Problem 80 - College Algebra and Calculus: An Applied Approach

Question

Find \((a)|A|\), (b) \(|B|\), (c) \(A B\), and \((d)\) \(|A B|\). $$ A=\left[\begin{array}{rrr} 3 & 2 & 0 \\ -1 & -3 & 4 \\ -2 & 0 & 1 \end{array}\right], \quad B=\left[\begin{array}{rrr} -3 & 0 & 1 \\ 0 & 2 & -1 \\ -2 & -1 & 1 \end{array}\right] $$

Step-by-Step Solution

Answer

|A| = -15, |B| = -4, AB = [-6, 4, -1; -8, -7, 5; -8 , -1, 1], |AB| = -110

Step 1: Compute the determinant of A

The determinant of a 3x3 matrix can be calculated using the rule of Sarrus. To calculate determinant of A, write first two columns to the right side of the matrix. Then, sum the product of the diagonals from top left to bottom right and subtract the product of the diagonals from top right to bottom left. \[|A| = 3*(-3)*1 + 2*4*(-2) + 0*(-1)*(-2) - (-2)*(-3)*1 - 0*(-1)*3 - 2*4*(-2) = -15\]

Step 2: Compute the determinant of B

Similarly, you can calculate the determinant of B using the rule of Sarrus.\[|B| = -3*2*1 + 0*(-1)*(-2) + 1*0*1 - (-2)*2*1 - 0*0*1 - 1*(-1)*(-2) = -4\]

Step 3: Compute the matrix product AB

To find the product of two matrices, multiply entries in the rows of the first matrix by corresponding entries in the columns of the second matrix and summing those products. So,\[A B=\left[\begin{array}{rrr} -6 & 4 & -1 \ -8 & -7 & 5 \ -8 & -1 & 1 \end{array}\right]\]

Step 4: Compute the determinant of AB

Now that you have AB, you can take its determinant. Again, using rule of Sarrus.\[|AB| = -6*(-7)*1 + 4*5*(-8) + (-1)*(-8)*1 - 8*(-7)*1 - 4*(-8)*(-6) - 5*(-8)*1 = -110\]

Key Concepts

Matrix multiplication

Matrix multiplication is a fundamental operation in linear algebra. It involves multiplying two matrices together to produce a third matrix. For matrices to be multiplied, the number of columns in the first matrix must be equal to the number of rows in the second matrix. This ensures that the multiplication process is feasible and meaningful.
In our example:

• Matrix \(A\) has 3 columns and Matrix \(B\) has 3 rows, making them compatible for multiplication.
• The resulting matrix, \(AB\), will have the same number of rows as Matrix \(A\) and the same number of columns as Matrix \(B\).

To compute the elements of the product matrix, follow these steps:

• Take each row from Matrix \(A\) and each column from Matrix \(B\).
• Multiply corresponding elements from the row and column.
• Sum these products to get an entry in the resulting matrix \(AB\).

This process is repeated for all row-column pairs until the matrix is completely filled. It’s a methodical process that, when performed carefully, yields the matrix product accurately.

3x3 matrix

A 3x3 matrix is a square matrix consisting of three rows and three columns. Each element within this matrix is defined by its position given by its row and column number. The general form of a 3x3 matrix looks like this: \[\begin{bmatrix}a_{11} & a_{12} & a_{13} \a_{21} & a_{22} & a_{23} \a_{31} & a_{32} & a_{33} \end{bmatrix}\]In our exercise, both matrices \(A\) and \(B\) are 3x3 matrices, which means:

• Each matrix has 9 elements.
• They are suitable for operations such as determinant calculation and matrix multiplication.
• Determining their determinants largely depends on their arrangement of elements.

Understanding the structure of a 3x3 matrix is crucial for performing and comprehending the mathematical operations involved. It lays the groundwork for complex calculations and simplifications in various mathematical fields.

Sarrus' rule

Sarrus' Rule is a quick and handy method used to calculate the determinant of a 3x3 matrix. It's a visual technique that simplifies what could amount to more complex algebraic computations. To apply Sarrus' rule:

• First, rewrite the first two columns of the matrix adjacent to the original three columns.
• Calculate the sum of the products of the diagonals that run from the top-left to the bottom-right.
• Then, subtract the sum of the products of the diagonals that run from the top-right to the bottom-left.

For example, using Sarrus’ Rule for Matrix \(A\):

• Multiply along the diagonals: \(3 \cdot (-3) \cdot 1\), \(2 \cdot 4 \cdot (-2)\) and add them up.
• Subtract the diagonals running the opposite way: \((-2) \cdot (-3) \cdot 1\), etc.

This method not only saves time but reduces the chances of errors occurring in computations, especially useful for students learning determinants.