Problem 51 - Calculus for Biology and Medicine

Question

Suppose that $$ A=\left[\begin{array}{rr} -1 & 0 \\ 2 & -3 \end{array}\right] \text { and } D=\left[\begin{array}{l} -2 \\ -5 \end{array}\right] $$ Find \(X\) such that \(A X=D\) by (a) solving the associated system of linear equations and (b) using the inverse of \(A\).

Step-by-Step Solution

Answer

(a) \(X = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and (b) \(X = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) using the inverse.

Step 1: Set up the system of linear equations

The equation \( A X = D \) corresponds to the matrix equation \( \begin{pmatrix} -1 & 0 \ 2 & -3 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} -2 \ -5 \end{pmatrix} \). This can be written as two simultaneous linear equations:1. \(-x_1 = -2\)2. \(2x_1 - 3x_2 = -5\).

Step 2: Solve the first equation for $x_1$

From the first equation \(-x_1 = -2\), solve for \(x_1\) by multiplying both sides by -1: \[ x_1 = 2 \].

Step 3: Substitute $x_1$ in the second equation and solve for $x_2$

Substitute \(x_1 = 2\) into the second equation:\(2(2) - 3x_2 = -5\)\(4 - 3x_2 = -5\)Subtract 4 from both sides:\(-3x_2 = -9\)Divide both sides by -3:\(x_2 = 3\).

Step 4: Write the solution for part (a)

The solution to the system of equations is \( x_1 = 2 \) and \( x_2 = 3 \). Therefore, \( X = \begin{pmatrix} 2 \ 3 \end{pmatrix} \).

Step 5: Compute the inverse of A

The inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is given by\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]. Here, \( A = \begin{pmatrix} -1 & 0 \ 2 & -3 \end{pmatrix} \) has \( a = -1, b = 0, c = 2, d = -3 \). Calculate the determinant: \( ad - bc = (-1)(-3) -(0)(2) = 3 \).So, \( A^{-1} = \frac{1}{3} \begin{pmatrix} -3 & 0 \ -2 & -1 \end{pmatrix} = \begin{pmatrix} -1 & 0 \ \frac{-2}{3} & \frac{-1}{3} \end{pmatrix} \).

Step 6: Find $X$ using the inverse of $A$

Multiply both sides of \( A X = D \) by \( A^{-1} \):\( X = A^{-1} D \).So, \( X = \begin{pmatrix} -1 & 0 \ \frac{-2}{3} & \frac{-1}{3} \end{pmatrix} \begin{pmatrix} -2 \ -5 \end{pmatrix} \).Calculate the matrix multiplication:- Row 1: \(-1 \times -2 + 0 \times -5 = 2\)- Row 2: \(\frac{-2}{3} \times -2 + \frac{-1}{3} \times -5 = \frac{4}{3} - \frac{5}{3} = -\frac{1}{3}\).Thus, \( X = \begin{pmatrix} 2 \ 3 \end{pmatrix} \).

Step 7: Write the solution for part (b)

The solution using the inverse of \(A\) confirms that \(X = \begin{pmatrix} 2 \ 3 \end{pmatrix} \) consistent with part (a).

Key Concepts

Inverse Matrix

In linear algebra, an inverse matrix is a powerful concept used to solve systems of equations. If a square matrix \(A\) has an inverse, denoted \(A^{-1}\), such that \(AA^{-1} = I\), where \(I\) is the identity matrix, then \(A^{-1}\) effectively "undoes" the operation of \(A\). Not all matrices have an inverse, and a matrix must be square (same number of rows and columns) to qualify for inversion.

The formula for the inverse of a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\) is given by:

• Calculate the determinant: \(ad - bc\). If the determinant is zero, the matrix is not invertible.
• The inverse is \(A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\).

In the example, the matrix \(A = \begin{pmatrix} -1 & 0 \ 2 & -3 \end{pmatrix}\) has a determinant of 3, meaning it is invertible, and its inverse is \(\begin{pmatrix} -1 & 0 \ -\frac{2}{3} & -\frac{1}{3} \end{pmatrix}\). This inversion allows us to solve for \(X\) in equations like \(AX = D\) by calculating \(X = A^{-1}D\).

System of Equations

Solving systems of linear equations is a frequent task in linear algebra. A system of equations involves multiple equations that are all satisfied by the same set of variable values.

For example, consider solving the system:

• \(-x_1 = -2\)
• \(2x_1 - 3x_2 = -5\)

To solve, we can first isolate one variable, as we did with \(x_1\) to get \(x_1 = 2\).

Next, substitute \(x_1\) into the second equation to find \(x_2\). This systematic approach finds the solution \((x_1, x_2)\), which for this example is \((2, 3)\).

Alternatively, systems of equations can be solved using matrix operations, which include writing the system in the form \(AX = D\) and using the inverse of \(A\) to solve for \(X\). This method can often be more efficient, especially with larger systems.

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, pivotal for working with systems of equations and transformations.

To multiply matrices, the number of columns in the first matrix must match the number of rows in the second matrix. The product of matrices \(A\) and \(B\) is a new matrix where each entry is the sum of the products of the corresponding rows of \(A\) and columns of \(B\).

In our exercise, we find the solution \(X = A^{-1}D\) by multiplying the inverse of \(A\) by matrix \(D\):

• Row 1: Multiply and sum \(-1 \times -2 + 0 \times -5\) to get 2.
• Row 2: Multiply and sum \(-\frac{2}{3} \times -2 + -\frac{1}{3} \times -5\) to get the result.

The calculation yields the vector \(X = \begin{pmatrix} 2 \ 3 \end{pmatrix}\), verifying the solution by both substitution and inverse matrix methods. Understanding matrix multiplication is crucial for efficiently solving linear equations and exploring further linear algebra applications.