Find matrix A such that A × $\begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix}$ = $\begin{bmatrix} 0 & -4 \ 10 & 3 \end{bmatrix}$.

Let B = $\begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix}$ Then, AB = $\begin{bmatrix} 0 & -4 \ 10 & 3 \end{bmatrix}$

Since AB exists, we have:

Number of columns of A = Number of rows in B = 2

Number of rows of A = Number of rows in AB = 2

Order of A is 2 × 2

Let A = $\begin{bmatrix} a & b \ c & d \end{bmatrix}$.

Then,

$\Rightarrow \begin{bmatrix} a & b \ c & d \end{bmatrix} \times \begin{bmatrix} 2 & 3 \ 4 & 5 \end{bmatrix} = \begin{bmatrix} 0 & -4 \ 10 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2a + 4b & 3a + 5b \ 2c + 4d & 3c + 5d \end{bmatrix} =\begin{bmatrix} 0 & -4 \ 10 & 3 \end{bmatrix}.$

Solving for a and b:

∴ 2a + 4b = 0

⇒ 2(a + 2b) = 0

⇒ a + 2b = 0

⇒ a = -2b ……(1)

∴ 3a + 5b = -4

Substituting value of a from equation(1) in 3a + 5b = -4, we get:

⇒ 3(-2b) + 5b = -4

⇒ -6b + 5b = -4

⇒ -b = -4

⇒ b = 4.

Substituting value of b in equation 1, we get,

⇒ a = -2(4)

⇒ a = -8.

Solving for c and d:

∴ 2c + 4d = 10

⇒ 2(c + 2d) = 10

⇒ c + 2d = 5

⇒ c = 5 - 2d …….(2)

∴ 3c + 5d = 3

Substituting value of c from equation (2) in 3c + 5d = 3, we get :

⇒ 3c + 5d = 3

⇒ 3(5 - 2d) + 5d = 3

⇒ 15 - 6d + 5d = 3

⇒ 15 - d = 3

⇒ d = 15 - 3

⇒ d = 12.

Substituting value of d in equation (2), we get,

⇒ c = 5 - 2(12)

⇒ c = 5 - 24

⇒ c = -19.

$\therefore A = \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} -8 & 4 \ -19 & 12 \end{bmatrix}$.

Hence, A = $\begin{bmatrix} -8 & 4 \ -19 & 12 \end{bmatrix}$.