Problem 65 - Calculus for Biology and Medicine

Question

Use the determinant to determine whether $$C=\left[\begin{array}{ll} 1 & 3 \\ 1 & 3 \end{array}\right]$$ is invertible. If it is invertible, compute its inverse. In either case, solve \(C X=\mathbf{0}\).

Step-by-Step Solution

Answer

Matrix \( C \) is not invertible, and the solutions to \( C X = \mathbf{0} \) are \( t \begin{bmatrix} -3 \\ 1 \end{bmatrix} \) for any real number \( t \).

Step 1: Understand Invertibility Condition

A square matrix is invertible if and only if its determinant is non-zero. Therefore, to determine if the matrix \( C \) is invertible, calculate its determinant.

Step 2: Calculate the Determinant of C

For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is \( ad - bc \). Thus, for \( C = \begin{bmatrix} 1 & 3 \ 1 & 3 \end{bmatrix} \), the determinant is computed as \( 1 \cdot 3 - 3 \cdot 1 = 0 \).

Step 3: Determine Invertibility

Since the determinant of \( C \) is 0, the matrix is not invertible. A matrix is invertible only if its determinant is non-zero.

Step 4: Solve the System \( C X = \mathbf{0} \)

Since \( C \) is not invertible, the system \( C X = \mathbf{0} \) may have non-trivial solutions. Set up the equation \( \begin{bmatrix} 1 & 3 \ 1 & 3 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \):This expands to the equations:1. \( x + 3y = 0 \)2. \( x + 3y = 0 \)Both equations are identical, indicating infinitely many solutions of the form \( x = -3y \). Choose \( y = t \), a parameter, then \( x = -3t \).Thus, the solutions take the form \( X = t \begin{bmatrix} -3 \ 1 \end{bmatrix} \).

Key Concepts

Determinant Calculation

The determinant is a critical concept when assessing the invertibility of a matrix. Specifically, for a 2x2 square matrix, the determinant is calculated using the formula:

• For a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is \( \det(A) = ad - bc \).

To check invertibility, we calculate this value. If the determinant is zero, the matrix is non-invertible. In our exercise, the matrix \( C = \begin{bmatrix} 1 & 3 \ 1 & 3 \end{bmatrix} \) has a determinant calculated as \( 1 \cdot 3 - 3 \cdot 1 = 0 \).
This result implies that the matrix has no inverse, as its determinant is zero. Understanding determinants is essential because they tell us about the singularity of a matrix. A zero determinant indicates linear dependence between the rows or columns of the matrix, pointing to its inability to reverse a transformation.

System of Linear Equations

When dealing with matrices, solving a system of linear equations can often involve matrix equations. The exercise presented here involves solving the equation \( C X = \mathbf{0} \), where \( C \) is our matrix and \( X \) is the vector of unknowns.
Given \( C = \begin{bmatrix} 1 & 3 \ 1 & 3 \end{bmatrix} \), this expands into the linear equations \( x + 3y = 0 \) and \( x + 3y = 0 \). Both equations are identical due to the non-invertibility of the matrix, which means they lack diversity. This scenario typically results in infinitely many solutions.

• The solutions are parameterized as \( x = -3y \).
• If we set \( y = t \), where \( t \) is any real number, then \( x = -3t \).

Thus, the solution to the system can be written in the form \( X = t \begin{bmatrix} -3 \ 1 \end{bmatrix} \), indicating a family of solutions rather than a unique one.

Non-Invertible Matrix

A matrix is described as non-invertible, or singular, when its determinant is zero. This attribute signifies that the matrix does not have an inverse, which has profound implications not only in pure mathematics but in real-world applications like solving systems of equations. When a matrix is non-invertible, it implies that the transformations represented by the matrix "collapse" dimensions.
In practical terms, this often means:

• The rows or columns of the matrix are linearly dependent, indicating they are not uniquely pointing in separate directions.
• Any system of equations it represents will either have no solution or infinitely many solutions.

For \( C = \begin{bmatrix} 1 & 3 \ 1 & 3 \end{bmatrix} \), it is non-invertible because its rows are scalar multiples of each other. Therefore, the system \( C X = \mathbf{0} \) is underdetermined, providing a line of solutions rather than a single intersection point. This is why understanding matrix invertibility is crucial in both theoretical mathematics and applied fields such as physics and engineering.