3. A fruit grower raises two crops, apples and peaches. The grower ships each of these crops to three different outlets.

In the matrix:

\[
[tex] A = \begin{bmatrix} 125 & 100 & 75 \\ 100 & 175 & 125 \end{bmatrix} [/tex]
\]

\[ [tex] a_{ij} [/tex] \] represents the number of units of crop $[tex] i [/tex]$ that the grower ships to outlet $[tex] j [/tex]$.

The matrix:

\[
[tex] B = \begin{bmatrix} \$3.5 & \$6.00 \end{bmatrix} [/tex]
\]

represents the profit per unit. Find the product $[tex] BA [/tex]$ and state what each entry of the matrix represents.

---

4. A corporation has three factories, each of which manufactures acoustic guitars and electric guitars.

In the matrix:

\[
[tex] A = \begin{bmatrix} 70 & 50 & 25 \\ 35 & 100 & 70 \end{bmatrix} [/tex]
\]

\[ [tex] a_{ij} [/tex] \] represents the number of guitars of type $[tex] i [/tex]$ produced at factory $[tex] j [/tex]$ in one day. Find the production levels when production increases by $[tex] 20\% [/tex]$.

---

5. Find the value of $[tex] x [/tex]$ for which the matrix is equal to its own inverse:

(a) \[
[tex] \begin{bmatrix} 3 & x \\ -2 & -3 \end{bmatrix} [/tex]
\]

(b) \[
[tex] \begin{bmatrix} 2 & x \\ -1 & -2 \end{bmatrix} [/tex]
\]

(c) \[
[tex] \begin{bmatrix} x & 2 \\ -3 & 4 \end{bmatrix} [/tex]
\]

---

6. If

\[
[tex] A = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} [/tex]
\]

then:

i. Show that $[tex] A = A^{-1} [/tex]$.

ii. Show that $[tex] A^n = \begin{bmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{bmatrix} [/tex]$.

---

7. If

\[
[tex] A = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} [/tex]
\]

and

\[
[tex] B = \begin{bmatrix} \cos(\phi) & \sin(\phi) \\ -\sin(\phi) & \cos(\phi) \end{bmatrix} [/tex]
\]

then show that

\[
[tex] AB = \begin{bmatrix} \cos(\theta+\phi) & \sin(\theta+\phi) \\ -\sin(\theta+\phi) & \cos(\theta+\phi) \end{bmatrix} [/tex]
\]

---

8. Determine the values of $[tex] \alpha [/tex]$ for which the matrix

\[
[tex] A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 2 & \alpha \end{bmatrix} [/tex]
\]

is invertible and find $[tex] A^{-1} [/tex]$.

---

9. Show that if $[tex] A [/tex]$ is invertible, then so is $[tex] A^m [/tex]$ for every positive integer $[tex] m [/tex]$; moreover, $[tex] (A^m)^{-1} = (A^{-1})^m [/tex]$.

---

10. If $[tex] A [/tex]$ and $[tex] B [/tex]$ are $[tex] n \times n [/tex]$ matrices with $[tex] A [/tex]$ invertible, then show that

\[
[tex] (A+B)A^{-1}(A-B) = (A-B)A^{-1}(A+B) [/tex]
\]

---

11. Solve the following systems of linear equations using Gaussian elimination:

(a)

\[
[tex] \begin{aligned}
& x_1 - x_2 + 2x_3 = 4 \\
& x_1 + x_3 = 6 \\
& 2x_1 - 3x_2 + 5x_3 = 4 \\
& 3x_1 + 2x_2 - x_3 = 1 \\
\end{aligned} [/tex]
\]

(b)

\[
[tex] \begin{aligned}
& x_1 - 2x_2 + 3x_3 = 9 \\
& -x_1 + 3x_2 = -4 \\
& 2x_1 - 5x_2 + 5x_3 = 17 \\
& 2x_1 + x_2 - x_3 + 2x_4 = -6 \\
& 3x_1 + 4x_2 + x_4 = 12 \\
& x_1 + 5x_2 + 2x_3 + 6x_4 = -3 \\
& 5x_1 + 2x_2 - x_3 - x_4 = 1 \\
\end{aligned} [/tex]
\]

---

12. Use Cramer's rule (if possible) to solve the following linear systems:

(a)

\[
[tex] \begin{aligned}
& x_1 + 2x_2 = 5 \\
& -x_1 + x_2 = 1 \\
\end{aligned} [/tex]
\]

To solve the linear system using Cramer's Rule, follow these steps:

We have the system of equations:

1. [tex]$ x_1 + 2x_2 = 5 $[/tex]
2. [tex]$-x_1 + x_2 = 1 $[/tex]

Step 1: Formulate the coefficient matrix and the determinant.

The coefficient matrix [tex]$ A $[/tex] is:

[tex]\[
A = \begin{bmatrix}
1 & 2 \\
-1 & 1
\end{bmatrix}
\][/tex]

The determinant of matrix [tex]$ A $[/tex] (denoted as [tex]$ \det(A) $[/tex]) is calculated as follows:

[tex]\[
\det(A) = (1)(1) - (-1)(2) = 1 + 2 = 3
\][/tex]

Since the determinant is non-zero ([tex]$ \det(A) = 3 $[/tex]), Cramer's Rule can be applied, indicating the system has a unique solution.

Step 2: Calculate the determinants for [tex]$ x_1 $[/tex] and [tex]$ x_2 $[/tex].

To find [tex]$ x_1 $[/tex], replace the first column of the coefficient matrix with the constants from the right side of the equations:

[tex]\[
A_{x_1} = \begin{bmatrix}
5 & 2 \\
1 & 1
\end{bmatrix}
\][/tex]

The determinant [tex]$ \det(A_{x_1}) $[/tex] is:

[tex]\[
\det(A_{x_1}) = (5)(1) - (1)(2) = 5 - 2 = 3
\][/tex]

To find [tex]$ x_2 $[/tex], replace the second column of the coefficient matrix with the constants:

[tex]\[
A_{x_2} = \begin{bmatrix}
1 & 5 \\
-1 & 1
\end{bmatrix}
\][/tex]

The determinant [tex]$ \det(A_{x_2}) $[/tex] is:

[tex]\[
\det(A_{x_2}) = (1)(1) - (-1)(5) = 1 + 5 = 6
\][/tex]

Step 3: Solve for [tex]$ x_1 $[/tex] and [tex]$ x_2 $[/tex] using Cramer's Rule.

According to Cramer's Rule:

[tex]\[
x_1 = \frac{\det(A_{x_1})}{\det(A)} = \frac{3}{3} = 1.0
\][/tex]

[tex]\[
x_2 = \frac{\det(A_{x_2})}{\det(A)} = \frac{6}{3} = 2.0
\][/tex]

Thus, the solution to the system of equations is:
[tex]$ x_1 = 1.0 $[/tex] and [tex]$ x_2 = 2.0 $[/tex].