Problem 19 - College Algebra

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 0 s below the \(I\) s, are shown. Fill in the missing numbers in the steps that are shown. $$ \begin{array}{rrrr|r} \left[\begin{array}{rrr|r} 1 & -1 & 1 & 8 \\ 2 & 3 & -1 & -2 \\ 3 & -2 & -9 & 9 \end{array}\right] \rightarrow & & \left[\begin{array}{rrr|r} 1 & -1 & 1 & 8 \\ 0 & 5 & \square & \square \\ 0 & 1 & \square & \square \end{array}\right] \\ & \rightarrow & \left[\begin{array}{rrr} 1 & -1 & 1 & 8 \\ 0 & 1 & \square & \square \\ 0 & 1 & \square & \square \end{array}\right] \end{array} $$

Step-by-Step Solution

Answer

The missing numbers in the second matrix are -3 and -18 in the second row and -12 and -15 in the third row, respectively. In the third matrix, the missing numbers in third row are -9 and 3 respectively.

Step 1: Identify the row operations used

The operation going from the first matrix to the second one is subtracting multiples of the first row from the two subsequent rows. By subtracting 2 times first row from the second and subtracting 3 times the first row from the third, zeros are obtained in the first column of the second and third row. In order to find the missing values, one can apply those same operations to the elements which are not yet eliminated in second and third row.

Step 2: Apply the row operations to the second row

In order to find the missing number under the ‘square’ sign in the second row, the operations identified in step 1 are applied to the elements in the original matrix. The new value of the third element in the second row equals the old value (which is -1) minus 2 times the corresponding value in the first row (which is 1): -1 - 2*1 = -3. The new number in the last column of second row results to be -2 - 2*8 = -18. Therefore, in the second matrix, the second row should read '0 5 -3 -18'.

Step 3: Apply the row operations to the third row

Similarly, the operations identified in step 1 are applied to the elements in the third row. The new value of the third element is the old value (-9) minus 3 times the corresponding value in the first row (which is 1): -9 - 3*1 = -12. The new number in the last column of the third row is the old value (9) minus 3 times the corresponding value in the first row (which is 8): 9 - 3*8 = -15. Therefore, in the second matrix, the third row should read '0 1 -12 -15'.

Step 4: Decreasing the second row from third row

The operation going from the second matrix to the third one appears to be decreasing the second row from the third one, which makes the second element of the third row becomes zero without influencing the first row and the first and second elements of the second row. Unsure about the two missing numbers, the same operation can be used to determine these numbers. The third element in the third row is -12 minus -3 (noting the double negative becomes a plus): -12 + 3 = -9. The number in the last column of the third row is -15 minus -18: -15 + 18 = 3. Therefore, the third row in the third matrix should read '0 0 -9 3'.

Key Concepts

Row Operations in Matrices

Matrix row operations are fundamental tools in linear algebra used for simplifying matrices and solving systems of linear equations. They consist of three permissible moves:

• Swapping two rows,
• Multiplying a row by a nonzero scalar,
• Adding or subtracting a multiple of one row to another row.

In the context of achieving a row-echelon form, these operations aim to create zeros below the leading coefficient (often referred to as the pivot) in each row to simplify the matrix gradually. This is precisely what the exercise demonstrates.
Let's take a closer look at how these operations are applied. From the given exercise, we see that the transformation from the first matrix to the second involves subtraction of multiples of the first row from the subsequent rows. Specifically, the second and third rows are manipulated to introduce zeros in the first column, serving as a step towards row-echelon form.
Understanding and applying these basic row operations is essential in linear algebra, as they are the building blocks of more complex procedures like Gaussian elimination. Each step should be performed carefully to ensure that the matrix is being simplified correctly without altering the original system's solution.

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations, and it's closely related to matrix row operations. It consists of two main phases:

• Forward elimination, and
• Back substitution.

In the forward elimination phase, the goal is to convert the system's matrix into an upper triangular or row-echelon form where all elements below the diagonal are zero. This is achieved through the strategic application of row operations that the exercise demonstrates. Once this form is attained, the back substitution phase starts, solving for the variables beginning with the last row moving upwards.
The exercise has specifically illustrated part of the forward elimination phase, with the goal to simplify the matrix by transforming it to row-echelon form. The steps provided show the progress as zeros are placed beneath the pivots, one column at a time, utilizing row operations. After applying these operations to the given system of equations, the matrix is step by step brought closer to a state from which solutions can be easily deduced.

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. Mathematically, a linear equation represents a straight line, and a system is the overlap of such lines; their points of intersection represent the solutions of the system.
The main objective in solving such systems is to find values for the variables that satisfy all the equations simultaneously. To do this, methods such as substitution, graphing, and elimination (such as Gaussian elimination) are used.
The matrix from the exercise represents a system of linear equations. Each row corresponds to an equation and the columns represent the coefficients of the variables in these equations. The vertical bar separates the coefficients from the solutions in the last column. Row operations help us simplify this system without changing the solutions, hence providing a clearer pathway to finding the intersection points or, in algebraic terms, the values of the variables.
By applying the Gaussian elimination method on the system of equations from the exercise, we systematically work towards these solutions. Understanding these methods is essential because systems of linear equations form the bedrock for various applied mathematics fields, including engineering, physics, computer science, and economics, to model and solve real-world problems.