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Chapter 4: Problem 15

Cramer's Rule breaks down when det \(A=0\). Example (a) has no solution, whereas (b) has infinitely many. What are the ratios \(x_{j}=\operatorname{det} B_{j} / \operatorname{det} A\) ? (a) \(2 x_{1}+3 x_{2}=1\) \(4 x_{1}+6 x_{2}=1\) (parallel lines) (b) \(\begin{aligned} 2 x_{1}+3 x_{2} &=1 \\ 4 x_{1}+6 x_{2} &=2 \end{aligned}\) (same line)

Short Answer

Expert verified
Both systems yield undefined ratios due to \( \text{det}(A) = 0 \), making Cramer's Rule inapplicable.

Step by step solution

01

Formulate the coefficient matrix

For both examples, we start by writing the coefficient matrix \( A \). Given the equations in example (a) and example (b), the coefficients of \( x_1 \) and \( x_2 \) are the same:\[A = \begin{pmatrix} 2 & 3 \ 4 & 6 \end{pmatrix}\]
02

Compute the determinant of the coefficient matrix

The determinant of matrix \( A \) is calculated as: \[ \text{det}(A) = \begin{vmatrix} 2 & 3 \ 4 & 6 \end{vmatrix} = (2)(6) - (4)(3) = 12 - 12 = 0 \] Since the determinant is zero, Cramer's Rule cannot be applied.
03

Understand the implications of det(A) = 0 for the systems

When the determinant of \( A \) is zero, it means the system of equations is either inconsistent with no solutions or it has infinitely many solutions. This will depend on whether the lines are the same or parallel.
04

Analyze system (a)

In system (a), the equations are:\[2x_1 + 3x_2 = 1 \4x_1 + 6x_2 = 1\]These two equations represent parallel lines, as they have the same slope but different intercepts, which means they will never intersect. Therefore, system (a) has no solutions.
05

Analyze system (b)

In system (b), the equations are:\[2x_1 + 3x_2 = 1 \4x_1 + 6x_2 = 2\]These are multiples of each other (the second equation is simply twice the first), meaning they represent the same line. Therefore, system (b) has infinitely many solutions as every point on the line satisfies both equations.
06

Compute the ratios \(x_{j}\) for both systems

For system (a), since \( \text{det}(A) = 0 \), the ratios \( x_1 = \frac{\text{det}(B_1)}{\text{det}(A)} \) and \( x_2 = \frac{\text{det}(B_2)}{\text{det}(A)} \) are undefined.For system (b), similarly, the ratios are undefined because the determinant of \( A \) is zero, so \( x_1 \) and \( x_2 \) cannot be determined uniquely using Cramer's Rule.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant of a Matrix
The determinant of a matrix plays a crucial role in understanding systems of linear equations. Think of it as a number that tells us about the matrix's properties. For a 2x2 matrix like \[ A = \begin{pmatrix} 2 & 3 \ 4 & 6 \end{pmatrix} \]the determinant is calculated by:
  • Multiplying the top left and bottom right diagonals: \((2)(6)\)
  • Subtracting the product of the other diagonal: \((4)(3)\)
So, \[ \text{det}(A) = 12 - 12 = 0 \]When the determinant is zero, the matrix does not have an inverse. This specific situation indicates that there is a problem with solving equations using this matrix. In systems of linear equations, it helps identify whether Cramer's Rule can be applied. When it is zero, we need to check for parallel or identical lines.
System of Linear Equations
A system of linear equations is a collection of two or more equations with the same set of variables. These systems can represent various real-world phenomena such as intersections of lines. For instance, the equations in our examples:
  • (a) \(2x_1 + 3x_2 = 1\) and \(4x_1 + 6x_2 = 1\)
  • (b) \(2x_1 + 3x_2 = 1\) and \(4x_1 + 6x_2 = 2\)
demonstrate different geometrical situations. Each equation represents a line, and their graphs can either intersect at one point, overlap entirely, or not touch at all. The nature of their intersection is crucial to find solutions. When both lines coincide or are parallel, the determinant will be zero, indicating no unique solution exists.
Infinite Solutions
Infinite solutions arise when every point on one line is also a point on another. In system (b), the equations \(2x_1 + 3x_2 = 1\) and \(4x_1 + 6x_2 = 2\) actually define the same line. The second equation is simply the first one, multiplied by two. This means every solution of the first equation works for the second, leading to infinitely many solutions.
Here鈥檚 what happens with infinite solutions:
  • Each equation represents overlapping lines.
  • They coincide perfectly, sharing all points.
  • The determinant is zero, preventing the matrix from being invertible.
This situation means Cramer's Rule cannot find distinct solutions, but doesn't mean the system has no solutions. It has too many, making every point on the line a solution.

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Most popular questions from this chapter

In a 5 by 5 matrix, does a \(+\) sign or - sign go with \(a_{15} a_{24} a_{33} a_{42} a_{51}\) down the reverse diagonal? In other words, is \(P=(5,4,3,2,1)\) even or odd? The checkerboard pattern of \(\pm\) signs for cofactors does not give det \(P\).

(a) Draw the triangle with vertices \(A=(2,2), B=(-1,3)\), and \(C=(0,0)\). By regarding it as half of a parallelogram, explain why its area equals $$ \operatorname{area}(A B C)=\frac{1}{2} \operatorname{det}\left[\begin{array}{rr} 2 & 2 \\ -1 & 3 \end{array}\right] $$ (b) Move the third vertex to \(C=(1,-4)\) and justify the formula $$ \operatorname{area}(A B C)=\frac{1}{2} \operatorname{det}\left[\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right]=\frac{1}{2} \operatorname{det}\left[\begin{array}{rrr} 2 & 2 & 1 \\ -1 & 3 & 1 \\ 1 & -4 & 1 \end{array}\right] $$ Hint: Subtracting the last row from each of the others leaves $$ \operatorname{det}\left[\begin{array}{rrr} 2 & 2 & 1 \\ -1 & 3 & 1 \\ 1 & -4 & 1 \end{array}\right]=\operatorname{det}\left[\begin{array}{rrr} 1 & 6 & 0 \\ -2 & 7 & 0 \\ 1 & -4 & 1 \end{array}\right]=\operatorname{det}\left[\begin{array}{rr} 1 & 6 \\ -2 & 7 \end{array}\right] $$ Sketch \(A^{\prime}=(1,6), B^{\prime}=(-2,7), C^{\prime}=(0,0)\) and their relation to \(A, B, C\).

Show that the partial derivatives of \(\ln \left(\operatorname{det} A\right.\) ) give \(A^{-1}\) : $$ f(a, b, c, d)=\ln (a d-b c) \quad \text { leads to } \quad\left[\begin{array}{ll} \partial f / \partial a & \partial f / \partial c \\ \partial f / \partial b & \partial f / \partial d \end{array}\right]=A^{-1} $$

True or false, with reason if true and counterexample if false: (a) If \(A\) and \(B\) are identical except that \(b_{11}=2 a_{11}\), then \(\operatorname{det} B=2 \operatorname{det} A\). (b) The determinant is the product of the pivots. (c) If \(A\) is invertible and \(B\) is singular, then \(A+B\) is invertible. (d) If \(A\) is invertible and \(B\) is singular, then \(A B\) is singular. (c) The determinant of \(A B-B A\) is zero.

Find all the odd permutations of the numbers \(\\{1,2,3,4\\}\). They come from an odd number of exchanges and lead to det \(P=-1\).
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