Find the values of x, y, a and b if $\begin{bmatrix$}[r] x - 2 & y \ a + 2b & 3a - b \end{bmatrix} = \begin{bmatrix}[r] 3 & 1 \ 5 & 1 \end{bmatrix}.

Given, $\begin{bmatrix$}[r] x - 2 & y \ a + 2b & 3a - b \end{bmatrix} = \begin{bmatrix}[r] 3 & 1 \ 5 & 1 \end{bmatrix}.

By definition of equality of matrices, we get

x - 2 = 3 or x = 5,
y = 1,
a + 2b = 5       […Eq 1],
3a - b = 1        […Eq 2].

From Eq 1 we get,

⇒ a + 2b = 5
⇒ a = 5 - 2b

Putting this value of a in Eq 2,

⇒ 3(5 - 2b) - b = 1
⇒ 15 - 6b - b = 1
⇒ 7b = 14
⇒ b = 2.

Using value of b to find value of a,

⇒ a = 5 - 2b
⇒ a = 5 - 2(2)
⇒ a = 5 - 4
⇒ a = 1.

∴ x = 5, y = 1, a = 1 and b = 2.

Hence, the values are: x = 5, y = 1, a = 1 and b = 2.