Sure, let's solve the given system of linear equations using the given methods:
The system of equations is:
[tex]\[
3m + 2y = 1
\][/tex]
[tex]\[
4m - 5y = -14
\][/tex]
### a) Using Inverses
Step 1: Write the system in matrix form:
[tex]\[
A \mathbf{x} = \mathbf{b}
\][/tex]
where
[tex]\[
A = \begin{pmatrix}
3 & 2 \\
4 & -5 \\
\end{pmatrix}
\][/tex],
[tex]\[
\mathbf{x} = \begin{pmatrix}
m \\
y \\
\end{pmatrix}
\][/tex],
and
[tex]\[
\mathbf{b} = \begin{pmatrix}
1 \\
-14 \\
\end{pmatrix}.
\][/tex]
Step 2: Find the inverse [tex]\( A^{-1} \)[/tex] of matrix [tex]\( A \)[/tex]:
[tex]\[
A^{-1} = \begin{pmatrix}
0.2173913 & 0.08695652 \\
0.17391304 & -0.13043478 \\
\end{pmatrix}
\][/tex]
Step 3: Multiply the inverse of the matrix [tex]\( A \)[/tex] by the vector [tex]\( \mathbf{b} \)[/tex] to get [tex]\( \mathbf{x} \)[/tex]:
[tex]\[
\mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix}
-1 \\
2 \\
\end{pmatrix}
\][/tex]
So, the solution using the inverse method is:
[tex]\[
m = -1, \quad y = 2
\][/tex]
### b) Using Determinants (Cramer's Rule)
Step 1: Calculate the determinant of matrix [tex]\( A \)[/tex]:
[tex]\[
\text{det}(A) = \begin{vmatrix}
3 & 2 \\
4 & -5 \\
\end{vmatrix} = (3 \cdot (-5)) - (2 \cdot 4) = -15 - 8 = -23
\][/tex]
Step 2: Compute the determinant of matrix obtained by replacing the first column of [tex]\( A \)[/tex] with [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \text{det}(A_m) \)[/tex]:
[tex]\[
A_m = \begin{pmatrix}
1 & 2 \\
-14 & -5 \\
\end{pmatrix}
\][/tex]
[tex]\[
\text{det}(A_m) = \begin{vmatrix}
1 & 2 \\
-14 & -5 \\
\end{vmatrix} = (1 \cdot (-5)) - (2 \cdot (-14)) = -5 + 28 = 23
\][/tex]
Step 3: Compute the determinant of matrix obtained by replacing the second column of [tex]\( A \)[/tex] with [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \text{det}(A_y) \)[/tex]:
[tex]\[
A_y = \begin{pmatrix}
3 & 1 \\
4 & -14 \\
\end{pmatrix}
\][/tex]
[tex]\[
\text{det}(A_y) = \begin{vmatrix}
3 & 1 \\
4 & -14 \\
\end{vmatrix} = (3 \cdot (-14)) - (1 \cdot 4) = -42 - 4 = -46
\][/tex]
Step 4: Use Cramer's Rule to find [tex]\( m \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
m = \frac{\text{det}(A_m)}{\text{det}(A)} = \frac{23}{-23} = -1
\][/tex]
[tex]\[
y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-46}{-23} = 2
\][/tex]
So, the solution using determinants (Cramer's Rule) is:
[tex]\[
m = -1, \quad y = 2
\][/tex]
In both methods, we obtained the same solution:
[tex]\[
m = -1, \quad y = 2
\][/tex]
