Problem 66 - Calculus for Biology and Medicine

Question

Use the determinant to determine whether $$ D=\left[\begin{array}{ll} -3 & 6 \\ -4 & 8 \end{array}\right] $$ is invertible. If it is invertible, compute its inverse. In either case, solve \(D X=\mathbf{0}\).

Step-by-Step Solution

Answer

Matrix \( D \) is not invertible because its determinant is 0. The solution to \( DX = \mathbf{0} \) is \( X = t \begin{bmatrix} 2 \\ 1 \end{bmatrix} \) for any real number \( t \).

Step 1: Calculate the Determinant

To determine if matrix \( D \) is invertible, we first calculate its determinant. The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( ad - bc \).For the matrix \( D = \begin{bmatrix} -3 & 6 \ -4 & 8 \end{bmatrix} \), we calculate its determinant:\[\text{det}(D) = (-3)(8) - (6)(-4) = -24 + 24 = 0.\]Since the determinant is 0, \( D \) is not invertible.

Step 2: Solve the Homogeneous System

Since \( D \) is not invertible, it means the homogeneous system \( DX = \mathbf{0} \) has non-trivial solutions. Let's find solutions for:\[\begin{bmatrix} -3 & 6 \ -4 & 8 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]Writing it out in terms of equations:1. \(-3x_1 + 6x_2 = 0\)2. \(-4x_1 + 8x_2 = 0\)Note that both equations are multiples of each other. From the first equation, \(-3x_1 + 6x_2 = 0\), we can express it as:\( x_1 = 2x_2 \).Let \( x_2 = t \) (where \( t \) is any real number), then:\( x_1 = 2t \).Thus, the solution set is:\[X = t \begin{bmatrix} 2 \ 1 \end{bmatrix}, \quad \text{for any } t \in \mathbb{R}.\]

Key Concepts

Matrix Invertibility

A matrix is considered **invertible** if there is another matrix that, when multiplied together, results in the identity matrix. To check if a matrix is invertible, we calculate its determinant. If the determinant is zero, the matrix is not invertible. This means it cannot be "reversed" to return the original values when multiplied by another matrix.

For a 2x2 matrix \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\], the determinant is computed using the formula:\[ad - bc\].
For our specific matrix, \[D=\begin{bmatrix} -3 & 6 \ -4 & 8 \end{bmatrix}\], the determinant calculation was:

• \((-3)(8) = -24\)
• \((6)(-4) = -24\)

Their sum is zero, confirming that **Matrix D is not invertible**.

Homogeneous Systems of Equations

A **homogeneous system** of equations is a system where all the constants on the right-hand side of the equations are zero. Mathematically, it is represented as:\[AX = \mathbf{0}\] where \(A\) is a matrix and \(X\) is a vector.

In the exercise, we dealt with the system:

• \(-3x_1 + 6x_2 = 0\)
• \(-4x_1 + 8x_2 = 0\)

This setup often arises when you need to find out if the solution to such a system is just the trivial solution (where every variable is zero), or if there are more (non-trivial) solutions. Because invertibility of the coefficient matrix affects the type of solutions, it's essential to calculate it first. In this case, since the determinant is zero, the system potentially has non-trivial solutions. This happens because the rows of the matrix are not independent, as evidenced by the multiplication factor between the two equations.

Non-Trivial Solutions

When discussing linear systems, a **non-trivial solution** is any solution other than the zero solution. For homogeneous systems, zero (the trivial solution) is always a solution, which arises from setting all variables to zero.

Finding non-trivial solutions requires that the matrix associated with the system has rows (or columns) that are linearly dependent, meaning one is a multiple of another. This is what we encountered with our matrix \(D\).

To find these solutions, express one variable in terms of another. Taking the equation \(-3x_1 + 6x_2 = 0\), solve for \(x_1\) in terms of \(x_2\):

• \(x_1 = 2x_2\)

By letting \(x_2 = t\) (where \(t\) is any real number), the solutions form:

• \(x_1 = 2t\)

The solutions thus form a family represented by:\[X = t \begin{bmatrix} 2 \ 1 \end{bmatrix}\]This means any vector that scales by \(2\) in the first dimension and by \(1\) in the second dimension is a valid, non-trivial solution, highlighting the dependency of rows in the matrix.