Determinants and Cramer's Rule: Videos & Practice Problems

In the study of matrices, one important concept is the determinant, which is a scalar value that can be calculated from a square matrix. Understanding how to compute the determinant is essential, as it plays a crucial role in solving systems of equations and analyzing matrix properties.

For a 2x2 matrix, the determinant can be calculated using a straightforward formula. Given a matrix represented as:

\[\begin{pmatrix}a & b \\c & d\end{pmatrix}\]

the determinant is calculated using the formula:

\[\text{det}(A) = ad - bc\]

Here, \(a\) and \(d\) are the elements on the main diagonal (from the top left to the bottom right), while \(b\) and \(c\) are the elements on the other diagonal (from the top right to the bottom left). The process involves multiplying the elements of each diagonal and then subtracting the product of the second diagonal from the product of the first.

For example, consider the matrix:

\[\begin{pmatrix}3 & 2 \\5 & 4\end{pmatrix}\]

To find the determinant, we calculate:

\[\text{det}(A) = (3 \times 4) - (2 \times 5) = 12 - 10 = 2\]

This shows that the determinant of this matrix is 2.

Another example involves the matrix:

\[\begin{pmatrix}8 & 4 \\5 & 0\end{pmatrix}\]

Applying the determinant formula gives:

\[\text{det}(A) = (8 \times 0) - (4 \times 5) = 0 - 20 = -20\]

Thus, the determinant of this matrix is -20.

When working with matrices that include negative numbers, such as:

\[\begin{pmatrix}-3 & -7 \\-2 & 1\end{pmatrix}\]

the calculation proceeds similarly:

\[\text{det}(B) = (-3 \times 1) - (-7 \times -2) = -3 - 14 = -17\]

In this case, the determinant is -17. Understanding these calculations is vital for further applications in linear algebra, including finding inverses and solving linear systems.

Cramer's Rule is a powerful method for solving a system of two linear equations with two unknowns. This technique utilizes determinants of matrices to find the values of the unknowns directly. To illustrate this, consider the system of equations:

1. \(2x + y = 5\)

2. \(-4x + 6y = -2\)

To apply Cramer's Rule, we first represent the coefficients and constants in an augmented matrix format:

\[\begin{bmatrix}2 & 1 & | & 5 \\-4 & 6 & | & -2\end{bmatrix}\]

In this matrix, the first column contains the coefficients of \(x\), the second column contains the coefficients of \(y\), and the third column (after the bar) contains the constants from the right side of the equations.

Cramer's Rule states that to find the value of \(x\) and \(y\), we need to calculate the determinants of specific matrices. The determinant for \(x\) is calculated by replacing the \(x\) coefficients with the constants from the right side of the equations:

\[D_x = \begin{vmatrix}5 & 1 \\-2 & 6\end{vmatrix} = (5)(6) - (1)(-2) = 30 + 2 = 32\]

The determinant for the denominator, which is the determinant of the original coefficients, is:

\[D = \begin{vmatrix}2 & 1 \\-4 & 6\end{vmatrix} = (2)(6) - (1)(-4) = 12 + 4 = 16\]

Thus, the value of \(x\) can be found using the formula:

\[x = \frac{D_x}{D} = \frac{32}{16} = 2\]

Next, we find \(y\) by replacing the \(y\) coefficients with the constants:

\[D_y = \begin{vmatrix}2 & 5 \\-4 & -2\end{vmatrix} = (2)(-2) - (5)(-4) = -4 + 20 = 16\]

Since the denominator remains the same, we can calculate \(y\) as follows:

\[y = \frac{D_y}{D} = \frac{16}{16} = 1\]

In conclusion, the solution to the system of equations is \(x = 2\) and \(y = 1\). This method not only provides a systematic approach to solving linear equations but also reinforces the concept of determinants in linear algebra.

To calculate the determinant of a 3 by 3 matrix, we build upon the method used for 2 by 2 matrices, which involves multiplying the diagonals and subtracting. However, the process for 3 by 3 matrices is slightly more complex, as it requires calculating several 2 by 2 determinants.

