Problem 9

Question

Solve the given system of equations by Cramer's rule. $$ \begin{array}{r} u+2 v+\quad w=8 \\ 2 u-2 v+2 w=7 \\ u-4 v+3 w=1 \end{array} $$

Step-by-Step Solution

Verified

Answer

$ u = 3, v = \frac{9}{8}, w = \frac{3}{4} $.

1Step 1: Write the System of Equations in Matrix Form

Convert the given system of equations into a matrix form: $ A\mathbf{x} = \mathbf{b} $, where $ A $ is the coefficient matrix, $ \mathbf{x} $ is the column matrix of variables, and $ \mathbf{b} $ is the column matrix of constants.So, the system:\[\begin{align*}1u + 2v + 1w &= 8, \2u - 2v + 2w &= 7, \1u - 4v + 3w &= 1,\end{align*}\]can be written as:\[A = \begin{bmatrix} 1 & 2 & 1 \ 2 & -2 & 2 \ 1 & -4 & 3 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} u \ v \ w \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 8 \ 7 \ 1 \end{bmatrix}.\]

2Step 2: Compute the Determinant of the Coefficient Matrix

Calculate $ \det(A) $, the determinant of the matrix $ A $:\[\begin{vmatrix} 1 & 2 & 1 \ 2 & -2 & 2 \ 1 & -4 & 3 \end{vmatrix} = 1((-2)(3) - (-4)(2)) - 2((2)(3) - (1)(2)) + 1((2)(-4) - (1)(2)).\]Solving this gives:\[\det(A) = 1(-6 + 8) - 2(6 - 2) + 1(-8 - 2) = 2 - 8 - 10 = -16.\]

3Step 3: Compute Determinant for Variables

For each variable, compute the determinant by replacing the corresponding column of $ A $ with $ \mathbf{b} $.

4Step 3.1: Compute Determinant for u

Replace the first column of $ A $ with $ \mathbf{b} $:\[A_u = \begin{bmatrix} 8 & 2 & 1 \ 7 & -2 & 2 \ 1 & -4 & 3 \end{bmatrix}.\]Compute $ \det(A_u) $:\[= 8((-2)(3) - (-4)(2)) - 2((7)(3) - (1)(2)) + 1((7)(-4) - (-2)(1)).\]Solving this gives:\[\det(A_u) = 8(-6 + 8) - 2(21 - 2) + 1(-28 + 2) = 16 - 38 - 26 = -48.\]

5Step 3.2: Compute Determinant for v

Replace the second column of $ A $ with $ \mathbf{b} $:\[A_v = \begin{bmatrix} 1 & 8 & 1 \ 2 & 7 & 2 \ 1 & 1 & 3 \end{bmatrix}.\]Compute $ \det(A_v) $:\[= 1((7)(3) - (1)(2)) - 8((2)(3) - (1)(2)) + 1((2)(1) - (7)(1)).\]Solving this gives:\[\det(A_v) = 1(21 - 2) - 8(6 - 2) + 1(2 - 7) = 19 - 32 - 5 = -18.\]

6Step 3.3: Compute Determinant for w

Replace the third column of $ A $ with $ \mathbf{b} $:\[A_w = \begin{bmatrix} 1 & 2 & 8 \ 2 & -2 & 7 \ 1 & -4 & 1 \end{bmatrix}.\]Compute $ \det(A_w) $:\[= 1((-2)(1) - (-4)(7)) - 2((2)(1) - (1)(7)) + 8((2)(-4) - (1)(-2)).\]Solving this gives:\[\det(A_w) = 1(-2 + 28) - 2(2 - 7) + 8(-8 + 2) = 26 + 10 - 48 = -12.\]

7Step 4: Solve for Each Variable Using Cramer's Rule

Using Cramer's Rule, solve each variable:\[u = \frac{\det(A_u)}{\det(A)} = \frac{-48}{-16} = 3,\]\[v = \frac{\det(A_v)}{\det(A)} = \frac{-18}{-16} = \frac{9}{8},\]\[w = \frac{\det(A_w)}{\det(A)} = \frac{-12}{-16} = \frac{3}{4}.\]

