Problem 64 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

Matrix B is invertible, with an inverse \( \begin{bmatrix} -1 & 1 \\ 2 & -1 \end{bmatrix} \). The solution to \( B\mathbf{X} = \mathbf{0} \) is \( \mathbf{X} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)."

Step 1: Determine Invertibility Using the Determinant

First, calculate the determinant of matrix \( B \). The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( ad - bc \). For matrix \( B \), the determinant is \( (1)(1) - (1)(2) = 1 - 2 = -1 \). Since the determinant is not zero, matrix \( B \) is invertible.

Step 2: Compute the Inverse of Matrix B

The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). For \( B \), substitute the values: \( a = 1 \), \( b = 1 \), \( c = 2 \), \( d = 1 \) and compute the inverse. It becomes \( \frac{1}{-1} \begin{bmatrix} 1 & -1 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 1 \ 2 & -1 \end{bmatrix} \).

Step 3: Solve the System BX=0

Since \( B \) is invertible, the only solution to \( B \mathbf{X} = \mathbf{0} \) is the trivial solution \( \mathbf{X} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). Use the inverse we found to verify the solution: \( B^{-1}B \mathbf{X} = B^{-1}\mathbf{0} \rightarrow \mathbf{X} = \mathbf{0} \).

Key Concepts

Determinant

The determinant of a matrix is a scalar value that is crucial in understanding various properties of the matrix, including its invertibility. In simple terms, the determinant gives us a number that summarizes certain properties of a matrix. If a matrix has a non-zero determinant, it is invertible or non-singular. This means that there exists another matrix, called the inverse, which when multiplied by the original matrix, results in the identity matrix.

For a 2x2 matrix, the determinant is calculated using the formula:

• Determinant = \(ad - bc\), where \(a, b, c,\) and \(d\) are elements of the matrix.

In the given exercise, the matrix \(B\) is invertible because its determinant is calculated to be -1. Since -1 is not zero, matrix \(B\) is non-singular, confirming its invertibility.

Inverse of a Matrix

The inverse of a matrix is a matrix that, when multiplied with the original matrix, yields the identity matrix. Finding the inverse is essential for solving linear equations and understanding transformations in linear algebra.

For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by:

• \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).

This formula shows that the determinant \( ad-bc \) acts as a divisor. So, the matrix's determinant cannot be zero when finding the inverse. If it was, division by zero would occur, which is impossible.

In the exercise, the inverse of matrix \(B\) is computed using this method, resulting in:

• Inverse of \(B\) = \( \begin{bmatrix} -1 & 1 \ 2 & -1 \end{bmatrix} \).

Trivial Solution

In linear algebra, when solving equations like \(BX = \mathbf{0}\), where \(B\) is a matrix and \(X\) is a vector, solutions can be either trivial or non-trivial.

A trivial solution is when the vector \(X\) is simply the zero vector. That means every element of \(X\) is zero, such as \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This solution is always possible for homogeneous systems, regardless of the matrix's properties.

In the context of the exercise, since matrix \(B\) is invertible, the only solution to \(BX = \mathbf{0}\) is the trivial one. The invertibility of \(B\) ensures there are no non-zero solutions, as verified by calculating \(B^{-1}B\); it leaves us with \(X = \mathbf{0}\). This demonstrates the singular or unique nature of the trivial solution in such cases where the matrix is invertible.