Given P = a & b c & d , M = 3 & -2 -4 & 3 , and N = 5 & 4 -2 & 6 .

Given $P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $M = \begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix}$, and $N = \begin{pmatrix} 5 & 4 \\ -2 & 6 \end{pmatrix}$.
The equation is $PM = N$.

Equating the corresponding elements gives two systems of equations:
For $a$ and $b$:

1. $3a - 4b = 5$
2. $-2a + 3b = 4$

For $c$ and $d$: 3) $3c - 4d = -2$ 4) $-2c + 3d = 6$

Multiply equation (1) by 2 and equation (2) by 3:
$2(3a - 4b) = 2(5) \implies 6a - 8b = 10$
$3(-2a + 3b) = 3(4) \implies -6a + 9b = 12$
Add the two new equations:
$(6a - 8b) + (-6a + 9b) = 10 + 12$
$b = 22$
Substitute $b = 22$ into equation (1):
$3a - 4(22) = 5$
$3a - 88 = 5$
$3a = 5 + 88$
$3a = 93$
$a = \frac{93}{3}$
$a = 31$

Multiply equation (3) by 2 and equation (4) by 3:
$2(3c - 4d) = 2(-2) \implies 6c - 8d = -4$
$3(-2c + 3d) = 3(6) \implies -6c + 9d = 18$
Add the two new equations:
$(6c - 8d) + (-6c + 9d) = -4 + 18$
$d = 14$
Substitute $d = 14$ into equation (3):
$3c - 4(14) = -2$
$3c - 56 = -2$
$3c = -2 + 56$
$3c = 54$
$c = \frac{54}{3}$
$c = 18$

The matrix $P$ is: