Problem 47 - Calculus for Biology and Medicine

Question

Let $$\boldsymbol{A}=\left[\begin{array}{rr} -\mathbf{1} & \mathbf{1} \\ \mathbf{2} & \mathbf{3} \end{array}\right], \quad \boldsymbol{B}=\left[\begin{array}{ll} \mathbf{2} & \mathbf{0} \\ \mathbf{3} & \mathbf{2} \end{array}\right]$$ Show that \(\left(A^{-1}\right)^{-1}=A\)

Step-by-Step Solution

Answer

The property \(\left(A^{-1}\right)^{-1} = A\) is confirmed by calculating \(A^{-1}\) and then its inverse, which equals \(A\).

Step 1: Understanding the Inverse of the Inverse

The problem is to show a property of matrix inverses, specifically that \(\left(A^{-1}\right)^{-1}=A\). This means that inverting a matrix and then taking the inverse again returns the original matrix. Therefore, if we can find \(A^{-1}\) and then invert it to return to \(A\), that will prove the statement.

Step 2: Calculate the Inverse of Matrix A

To find \(A^{-1}\), we need the formula for a 2x2 matrix inverse: \(A=\begin{bmatrix} a & b \ c & d \end{bmatrix}\), then \(A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). For \(A=\begin{bmatrix} -1 & 1 \ 2 & 3 \end{bmatrix}\), calculate:- Determinant = \(ad-bc = (-1)(3) - (2)(1) = -3 - 2 = -5\).- Inverse \(A^{-1} = \frac{1}{-5}\begin{bmatrix} 3 & -1 \ -2 & -1 \end{bmatrix} = \begin{bmatrix} -3/5 & 1/5 \ 2/5 & 1/5 \end{bmatrix}\).

Step 3: Calculate the Inverse of A's Inverse

Now, compute the inverse of \(A^{-1}\), which we computed as \(-3/5, 1/5, 2/5, 1/5\). The inverse of a general 2x2 matrix \(\begin{bmatrix} e & f \ g & h \end{bmatrix}\) is calculated similarly: \(\frac{1}{eh-fg}\begin{bmatrix} h & -f \ -g & e \end{bmatrix}\). - Here, determinant = \(eh - fg = \left(\frac{-3}{5}\right)\left(\frac{1}{5}\right) - \left(\frac{1}{5}\right)\left(\frac{2}{5}\right) = -\frac{3}{25} - \frac{2}{25} = -\frac{5}{25} = -\frac{1}{5}\).- Inverse \(\left(A^{-1}\right)^{-1} = \frac{1}{-\frac{1}{5}}\begin{bmatrix} \frac{1}{5} & -\frac{1}{5} \ -\frac{2}{5} & -\frac{3}{5} \end{bmatrix} = \begin{bmatrix} 1 & -1 \ -2 & -3 \end{bmatrix} = A\).

Step 4: Conclusion

After calculating the inverse of \(A^{-1}\), we've shown that reversing the inversion process retrieves the original matrix \(A\). Therefore, \(\left(A^{-1}\right)^{-1} = A\), confirming the property that taking the inverse twice brings us back to the original matrix.

Key Concepts

2x2 matrix

A 2x2 matrix is a square matrix with two rows and two columns. It is one of the simplest types of matrices, and yet, understanding its properties is critical for more complex matrix calculations. These matrices can be denoted in a general form as \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), where \(a\), \(b\), \(c\), and \(d\) are the elements of the matrix.
The simplicity of 2x2 matrices makes them ideal for illustrating core matrix operations, such as addition, multiplication, and finding determinants. In our specific example, we used the matrices:

• \(A=\begin{bmatrix} -1 & 1 \ 2 & 3 \end{bmatrix}\)
• \(B=\begin{bmatrix} 2 & 0 \ 3 & 2 \end{bmatrix}\)

Understanding these small matrices helps in building intuition for how larger, more complex matrices behave.

determinant calculation

The determinant of a matrix is a special number that can be calculated from its elements. For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is calculated using the formula:\[- (a \cdot d - b \cdot c) \]The determinant can tell us a lot about the matrix. Most importantly, if the determinant is zero, the matrix does not have an inverse, which means it cannot be used to solve certain systems of equations.
In our exercise, we calculated the determinant for matrix \(A\):

• \(- (1 \cdot 3 - 2 \cdot 1) = - (3 - 2) = -5 \)

This non-zero determinant implies that \(A\) is invertible, allowing us to find its inverse, which is crucial for confirming properties like \(\left(A^{-1}\right)^{-1} = A\).

properties of inverses

Matrix inversion is an operation that finds another matrix, called the inverse, which, when multiplied with the original matrix, yields the identity matrix. For 2x2 matrices, the inverse of \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by \(A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\).
One key property of matrix inverses that was examined in the exercise is \(\left(A^{-1}\right)^{-1} = A\). This means that if you find the inverse of a matrix and then find the inverse of that inverse, you will return to your original matrix.
Using matrix \(A\) from our example, we found:

• First, \(A^{-1} = \begin{bmatrix} -3/5 & 1/5 \ 2/5 & 1/5 \end{bmatrix}\)
• Then, \(\left(A^{-1}\right)^{-1} = \begin{bmatrix} 1 & -1 \ -2 & -3 \end{bmatrix} = A\)

This demonstrates the self-reversing property of inverses, essentially confirming that inverting twice brings you back to the starting matrix.