Math Problem Statement

Perform one step of row reduction, in order to calculate the values for x and y by back substitution. Then calculate the values for x and y. Also calculate the determinant of the original matrix. ( Note: since the determinant is unchanged by type 3 row operations, it will be easier to calculate the determinant of the row reduced matrix.)

\left[\begin{array}{cc} 7 &27\cr 9 &-11 \end{array}\right] \left\lbrack \begin{array}{c} x \ y \end{array} \right\rbrack

\left[\begin{array}{c} 5\cr 2 \end{array}\right]

\left.\vphantom{\begin{array}{c}!\strut\!\strut\!\strut\\end{array}}\right[ 7 27 \left.\vphantom{\begin{array}{c}!\strut\!\strut\!\strut\\end{array}}\right] 0 \left.\vphantom{\begin{array}{c}!\strut\!\strut\!\strut\\end{array}}\right[ x \left.\vphantom{\begin{array}{c}!\strut\!\strut\!\strut\\end{array}}\right] y

\left.\vphantom{\begin{array}{c}!\strut\!\strut\!\strut\\end{array}}\right[ 5 \left.\vphantom{\begin{array}{c}!\strut\!\strut\!\strut\\end{array}}\right]

x =

y =

\mbox{det} =

Solution

Given System of Equations:

We are given the matrix equation:

[ \left[\begin{array}{cc} 7 & 27 \ 9 & -11 \end{array}\right] \left[\begin{array}{c} x \ y \end{array}\right] = \left[\begin{array}{c} 5 \ 2 \end{array}\right] ]

Step 1: Row Reduction

To solve for xx and yy, we will perform row reduction (Gaussian elimination). The goal is to transform the system into an upper triangular form so that back substitution can be used to solve for xx and yy.

The system is:

7x+27y=57x + 27y = 5 9x11y=29x - 11y = 2

To eliminate xx from the second equation, we can perform the following row operation:

R2R2(97)R1R_2 \rightarrow R_2 - \left(\frac{9}{7}\right)R_1

Step-by-Step Row Operation:

  1. The first equation remains the same: 7x+27y=57x + 27y = 5

  2. Subtract 97\frac{9}{7} times the first row from the second row: 9x11y(97)(7x+27y)=2(97)(5)9x - 11y - \left(\frac{9}{7}\right)(7x + 27y) = 2 - \left(\frac{9}{7}\right)(5)

Simplify:

9x11y(9x+2437y)=24579x - 11y - \left(9x + \frac{243}{7}y\right) = 2 - \frac{45}{7}

This simplifies to:

9x11y9x2437y=1474579x - 11y - 9x - \frac{243}{7}y = \frac{14}{7} - \frac{45}{7}

11y2437y=317-11y - \frac{243}{7}y = \frac{-31}{7}

Multiply through by 7 to simplify:

77y243y=31-77y - 243y = -31

320y=31-320y = -31

Solve for yy:

y=31320=31320y = \frac{-31}{-320} = \frac{31}{320}

Step 2: Back Substitution

Now that we have y=31320y = \frac{31}{320}, we substitute this value into the first equation to solve for xx:

7x+27(31320)=57x + 27\left(\frac{31}{320}\right) = 5

Simplify:

7x+837320=57x + \frac{837}{320} = 5

Multiply both sides by 320 to eliminate the fraction:

2240x+837=16002240x + 837 = 1600

Solve for xx:

2240x=1600837=7632240x = 1600 - 837 = 763

x=7632240x = \frac{763}{2240}

Thus, the solution is: x=7632240,y=31320x = \frac{763}{2240}, \quad y = \frac{31}{320}

Step 3: Calculate the Determinant

The determinant of the original matrix is:

det=(7)(11)(9)(27)\text{det} = (7)(-11) - (9)(27)

det=77243=320\text{det} = -77 - 243 = -320

Final Answer:

x=7632240,y=31320,determinant=320x = \frac{763}{2240}, \quad y = \frac{31}{320}, \quad \text{determinant} = -320

Would you like further details or have any questions?

Here are some related questions:

  1. How does Gaussian elimination differ from row reduction?
  2. Can the determinant of a matrix ever be zero?
  3. How are type 3 row operations used in matrix reduction?
  4. What is the significance of the determinant in a system of equations?
  5. How do row operations affect the determinant of a matrix?

Tip: When the determinant of a matrix is zero, the system of equations has either no solution or infinitely many solutions.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Row Reduction
Gaussian Elimination
Determinants
Back Substitution

Formulas

Row Reduction: R2 -> R2 - (9/7)R1
Back Substitution
Determinant: det(A) = ad - bc

Theorems

Gaussian Elimination
Properties of Determinants

Suitable Grade Level

College level (Linear Algebra)