8Step 8: Conclusion: Write the Solution

The solution to the system of equations is:\[ u = 3, \quad v = \frac{9}{8}, \quad w = \frac{3}{4}.\]

Key Concepts

System of EquationsDeterminant CalculationMatrix AlgebraLinear Algebra Concepts

System of Equations

A system of equations is a collection of equations that you deal with all at once. Each equation has multiple variables, and the solution is the set of variable values that make all the equations true simultaneously. In our example, we have three equations with variables: $ u, v, $ and $ w $. These equations together form a system:\[\begin{array}{r}u+2 v+ w=8 \2 u-2 v+2 w=7 \u-4 v+3 w=1 \\end{array}\]Such systems can often be solved using various algebraic methods, and in this case, Cramer's rule provides an efficient tactic when the system meets certain criteria.

Determinant Calculation

Determining the solution involves calculating the determinant, a special number that can be computed from a square matrix. It plays a crucial role in solving a system of linear equations using Cramer's rule.

For a 3×3 matrix $ A $, the determinant $ \det(A) $ is calculated based on its elements. For our matrix:\[A = \begin{bmatrix} 1 & 2 & 1 \2 & -2 & 2 \1 & -4 & 3 \end{bmatrix}\]data- Calculate it as follows: - Multiply and subtract products in a specific pattern: - Take the primary diagonal product and subtract the other diagonals.
After calculation, we find $ \det(A) = -16 $. This determinant is vital because if it were zero, Cramer's rule wouldn’t apply.

Matrix Algebra

Matrix algebra is a fundamental part of linear algebra. It refers to a set of operations that can be performed on matrices - ordered grids of numbers. In this context, matrices provide a compact way to handle linear equations.

For our system, the matrix form is:

Coefficient matrix $ A $: $ \begin{bmatrix} 1 & 2 & 1 \2 & -2 & 2 \1 & -4 & 3 \end{bmatrix} $
Variable column matrix $ \mathbf{x} $: $ \begin{bmatrix} u \v \w \end{bmatrix} $
Constants matrix $ \mathbf{b} $: $ \begin{bmatrix} 8 \7 \1 \end{bmatrix} $

Matrix calculations enable us to handle multiple equations efficiently, and applications like Cramer's rule rely on these operations to find solutions.

Linear Algebra Concepts

Linear algebra covers the study of vectors and matrices, which are central to numerous mathematical and real-world applications, including systems of equations.

Cramer's rule, a part of linear algebra, is a straightforward method for solving square systems of linear equations—where the number of equations equals the number of unknowns when the determinant is non-zero.

Each variable in the system corresponds to a modified version of the original matrix, with one column swapped for the constants.

In this exercise, determinants from modified matrices were calculated to find the solution variables $ u, v, $ and $ w $. This approach helps understand the intrinsic relationship between matrices and their determinants in solving equations.

Problem 9

Other exercises in this chapter

Problem 9

Determine whether the given matrix is orthogonal. $$ \left(\begin{array}{rrr} 1 & -1 & 1 \\ 1 & -1 & -1 \\ 1 & 2 & 0 \end{array}\right) $$

View solution

Problem 9

To determine whether the given matrix is singular or nonsingular. $$ \left(\begin{array}{rrr} 1 & 2 & 3 \\ 0 & -4 & 2 \\ -1 & 5 & 1 \end{array}\right) $$

View solution

Problem 9

Determine whether the given matrices are equal. $$ \left(\begin{array}{cc} \sqrt{(-2)^{2}} & 1 \\ 2 & \frac{2}{8} \end{array}\right),\left(\begin{array}{rr} -2

View solution

Problem 9

Use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. $x_{1}-x_{2}-x_{3}=8$ \(x_{1}-x_{2}+x_{

View solution