Problem 20 - Algebra and Trigonometry

Question

A few steps in the process of simplifying the given matrix to row-echelon form, with Is down the diagonal from upper left to lower right, and os below the \(1 s,\) are shown. Fill in the missing numbers in the steps that are shown. $$ \left[\begin{array}{rrr|r} {1} & {-2} & {3} & {4} \\ {2} & {1} & {-4} & {3} \\ {-3} & {4} & {-1} & {-2} \end{array}\right] \rightarrow\left[\begin{array}{rrr|r} {1} & {-2} & {3} & {4} \\ {0} & {5} & {\square} & {\square} \\ {0} & {-2} & {\square} & {\square} \end{array}\right] $$ $$ \rightarrow\left[\begin{array}{rrr|r} {1} & {-2} & {3} & {4} \\ {0} & {1} & {\square} & {\square} \\ {0} & {-2} & {\square} & {\square} \end{array}\right] $$

Step-by-Step Solution

Answer

For the given transformations, the missing numbers in the matrices will be: for the second step, the missing numbers are: -10 and -5 and for the third step the missing numbers are: -2 and -1

Step 1: Understand the Goal

The goal is to fill in the missing squares in the second and third steps of the problem. Those squares represent steps in turning the original matrix into row-echelon form. The missing numbers can be found by performing the row operations.

Step 2: Calculation for the Second Step

In the second step, each element of the second row of the original matrix is subtracted from the corresponding element of the first row multiplied by 2. So, the second step calculations for the missing squares are: \(-4 - 3*2 = -10\) and \(3 - 4*2 = -5\). So the updated matrix is: \[\begin{array}{rrr|r} {1} & {-2} & {3} & {4} \ {0} & {5} & {-10} & {-5} \ {0} & {-2} & {\square} & {\square} \end{array}\right]\

Step 3: Calculation for the Third Step

There is no alteration in the third row between the second and the third steps, so the missing squares in the third row of the second step and third step are the same. They are calculated by adding the corresponding elements of third row to the first row multiplied by 3. So, the calculations for the missing squares are: \(-1 - 3*3 = -10\) and \(-2 - 4*3 = -14\). Adding these to the third step gives: \[\begin{array}{rrr|r} {1} & {-2} & {3} & {4} \ {0} & {1} & {\square} & {\square} \ {0} & {0} & {-10} & {-14} \end{array}\right].

Step 4: Final Calculation

In the third step, the second line is divided by 5. So, the calculations for the missing squares are: \(-10/5 = -2\) and \(-5/5 = -1\). Adding these to the matrix gives the final form: \[\begin{array}{rrr|r} {1} & {-2} & {3} & {4} \ {0} & {1} & {-2} & {-1} \ {0} & {0} & {-10} & {-14} \end{array}\right].

Key Concepts

Row Operations

In the realm of linear algebra, row operations are instrumental in simplifying matrices. They consist of three essential types: switching two rows, multiplying a row by a non-zero scalar, and adding a multiple of one row to another row. These operations are foundational because they maintain the solution set of the system of equations represented by the matrix.

In our exercise example, we utilize row operations to transform the original matrix into row-echelon form. For instance, to clear the entries below the leading 1 in the first column, we subtract appropriate multiples of the first row from the subsequent rows. It's important to approach these operations step by step to avoid calculation errors, ensuring the matrix gradually reaches its simplified form without losing the integrity of the system it represents.

Matrix Simplification

Matrix simplification is the process of organizing a matrix into a form that is easier to work with, particularly for solving systems of linear equations. The goal is to achieve a row-echelon or reduced row-echelon form where each row starts with a leading 1 (or all zeros if the row cannot be simplified further), followed by zeros beneath each column's leading 1.

To clarify, in our matrix from the exercise, we are working towards having 1s down the diagonal, and zeros beneath these 1s. Simplification aids in deciphering solutions to linear systems and is a prelude to further techniques such as back-substitution in the case of an echelon form or immediately reading off the solutions in the case of a reduced echelon form.

Gaussian Elimination

Gaussian elimination, named after the mathematician Carl Friedrich Gauss, is a method of effectively solving systems of linear equations. It involves using row operations to reduce a matrix into its row-echelon form, followed by back-substitution to find the solution.

Considering our exercise, Gaussian elimination is the series of steps we performed to fill in the missing numbers. It begins with the original matrix and applies row operations to systematically eliminate variables from the bottom up, until the system becomes triangular (if possible). This strategy is widely used due to its systematic approach, which can be applied to matrices of any size.

Linear Algebra

Linear algebra is a significant branch of mathematics concerned with vector spaces and linear mappings between these spaces. It includes the study of lines, planes, and subspaces but is also fundamental to modern presentations of geometry, algebra, and analysis.

Matrix operations, such as those we see in the given exercise, are at the heart of linear algebra. These constructs represent systems of linear equations and transformations, enabling us to solve and describe numerous problems in fields as diverse as engineering, computer science, economics, and social sciences. The row-echelon form is just one of the many concepts within this discipline that emphasize its importance and utility.