Problem 69 - Precalculus with Limits

Question

In Exercises 63-70, find (a) \(|A|\), (b) \(|B|\), (c) \(AB\), and (d) \(|AB|\). \(A = \left[ \begin{array}{r} -1 && 2 && 1 \\ 1 && 0 && 1 \\ 0 && 1 && 0 \end{array} \right]\), \(B = \left[ \begin{array}{r} -1 && 0 && 0 \\ 0 && 2 && 0 \\ 0 && 0 && 3 \end{array} \right]\)

Step-by-Step Solution

Answer

\(|A| = 1\), \n\n |B| = -6, \n\n AB = \(\left[ \begin{array}{r} 1 && 4 && 3 \ -1 && 0 && 3 \ 0 && 2 && 0 \end{array} \right]\), \n\n and |AB| = -9.

Step 1: Determinant of Matrix A (\(|A|\))

The determinant of a 3x3 matrix can be calculated using the rule of Sarrus. \nFor matrix A, this gives \(-1*0*0 + 2*1*0 + 1*1*1 - 1*2*1 - 0*1*0 - 1*0*0 = 1\).

Step 2: Determinant of Matrix B (\(|B|\))

For matrix B, the determinant is simple because it's a diagonal matrix. The determinant is the product of diagonal entries, which is \(-1*2*3 = -6\).

Step 3: Matrix Multiplication (\(AB\))

Matrix A is multiplied by Matrix B by multiplying corresponding rows from A with corresponding columns from B and summing the results. This results in \n\(\left[ \begin{array}{r} -1*(-1) + 2*0 + 1*0 && -1*0 + 2*2 + 1*0 && -1*0 + 2*0 + 1*3 \ 1*(-1) + 0*0 + 1*0 && 1*0 + 0*2 + 1*0 && 1*0 + 0*0 + 1*3 \ 0*(-1) + 1*0 + 0*0 && 0*0 + 1*2 + 0*0 && 0*0 + 1*0 + 0*3 \end{array} \right]\), which simplifies to \(\left[ \begin{array}{r} 1 && 4 && 3 \ -1 && 0 && 3 \ 0 && 2 && 0 \end{array} \right]\).

Step 4: Determinant of Product Matrix (\(|AB|\))

Finally, to find \(|AB|\), apply the rule of Sarrus to the matrix obtained in Step 3, yielding \((1*0*0) + (4*3*0) + (3*-1*2) - (1*3*2) - (4*-1*0) - (3*0*0) = -9\).

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements. It's a useful quantity in linear algebra, helping us understand whether a matrix is invertible, which affects the solutions to systems of equations associated with the matrix. For a 3x3 matrix like the ones given in the exercise, we calculate the determinant using specific rules, such as the rule of Sarrus.

• To apply the rule of Sarrus, we first need to involve three diagonals leading from row to column (top-left to bottom-right) and three diagonals leading from column to row (top-right to bottom-left).
• Calculate the sum of the products of these diagonals. For example, for matrix A in this exercise, the diagonal products are:
  • First diagonal: \(-1\times 0\times 0\)
  • Second diagonal: \(2\times 1\times 0\)
  • Third diagonal: \(1\times 0\times 1\)
• Then, subtract the product of the diagonal elements calculated in the reverse diagonal order (bottom-left to top-right):
  • First reverse diagonal: \(1\times 2\times 1\)
  • Second reverse diagonal: \(0\times 1\times 0\)
  • Third reverse diagonal: \(1\times 0\times 0\)

This method offers a straightforward way to determine the determinant of a 3x3 matrix, giving us the measure to evaluate matrix properties like invertibility.

3x3 Matrix

A 3x3 matrix is a grid of numbers with three rows and three columns. This matrix size is a fundamental building block in linear algebra and beyond. Each entry in a 3x3 matrix helps determine its properties like determinant, trace, and inverses. Working with a 3x3 matrix:

• The layout is usually written as \[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \a_{21} & a_{22} & a_{23} \a_{31} & a_{32} & a_{33} \\end{array}\]where each up symbol \(a_{ij}\) represents the entry at row \(i\) and column \(j\).
• A 3x3 matrix can transform vectors, solve systems of linear equations, and much more.
• Special types of 3x3 matrices, such as diagonal or identity matrices, have unique properties making calculations easier.

In this exercise, matrix A and matrix B have specific arrangements that determine the outcome when performing operations such as finding determinant or matrix multiplication.

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra involving multiplying rows and columns from two matrices and summing the results. For two matrices to be multiplied, their dimensions must conform to the rule: the number of columns in the first matrix must equal the number of rows in the second matrix. In the given exercise, matrix A (3x3) and matrix B (3x3) satisfy this condition.

• To multiply matrices, take each row from the first matrix and multiply it by each column from the second matrix.
• The multiplication requires multiplying corresponding elements and then adding them together.
• The resulting matrix, product AB, will have the same number of rows as the first matrix and the same number of columns as the second matrix. For 3x3 matrices, the result is another 3x3 matrix.
• The computation steps for product matrix AB, for example, involve:
  • Row1*Col1: \((-1 \times -1) + (2 \times 0) + (1 \times 0)\)
  • Row1*Col2: \((-1 \times 0) + (2 \times 2) + (1 \times 0)\)
  • Continue this way for all row-column combinations resulting in the full matrix.

The multiplication process reveals how matrix product results in new transformations, which is useful in modeling and computations across varied applications.