We are given a 4x4 matrix $A$ and asked to find its determinant $|A|$ and the (3,4) entry of its inverse matrix $A^{-1}$. The matrix $A$ is given by $A = \begin{pmatrix} 3 & -1 & 2 & 1 \\ -2 & 4 & 1 & -1 \\ 6 & -5 & -2 & 2 \\ -3 & 7 & -2 & -5 \end{pmatrix}$.

1. Problem Description

We are given a 4x4 matrix $A$ and asked to find its determinant $|A|$ and the (3,4) entry of its inverse matrix $A^{-1}$.

The matrix $A$ is given by

$A = \begin{pmatrix} 3 & -1 & 2 & 1 \\ -2 & 4 & 1 & -1 \\ 6 & -5 & -2 & 2 \\ -3 & 7 & -2 & -5 \end{pmatrix}$.

2. Solution Steps

(1) To find the determinant of $A$, we can use cofactor expansion. Let's expand along the first row:

$|A| = 3C_{11} + (-1)C_{12} + 2C_{13} + 1C_{14}$

where $C_{ij} = (-1)^{i+j}M_{ij}$, and $M_{ij}$ is the determinant of the 3x3 matrix formed by removing the $i$-th row and $j$-th column.

$M_{11} = \begin{vmatrix} 4 & 1 & -1 \\ -5 & -2 & 2 \\ 7 & -2 & -5 \end{vmatrix} = 4(10+4) - 1(25+14) + (-1)(10+14) = 56 - 39 - 24 = -7$

$M_{12} = \begin{vmatrix} -2 & 1 & -1 \\ 6 & -2 & 2 \\ -3 & -2 & -5 \end{vmatrix} = -2(10+4) - 1(-30+6) + (-1)(-12-6) = -28 + 24 + 18 = 14$

$M_{13} = \begin{vmatrix} -2 & 4 & -1 \\ 6 & -5 & 2 \\ -3 & 7 & -5 \end{vmatrix} = -2(25-14) - 4(-30+6) + (-1)(42-15) = -22 + 96 - 27 = 47$

$M_{14} = \begin{vmatrix} -2 & 4 & 1 \\ 6 & -5 & -2 \\ -3 & 7 & -2 \end{vmatrix} = -2(10+14) - 4(-12-6) + 1(42-15) = -48 + 72 + 27 = 51$

$C_{11} = (-1)^{1+1}M_{11} = -7$

$C_{12} = (-1)^{1+2}M_{12} = -14$

$C_{13} = (-1)^{1+3}M_{13} = 47$

$C_{14} = (-1)^{1+4}M_{14} = -51$

$|A| = 3(-7) - 1(-14) + 2(47) + 1(-51) = -21 + 14 + 94 - 51 = 36$

(2) To find the (3,4) entry of $A^{-1}$, we use the formula $A^{-1} = \frac{1}{|A|}adj(A)$, where $adj(A)$ is the adjugate of $A$. The (3,4) entry of $A^{-1}$ is given by $\frac{1}{|A|}C_{43}$, where $C_{43}$ is the (4,3) cofactor of $A$.

$C_{43} = (-1)^{4+3} M_{43} = (-1)^7 M_{43} = -M_{43}$

$M_{43} = \begin{vmatrix} 3 & -1 & 1 \\ -2 & 4 & -1 \\ 6 & -5 & 2 \end{vmatrix} = 3(8-5) - (-1)(-4+6) + 1(10-24) = 3(3) + 1(2) + 1(-14) = 9 + 2 - 14 = -3$

$C_{43} = -(-3) = 3$

$(A^{-1})_{34} = \frac{1}{|A|}C_{43} = \frac{1}{36}(3) = \frac{1}{12}$

3. Final Answer

(1) $|A| = 36$

(2) $(A^{-1})_{34} = \frac{1}{12}$