Let A be a matrix such that A·[1 2; 0 3] is a scalar matrix and |3A| = 108. Then A² equals :

A is a matrix such that A·[1 2; 0 3] is a scalar matrix and |3A| = 108
Let the scalar matrix be [k 0; 0 k] ⇒ A·[1 2; 0 3] = [k 0; 0 k] ⇒ A = [k 0; 0 k][1 2; 0 3]⁻¹. [∵ AB = C ⇒ A = CB⁻¹]
Let B = [1 2; 0 3]
Now, |B| = 3
Then, B⁻¹ = (1/|B|) Co-factor matrix of B ⇒ A = (1/3)[k 0; 0 k][3 -2; 0 1] ⇒ A = [k 0; 0 k][1 -2/3; 0 1/3] ⇒ A = [k -2k/3; 0 k/3] (i)
|3A| = 108 [Given] ⇒ 108 = |3A| = 3ⁿ|A| = |3k -2k; 0 k| ⇒ 3k² = 108 ⇒ k² = 36 ⇒ k = ±6
Take k = 6 ⇒ A = [6 -4; 0 2] .. From (i) ⇒ A² = [6 -4; 0 2][6 -4; 0 2] ⇒ A² = [36 -24; 0 4]
For k = -6 ⇒ A = [-6 4; 0 -2] .. From (i) ⇒ A² = [-6 4; 0 -2][-6 4; 0 -2] ⇒ A² = [36 -24; 0 4]
Hence, A² = [36 0; 0 4]