\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)A = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) 

We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix.

So,

\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) is 2 × 2 matrix, and 
\( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) is 3 × 2 matrix

Now,

In order to get a 3 × 2 matrix as solution 2 × 2 matrix should be multiplied by 2 × 3 matrix. Hence matrix A is 2 × 3 matrix.

Let,

A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) 

So the given question becomes,

\( \begin{bmatrix}1 & 1 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) 

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

⇒ \( \begin{bmatrix}a+d &b+e &c+f \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}3 &3 & 5 \\[0.3em]1& 0 &1 \\[0.3em]\end{bmatrix}\) 

To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e.,

d = 1, e = 0, f = 1

a + d = 3

⇒ a + 1 = 3

⇒ a = 2

b + e = 3 

⇒ b + 0 = 3

⇒ b = 3

c + f = 5 

⇒ c + 1 = 5 

⇒ c = 4

Now,

Substituting these values in matrix A, we get

A = \( \begin{bmatrix}a &b &c \\[0.3em]d& e &f \\[0.3em]\end{bmatrix}\) 

=  \( \begin{bmatrix}2 &3 &4 \\[0.3em]1& 0 &1\\[0.3em]\end{bmatrix}\) is the matrix A.