Using elementary row transformations, find the inverse of the matrix A = [[1, 2, 3], [2, 5, 7], [6, 8, 9]]

Let A = [[1, 2, 3], [2, 5, 7], [6, 8, 9]]. We know that A = IA. Therefore,
[[1, 2, 3], [2, 5, 7], [6, 8, 9]] = [[1, 0, 0], [0, 1, 0], [0, 0, 1]]A
Applying R3 → R3 - 3R1, we have
[[1, 2, 3], [2, 5, 7], [0, 2, 0]] = [[1, 0, 0], [0, 1, 0], [-3, 0, 1]]A
Applying R3 → R3 - R2, we have
[[1, 2, 3], [2, 5, 7], [0, -3, -7]] = [[1, 0, 0], [0, 1, 0], [-3, -1, 1]]A
Applying R2 → R2 - 2R1,
[[1, 2, 3], [0, 1, 1], [0, -3, -7]] = [[1, 0, 0], [-2, 1, 0], [-3, -1, 1]]A
Applying R1 → R1 - 2R2,
[[1, 0, 1], [0, 1, 1], [0, -3, -7]] = [[5, -2, 0], [-2, 1, 0], [-3, -1, 1]]A
Applying R3 → R3 + 3R2,
[[1, 0, 1], [0, 1, 1], [0, 0, -4]] = [[5, -2, 0], [-2, 1, 0], [-9, 2, 1]]A
Applying R3 → -1/4 R3,
[[1, 0, 1], [0, 1, 1], [0, 0, 1]] = [[5, -2, 0], [-2, 1, 0], [9/4, -1/2, -1/4]]A
Applying R1 → R1 - R3,
[[1, 0, 0], [0, 1, 1], [0, 0, 1]] = [[11/4, -3/2, 1/4], [-2, 1, 0], [9/4, -1/2, -1/4]]A
Applying R2 → R2 - R3,
[[1, 0, 0], [0, 1, 0], [0, 0, 1]] = [[11/4, -3/2, 1/4], [-17/4, 3/2, 1/4], [9/4, -1/2, -1/4]]A
Therefore, A⁻¹ = [[11/4, -3/2, 1/4], [-17/4, 3/2, 1/4], [9/4, -1/2, -1/4]]