Problem 57 - Calculus for Biology and Medicine

Question

Suppose that $$ A=\left[\begin{array}{ll} 2 & 4 \\ 3 & 6 \end{array}\right] $$ (a) Compute det \(A\). Is \(A\) invertible? (b) Suppose that $$ X=\left[\begin{array}{l} x \\ y \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{l} b_{1} \\ b_{2} \end{array}\right] $$ Write \(A X=B\) as a system of linear equations. (c) Show that if $$ B=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] $$ then $$ A X=B $$ has infinitely many solutions. Graph the two straight lines associated with the corresponding system of linear equations, and explain why the system has infinitely many solutions. (d) Find a column vector $$ B=\left[\begin{array}{l} b_{1} \\ b_{2} \end{array}\right] $$ so that $$A X=B$$ has no solutions.

Step-by-Step Solution

Answer

(a) det(A) = 0, A is not invertible. (b) System: \(2x + 4y = b_1\), \(3x + 6y = b_2\). (c) For \(B = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\), infinite solutions exist as equations are the same line. (d) No solutions for \(B = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\).

Step 1: Compute Determinant of A

To check if a matrix is invertible, we first need to calculate the determinant of matrix \(A\), denoted as \(\det(A)\). For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is calculated as \(\det(A) = ad - bc\). For matrix \(A = \begin{bmatrix} 2 & 4 \ 3 & 6 \end{bmatrix}\), we have:\[det(A) = (2)(6) - (4)(3) = 12 - 12 = 0.\]Since the determinant is zero, matrix \(A\) is not invertible.

Step 2: Write System of Equations

Given the matrix equation \(AX = B\), where \(X = \begin{bmatrix} x \ y \end{bmatrix}\) and \(B = \begin{bmatrix} b_1 \ b_2 \end{bmatrix}\), we can express this as a system of linear equations by multiplying \(A\) and \(X\):\[AX = \begin{bmatrix} 2x + 4y \ 3x + 6y \end{bmatrix} = \begin{bmatrix} b_1 \ b_2 \end{bmatrix}.\]Therefore, the system of linear equations is:\[\begin{align*}2x + 4y &= b_1, \3x + 6y &= b_2.\end{align*}\]

Step 3: Show Infinite Solutions for Specific B

If we take \( B = \begin{bmatrix} 2 \ 3 \end{bmatrix} \), the system of equations becomes:\[\begin{align*}2x + 4y &= 2, \3x + 6y &= 3.\end{align*}\]Both equations are scalar multiples of each other, specifically the second equation is 1.5 times the first equation. Thus, they represent the same line, leading to infinitely many solutions.

Step 4: Graph the Lines for Infinite Solutions

The equations \(2x + 4y = 2\) and \(3x + 6y = 3\) simplify to the same line \(x + 2y = 1\). When graphed, they appear as a single line. Multiple solutions exist because each point on the line is a solution.

Step 5: Find B for No Solutions

For \(AX = B\) to have no solutions, \(B\) must not satisfy the linear dependency. Use a vector \(B = \begin{bmatrix} b_1 \ b_2 \end{bmatrix}\) that cannot satisfy \(b_1 = 2b_2/3\). For instance, take \(B = \begin{bmatrix} 1 \ 0 \end{bmatrix}\). This yields:\[\begin{align*}2x + 4y &= 1, \3x + 6y &= 0.\end{align*}\]Here, both lines are parallel and do not intersect, indicating no solutions.

Key Concepts

Determinant Calculation

Matrix invertibility is a key concept in linear algebra, frequently determined by calculating the matrix's determinant. For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is obtained using the formula: \[ \det(A) = ad - bc. \] When the determinant of a matrix is zero, the matrix cannot be inverted. For the given matrix \( A = \begin{bmatrix} 2 & 4 \ 3 & 6 \end{bmatrix} \), the determinant calculation \[ \det(A) = 2 \times 6 - 4 \times 3 = 0 \] shows that its value is zero. Because of this, matrix \( A \) is not invertible. Determinants help in analyzing matrix properties and solving linear systems. A zero determinant indicates that the matrix doesn't have a unique solution, highlighting its importance in both theoretical and applied mathematics.

System of Linear Equations

A system of linear equations can be derived from a matrix equation like \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable vector, and \( B \) is the constant vector. In this scenario, for the matrix \( A = \begin{bmatrix} 2 & 4 \ 3 & 6 \end{bmatrix} \) and vectors \( X = \begin{bmatrix} x \ y \end{bmatrix} \), \( B = \begin{bmatrix} b_1 \ b_2 \end{bmatrix} \), the equations are formed by multiplying \( A \) with \( X \):

• \( 2x + 4y = b_1 \)
• \( 3x + 6y = b_2 \)

These two linear equations describe the relationships between the variables \( x \) and \( y \) with the constants \( b_1 \) and \( b_2 \). Systems of equations like these are essential in fields requiring solutions to simultaneously occurring conditions, such as engineering and physics.

Infinite Solutions in Linear Algebra

In linear algebra, a system of equations may sometimes have infinitely many solutions. This occurs when the provided equations represent the same line, plane, or hyperplane in an n-dimensional space. Given the exercise with matrix \( A \), if \( B = \begin{bmatrix} 2 \ 3 \end{bmatrix} \), the equations

• \( 2x + 4y = 2 \)
• \( 3x + 6y = 3 \)

are essentially the same because the second is a scalar multiple of the first. This means both equations graph to the line \( x + 2y = 1 \). Hence, every point on this line is a solution, indicating infinitely many solutions. Situations like these are fundamental for understanding solution spaces and vector dependencies.