Problem 52 - Calculus for Biology and Medicine

Question

(a) Show that if \(X=A X+D\), then $$ X=(I-A)^{-1} D $$ provided that \(I-A\) is invertible. (b) Suppose that $$ A=\left[\begin{array}{rr} 3 & 2 \\ 0 & -1 \end{array}\right] \text { and } \quad D=\left[\begin{array}{r} -2 \\ 2 \end{array}\right] $$ Compute \((I-A)^{-1}\), and use your result in (a) to compute \(X\).

Step-by-Step Solution

Answer

(a) Proved; (b) \(X = \begin{bmatrix} 0 \\ -1 \end{bmatrix}\).

Step 1: Rearrange the Equation

Start with the equation \(X = AX + D\). We want to isolate \(X\) on one side. First, move the \(AX\) term to the other side: \(X - AX = D\).

Step 2: Factor Out \(X\)

Now the equation \(X - AX = D\) can be factored to the form \((I - A)X = D\), where \(I\) is the identity matrix. This rearrangement emphasizes that \(X\) is multiplied by \(I - A\).

Step 3: Solve for \(X\)

Assume \((I - A)\) is invertible, meaning there is an inverse matrix \((I - A)^{-1}\) such that \((I - A)^{-1}(I - A) = I\). Multiply both sides of the equation \((I - A)X = D\) by \((I - A)^{-1}\) to isolate \(X\): \(X = (I - A)^{-1}D\), demonstrating the formula is valid.

Step 4: Calculate \(I - A\)

Given the matrix \(A = \begin{bmatrix} 3 & 2 \ 0 & -1 \end{bmatrix}\), first, find the identity matrix of the same dimension as \(A\), which is \(I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\). Calculate \(I - A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 2 \ 0 & -1 \end{bmatrix} = \begin{bmatrix} -2 & -2 \ 0 & 2 \end{bmatrix}\).

Step 5: Calculate the Inverse \((I - A)^{-1}\)

For the matrix \(B = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse is \(B^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). For our matrix, \((-2)(2)-(0)(-2) = -4\). The inverse is therefore \(\frac{1}{-4}\begin{bmatrix} 2 & 2 \ 0 & -2 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & -\frac{1}{2} \ 0 & -\frac{1}{2} \end{bmatrix}\).

Step 6: Compute \(X\)

Now use \(X = (I - A)^{-1}D\) with \((I - A)^{-1} = \begin{bmatrix} -\frac{1}{2} & -\frac{1}{2} \ 0 & -\frac{1}{2} \end{bmatrix}\) and \(D = \begin{bmatrix} -2 \ 2 \end{bmatrix}\). Multiply these matrices: \(X = \begin{bmatrix} -\frac{1}{2} & -\frac{1}{2} \ 0 & -\frac{1}{2} \end{bmatrix}\begin{bmatrix} -2 \ 2 \end{bmatrix} = \begin{bmatrix} (-\frac{1}{2})(-2) + (-\frac{1}{2})(2) \ 0(-2) + (-\frac{1}{2})(2) \end{bmatrix} = \begin{bmatrix} 0 \ -1 \end{bmatrix}\).

Key Concepts

Matrix Inversion

Matrix inversion is a helpful concept in linear algebra, particularly when solving systems of equations involving matrices. We talk about matrix inversion when we find a matrix that undoes the multiplication effect of a given square matrix. In simpler terms, for a matrix \(A\), its inverse \(A^{-1}\) such that \(A \cdot A^{-1} = I\), where \(I\) is the identity matrix. This operation is similar to finding the reciprocal of a number. Not all matrices have an inverse; a matrix must be non-singular, meaning it has a non-zero determinant.

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

• First, compute the determinant: \(ad-bc\).
• If the determinant is non-zero, the inverse is \(\frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\).

Let's consider an example. Given the matrix \( I - A = \begin{bmatrix} -2 & -2 \ 0 & 2 \end{bmatrix} \), we calculated the inverse by first finding the determinant \((-2)(2)-(0)(-2) = -4\). Therefore, the inverse is \(\begin{bmatrix} -\frac{1}{2} & -\frac{1}{2} \ 0 & -\frac{1}{2} \end{bmatrix}\). This matrix, when multiplied with \(I - A\), gives the identity matrix, fulfilling the inversion property.

Matrix Multiplication

Matrix multiplication is the process of multiplying two matrices to produce another matrix. It is a core operation in matrix algebra and has many applications, such as transformations in graphics and solutions in linear equations.

When multiplying two matrices, the number of columns in the first matrix must match the number of rows in the second matrix. For example, if we have a matrix \(A\) of dimensions \(m \times n\) and a matrix \(B\) of dimensions \(n \times p\), the resulting matrix \(C = AB\) will have dimensions \(m \times p\).

• The element at the position \(c_{ij}\) in the resulting matrix \(C\) is the sum of products of the elements from the \(i\)-th row of \(A\) and the \(j\)-th column of \(B\).
• Each element is computed as: \(c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{in}b_{nj}\).

In the context of our original exercise, after computing the inverse of \(I-A\), we use matrix multiplication to find \(X\). For example, with \( (I-A)^{-1} = \begin{bmatrix} -\frac{1}{2} & -\frac{1}{2} \ 0 & -\frac{1}{2} \end{bmatrix}\) and \(D = \begin{bmatrix} -2 \ 2 \end{bmatrix}\), the resulting matrix \(X\) is calculated through multiplication, yielding \(X = \begin{bmatrix} 0 \ -1 \end{bmatrix}\).

Identity Matrix

The identity matrix is a special kind of square matrix, which acts as the multiplicative identity in the matrix world. It is the equivalent of the number 1 for regular numbers in multiplication.

For an identity matrix \(I\), all the entries on the main diagonal (from the top left to the bottom right) are 1s, and all other entries are 0s. For any square matrix \(A\), multiplying it by the identity matrix leaves \(A\) unchanged, i.e., \(AI = IA = A\). This property is incredibly useful when dealing with matrix equations, especially when isolating variables in matrix expressions, such as finding the inverse or solving equations.

• For a 2x2 identity matrix, it looks like \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\).
• Its role is crucial in the formula \((I - A)X = D\), as we need \(I-A\) to invert \((I-A)\) and solve for \(X\).

In our exercise, the identity matrix \(I\) helps define the equation \((I - A)X = D\) and emphasizes the significance of being able to solve for \(X\) uniquely when \(I-A\) is invertible. This ensures the solutions we find are valid and applicable.