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The determinant of a matrix is a special number that can be calculated for square matrices. It provides important information about the matrix, such as whether it is invertible. For a simple 2x2 matrix, calculating the determinant is relatively straightforward.

A 2x2 matrix has the following form:

• Let's consider a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \).

The determinant is calculated using the formula:

• \( \text{det}(A) = ad - bc \).

In our example exercise, the matrix \( A \) was \( \begin{pmatrix} 2 & -3 \ 4 & 1 \end{pmatrix} \).

So, the determinant is \( 2 \times 1 - (-3) \times 4 = 2 + 12 = 14 \).

Since the determinant is not zero (it's 14), we can conclude that the matrix is invertible. An invertible matrix is key to solving a system of linear equations using matrix methods.

Solving linear equations using matrices involves expressing a system of equations in matrix form. This approach can be efficient, especially with more complex systems.

In the original exercise, the system of equations was:

• \( 2x - 3y = 8 \)
• \( 4x + y = 2 \)

These can be expressed as a matrix equation \( A\mathbf{x} = \mathbf{b} \), where:

• \( A \) is the matrix of coefficients, \( \begin{pmatrix} 2 & -3 \ 4 & 1 \end{pmatrix} \).
• \( \mathbf{x} \) is the vector of variables, \( \begin{pmatrix} x \ y \end{pmatrix} \).
• \( \mathbf{b} \) is the constant vector, \( \begin{pmatrix} 8 \ 2 \end{pmatrix} \).

To solve for \( \mathbf{x} \), the inverse of the matrix \( A \) is needed, as this allows you to isolate \( \mathbf{x} \ \) by multiplying both sides by the inverse of \( A \), leading to:

• \( \mathbf{x} = A^{-1} \mathbf{b} \).

Using the inverse matrix method offers a systematic approach and can simplify computations for larger systems.

Matrix multiplication is a fundamental operation in linear algebra, crucial for solving systems of equations and other applications. It involves the multiplication of two matrices to form a new matrix.

When multiplying a matrix by a vector, the operation results in another vector. Here’s how it works:

• If you have a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \)
• and a vector \( \mathbf{b} = \begin{pmatrix} e \ f \end{pmatrix} \),

you can multiply them as follows:

• \( A\mathbf{b} = \begin{pmatrix} ae + bf \ ce + df \end{pmatrix} \).

In our solution, the inverse matrix \( A^{-1} = \frac{1}{14} \begin{pmatrix} 1 & 3 \ -4 & 2 \end{pmatrix} \) was multiplied by the constant vector \( \mathbf{b} = \begin{pmatrix} 8 \ 2 \end{pmatrix} \).

This matrix multiplication was performed as:

• \( \begin{pmatrix} 1 \cdot 8 + 3 \cdot 2 \ -4 \cdot 8 + 2 \cdot 2 \end{pmatrix} = \begin{pmatrix} 14 \ -28 + 4 \end{pmatrix} = \begin{pmatrix} 1 \ -\frac{12}{14} \end{pmatrix} \),

yielding the solution to the system. Understanding how matrices multiply provides a deeper insight into solving equations through matrices.