If `Delta(r)=|{:(r,r+1),(r+3,r+4):}|` then expand it

Answer

To expand the determinant \(\Delta(r) = \begin{vmatrix} r & r+1 \\ r+3 & r+4 \end{vmatrix}\), we will follow these steps:

### Step 1: Write the determinant
We start with the determinant:
\[ \Delta(r) = \begin{vmatrix} r & r+1 \\ r+3 & r+4 \end{vmatrix} \]

### Step 2: Apply the determinant formula
The formula for the determinant of a \(2 \times 2\) matrix \(\begin{vmatrix} a & b \\ c & d \end{vmatrix}\) is given by \(ad - bc\). Here, we identify:
- \(a = r\)
- \(b = r + 1\)
- \(c = r + 3\)
- \(d = r + 4\)

Thus, we can write:
\[ \Delta(r) = r \cdot (r + 4) - (r + 1) \cdot (r + 3) \]

### Step 3: Expand the terms
Now we expand both terms:
1. For the first term:
\[ r \cdot (r + 4) = r^2 + 4r \]
2. For the second term:
\[ (r + 1) \cdot (r + 3) = r^2 + 3r + r + 3 = r^2 + 4r + 3 \]

### Step 4: Substitute back into the determinant expression
Now substituting back, we have:
\[ \Delta(r) = (r^2 + 4r) - (r^2 + 4r + 3) \]

### Step 5: Simplify the expression
Now, simplify the expression:
\[ \Delta(r) = r^2 + 4r - r^2 - 4r - 3 \]
Notice that \(r^2\) and \(4r\) cancel out:
\[ \Delta(r) = -3 \]

### Final Result
Thus, the expanded form of \(\Delta(r)\) is:
\[ \Delta(r) = -3 \]