For a 3 by 3 matrix represented as:

\[\begin{bmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{bmatrix}\]

the determinant can be calculated using the formula:

\[\text{det}(A) = a_{11} \cdot \text{det}\begin{bmatrix}a_{22} & a_{23} \\a_{32} & a_{33}\end{bmatrix} - a_{12} \cdot \text{det}\begin{bmatrix}a_{21} & a_{23} \\a_{31} & a_{33}\end{bmatrix} + a_{13} \cdot \text{det}\begin{bmatrix}a_{21} & a_{22} \\a_{31} & a_{32}\end{bmatrix}\]

In this formula, the signs alternate: the first term is positive, the second is negative, and the third is positive. This pattern is crucial for correctly calculating the determinant.

To illustrate this, consider the matrix:

\[\begin{bmatrix}3 & 1 & 0 \\2 & -3 & -1 \\4 & 6 & 0\end{bmatrix}\]

We start by calculating the determinant using the first row:

1. For the first term, take \(3\) and calculate the determinant of the 2 by 2 matrix formed by eliminating the first row and first column:

\[\text{det}\begin{bmatrix}-3 & -1 \\6 & 0\end{bmatrix} = (-3) \cdot 0 - (-1) \cdot 6 = 6\]

2. For the second term, take \(1\) and calculate the determinant of the 2 by 2 matrix formed by eliminating the first row and second column:

\[\text{det}\begin{bmatrix}2 & -1 \\4 & 0\end{bmatrix} = (2) \cdot 0 - (-1) \cdot 4 = 4\]

3. For the third term, take \(0\) and calculate the determinant of the 2 by 2 matrix formed by eliminating the first row and third column:

\[\text{det}\begin{bmatrix}2 & -3 \\4 & 6\end{bmatrix} = (2) \cdot 6 - (-3) \cdot 4 = 12 + 12 = 24\]

Now, substituting these values back into the determinant formula gives:

\[\text{det}(A) = 3 \cdot 6 - 1 \cdot 4 + 0 \cdot 24 = 18 - 4 + 0 = 14\]

Thus, the determinant of the 3 by 3 matrix is \(14\). This method effectively reduces the complexity of calculating a 3 by 3 determinant by breaking it down into manageable 2 by 2 determinants, allowing for a systematic approach to finding the solution.

To solve a system of three equations with three unknowns using Cramer's rule, we first need to understand the structure of the equations and how to represent them in matrix form. Cramer's rule provides a straightforward method to find the values of the unknowns by using determinants of matrices.

Given a system of equations, we can express it in the form of an augmented matrix. For example, if we have the equations:

\[\begin{align*}-5x - y + 4z &= 4 \\0x + 3y + 6z &= 21 \\x + y + z &= 6\end{align*}\]

We can create the augmented matrix by organizing the coefficients of the variables and the constants:

\[\begin{bmatrix}-5 & -1 & 4 & | & 4 \\0 & 3 & 6 & | & 21 \\1 & 1 & 1 & | & 6\end{bmatrix}\]

In Cramer's rule, we define \(D\) as the determinant of the coefficient matrix, which consists of the coefficients of \(x\), \(y\), and \(z\). If \(D\) is zero, the system has no unique solution. To calculate \(D\), we use the formula for the determinant of a 3x3 matrix:

\[D = a(ei - fh) - b(di - fg) + c(dh - eg)\end{align*}\]

Where \(a\), \(b\), and \(c\) are the elements of the first row, and \(d\), \(e\), \(f\), \(g\), \(h\), and \(i\) are the elements of the remaining rows. After calculating \(D\), we can find \(D_x\), \(D_y\), and \(D_z\) by replacing the respective columns of the coefficient matrix with the constants from the right side of the equations.

For example, to find \(D_x\), we replace the first column (the coefficients of \(x\)) with the constants:

\[D_x = \begin{vmatrix}4 & -1 & 4 \\21 & 3 & 6 \\6 & 1 & 1\end{vmatrix}\end{align*}\]

Similarly, for \(D_y\) and \(D_z\), we replace the second and third columns with the constants, respectively. The solutions for \(x\), \(y\), and \(z\) can then be calculated as follows:

\[x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}\end{align*}\]

After performing the calculations, we find the values of \(x\), \(y\), and \(z\) that satisfy the original system of equations. This method is efficient for solving linear systems and highlights the relationship between algebra and matrix theory.