Problem 63 - Calculus for Biology and Medicine

Question

Use the determinant to determine whether $$A=\left[\begin{array}{rr} 1 & -1 \\ 0 & 2 \end{array}\right]$$ is invertible. If it is invertible, compute its inverse. In either case, solve \(A X=\mathbf{0}\).

Step-by-Step Solution

Answer

Matrix \( A \) is invertible with the inverse \( \left[ \begin{array}{rr} 1 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{array} \right] \), and the solution to \( A X = \mathbf{0} \) is \( \mathbf{0} \).

Step 1: Calculate the Determinant of Matrix A

To determine if matrix \( A \) is invertible, we first find its determinant. For a \( 2 \times 2 \) matrix \( A = \left[ \begin{array}{rr} a & b \ c & d \end{array} \right] \), the determinant is computed as \( \text{det}(A) = ad - bc \). For matrix \( A = \left[ \begin{array}{rr} 1 & -1 \ 0 & 2 \end{array} \right] \), we substitute the values: \( \text{det}(A) = (1)(2) - (-1)(0) = 2 \). Since the determinant is not zero, matrix \( A \) is invertible.

Step 2: Calculate the Inverse of Matrix A

The inverse of a \( 2 \times 2 \) matrix \( A = \left[ \begin{array}{rr} a & b \ c & d \end{array} \right] \), can be found using the formula \( A^{-1} = \frac{1}{ad - bc} \left[ \begin{array}{rr} d & -b \ -c & a \end{array} \right] \). For \( A = \left[ \begin{array}{rr} 1 & -1 \ 0 & 2 \end{array} \right] \) with \( \text{det}(A) = 2 \), apply the formula: \( A^{-1} = \frac{1}{2} \left[ \begin{array}{rr} 2 & 1 \ 0 & 1 \end{array} \right] = \left[ \begin{array}{rr} 1 & \frac{1}{2} \ 0 & \frac{1}{2} \end{array} \right] \).

Step 3: Solve the Matrix Equation AX=0

Given \( A X = \mathbf{0} \), where \( \mathbf{0} \) is the zero vector, and \( A^{-1} \) exists, then \( X = A^{-1} \mathbf{0} \). Because any matrix multiplied by a zero vector yields a zero vector, the solution is \( X = \mathbf{0} \). This indicates the trivial solution where \( X \) is the zero vector.

Key Concepts

Determinant

When determining if a matrix is invertible, the determinant plays a crucial role. For a given matrix, its determinant can help us decide whether it's possible to find an inverse or not.
For a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( \text{det}(A) = ad - bc \).

• If the determinant is non-zero, the matrix is invertible.
• If the determinant is zero, the matrix isn’t invertible.

In our exercise, the matrix \( A = \begin{bmatrix} 1 & -1 \ 0 & 2 \end{bmatrix} \) was presented. Calculating its determinant, \( \text{det}(A) = 1 \times 2 - (-1) \times 0 = 2 \), shows that it's not zero.
This means matrix \( A \) is invertible, allowing us to proceed to find its inverse. The determinant, therefore, is a helpful indicator of whether a matrix can be inverted.

Matrix Equation

Matrix equations like \( A \mathbf{X} = \mathbf{0} \) are fundamental in linear algebra. Here, \( A \) represents a matrix, \( \mathbf{X} \) is a variable column vector, and \( \mathbf{0} \) is the zero vector. This particular form of matrix equation is essential for understanding different types of solutions.

• To solve the equation, if \( A \) is invertible, we can use its inverse \( A^{-1} \).

By multiplying both sides of the equation \( A \mathbf{X} = \mathbf{0} \) by \( A^{-1} \), we get:\[ A^{-1} A \mathbf{X} = A^{-1} \mathbf{0} \]This simplifies to \( \mathbf{X} = A^{-1} \mathbf{0} \). Since any matrix multiplied by a zero vector remains a zero vector, \( \mathbf{X} \) will be the zero vector.
Matrix equations demonstrate systems of linear equations and their solutions, and the above highlights the simplicity when faced with an equation involving a zero vector.

Trivial Solution

In the context of linear algebra, particularly when dealing with the matrix equation \( A \mathbf{X} = \mathbf{0} \), we often encounter the concept of the trivial solution.
The trivial solution is simply when the variable vector \( \mathbf{X} \) is zero, or \( \mathbf{X} = \mathbf{0} \).

• This occurs naturally when multiplying any matrix by a zero vector.
• The result is always a zero vector as well.

Even though many equations could have multiple potential solutions, the trivial solution is the most straightforward. It serves as a baseline or default solution.
Understanding trivial solutions is key to recognizing basic patterns in more complex equations and systems. It tells us that regardless of the complexity of \( A \), the zero vector remains constant in its behavior when multiplied, highlighting the foundational principle of multiplication with zero.