A_5 = \begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix} \):

1. Calculate the determinant using cofactor expansion:
\[ \text{det}(A_5) = 0 \cdot \text{det}\begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} - 1 \cdot \text{det}\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix} + 2 \cdot \text{det}\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} \]
2. Calculate the 2x2 determinants:
\[ \text{det}\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix} = (1)(1) - (3)(3) = 1 - 9 = -8 \]
\[ \text{det}\begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} = (1)(1) - (2)(3) = 1 - 6 = -5 \]
3. Substitute back:
\[ \text{det}(A_5) = 0 - (-1)(-8) + 2(-5) = 0 - 8 - 10 = -18 \]
4. Since \( \text{det}(A_5) = -18 \neq 0 \), the matrix is **invertible**.

### Summary of Results:
- (i) Invertible
- (ii) Invertible
- (iii) Not Invertible
- (iv) Invertible
- (v) Invertible