We are given that,

\(\begin{bmatrix} a+b&2 \\[0.3em] 5 & b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&5 \\[0.3em] 2 & 2 \\[0.3em] \end{bmatrix}\)

We need to find the value of x. 

We know by the property of matrices,

 \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies, 

a₁₁ = b₁₁, 

a₁₂ = b₁₂, 

a₂₁ = b₂₁ and 

a₂₂ = b₂₂ 

So, if we have

 \(\begin{bmatrix} a+b&2 \\[0.3em] 5 & b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 6&5 \\[0.3em] 2 & 2 \\[0.3em] \end{bmatrix}\)

Corresponding elements of two elements are equal.

That is,

a + b = 6 …(i)

b = 4 …(ii) 

To solve for a, 

We have equations (i) and (ii).

We can’t solve for a using only equation (i) as equation (i) contains a as well as b. 

We need to find the value of b from equation (ii) first.

From equation (ii),

b = 4

Substituting the value of b = 4 in equation (i),

a + b = 6 

⇒ a + 4 = 6 

⇒ a = 6 – 4 

⇒ a = 2 

Thus, 

We get a = 2.