Evaluate the following :
(i) `[(4),(5)][7" "9]+[(4,0),(0,-5)]`
(ii) `["x y z"][(a,h,g),(h,b,f),(g,f,c)][(x),(y),(z)]`
(iii) `[(1,-1),(0,2),(2,3)]([(1,0,2),(2,0,1)]-[(0,1,2),(1,0,2)])`.

Answer

Let's evaluate the given expressions step by step.

### (i) Evaluate `[(4),(5)][(7, 9)] + [(4, 0), (0, -5)]`

1. **Matrix Multiplication**:
- We have a 2x1 matrix `[(4), (5)]` and a 1x2 matrix `[(7, 9)]`.
- The result will be a 2x2 matrix.
- Calculate the elements:
- First element: \(4 \times 7 = 28\)
- Second element: \(4 \times 9 = 36\)
- Third element: \(5 \times 7 = 35\)
- Fourth element: \(5 \times 9 = 45\)

So, the resulting matrix from multiplication is:
\[ \begin{bmatrix} 28 & 36 \\ 35 & 45 \end{bmatrix} \]

2. **Add the matrices**:
- Now, add the resulting matrix with `[(4, 0), (0, -5)]`:
\[ \begin{bmatrix} 28 & 36 \\ 35 & 45 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & -5 \end{bmatrix} = \begin{bmatrix} 28 + 4 & 36 + 0 \\ 35 + 0 & 45 - 5 \end{bmatrix} \]
- This results in:
\[ \begin{bmatrix} 32 & 36 \\ 35 & 40 \end{bmatrix} \]

### (ii) Evaluate `["x y z"][(a,h,g),(h,b,f),(g,f,c)][(x),(y),(z)]`

1. **Matrix Multiplication**:
- We have a 1x3 matrix `["x y z"]` and a 3x3 matrix `[(a,h,g),(h,b,f),(g,f,c)]`.
- The result will be a 1x3 matrix.
- Calculate the elements:
- First element: \(x \cdot a + y \cdot h + z \cdot g\)
- Second element: \(x \cdot h + y \cdot b + z \cdot f\)
- Third element: \(x \cdot g + y \cdot f + z \cdot c\)

So, the resulting matrix is:
\[ \begin{bmatrix} xa + yh + zg & xh + yb + zf & xg + yf + zc \end{bmatrix} \]

2. **Multiply with the column matrix**:
- Now, multiply the resulting 1x3 matrix with the column matrix `[(x),(y),(z)]`:
\[ \begin{bmatrix} xa + yh + zg & xh + yb + zf & xg + yf + zc \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]
- This results in:
\[ (xa + yh + zg)x + (xh + yb + zf)y + (xg + yf + zc)z \]

### (iii) Evaluate `[(1,-1),(0,2),(2,3)]([(1,0,2),(2,0,1)] - [(0,1,2),(1,0,2)])`

1. **Matrix Subtraction**:
- First, calculate the subtraction:
\[ \begin{bmatrix} (1-0) & (0-1) & (2-2) \\ (2-1) & (0-0) & (1-2) \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \end{bmatrix} \]

2. **Matrix Multiplication**:
- Now, multiply `[(1,-1),(0,2),(2,3)]` with the resulting matrix:
\[ \begin{bmatrix} 1 & -1 & 0 \\ 0 & 2 & 3 \end{bmatrix} \]
- The resulting matrix will be a 3x3 matrix:
- Calculate the elements:
- First row:
- \(1 \cdot 1 + (-1) \cdot 1 + 0 \cdot 1 = 0\)
- \(1 \cdot (-1) + (-1) \cdot 0 + 0 \cdot (-1) = -1\)
- \(1 \cdot 0 + (-1) \cdot (-1) + 0 \cdot (-1) = 1\)
- Second row:
- \(0 \cdot 1 + 2 \cdot 1 + 3 \cdot 1 = 5\)
- \(0 \cdot (-1) + 2 \cdot 0 + 3 \cdot (-1) = -3\)
- \(0 \cdot 0 + 2 \cdot (-1) + 3 \cdot (-1) = -5\)
- Third row:
- \(2 \cdot 1 + 3 \cdot 1 + 0 \cdot 1 = 5\)
- \(2 \cdot (-1) + 3 \cdot 0 + 0 \cdot (-1) = -2\)
- \(2 \cdot 0 + 3 \cdot (-1) + 0 \cdot (-1) = -3\)

So, the final result is:
\[ \begin{bmatrix} 0 & -1 & 1 \\ 5 & -3 & -5 \\ 5 & -2 & -3 \end{bmatrix} \]