Use the elementary row operations to prove that if [a d;b c] is invertible then its inverse is 1/ac-bd [c -d;-b a] find the conditions on a,b,c,d such that is true?

To prove that the inverse of the matrix [a d; b c] is 1/(ad - bc) [c -d; -b a], we need to show that the product of the two matrices is equal to the identity matrix.

Let A = [a d; b c] and B = 1/(ad - bc) [c -d; -b a]. Then, AB = [a d; b c] * 1/(ad - bc) [c -d; -b a] = 1/(ad - bc) [ac - bd; bc - bd] = 1/(ad - bc) [1 0; 0 1] = I, where I is the identity matrix.

Therefore, the inverse of [a d; b c] is indeed 1/(ad - bc) [c -d; -b a].

To find the conditions on a, b, c, and d such that this is true, we need to ensure that the determinant of the matrix [a d; b c] is non-zero. The determinant of a 2x2 matrix [a b; c d] is ad - bc. Therefore, for the inverse to exist, we must have ad - bc ≠ 0.

In other words, the condition for the matrix [a d; b c] to be invertible and for its inverse to be 1/(ad - bc) [c -d; -b a] is that ad - bc ≠ 0.

Elementary Row Operations:

- Swap two rows
- Multiply a row by a nonzero constant
- Add a multiple of one row to another row

Proof:

Step 1: Swap rows

Swap row 1 and row 2 to get:

```
[b c;a d]
```

Step 2: Multiply row 1 by 1/b

Multiply row 1 by 1/b to get:

```
[1 c/b;a d]
```

Step 3: Add -a times row 1 to row 2

Add -a times row 1 to row 2 to get:

```
[1 c/b;0 d-ac/b]
```

Step 4: Multiply row 2 by 1/(d-ac/b)

Multiply row 2 by 1/(d-ac/b) to get:

```
[1 c/b;0 1]
```

Step 5: Add -c/b times row 1 to row 2

Add -c/b times row 1 to row 2 to get:

```
[1 0;0 1]
```

Therefore, the inverse of [a d;b c] is:

```
1/ac-bd [c -d;-b a]
```

Conditions for Invertibility:

The matrix [a d;b c] is invertible if and only if its determinant is nonzero. The determinant is given by:

```
det([a d;b c]) = ac - bd
```

Therefore, the matrix is invertible if and only if:

```
ac - bd ≠ 0
```

Which is equivalent to:

```
a ≠ 0 and c ≠ 0
```