Question

Find a \(2 \times 2\) matrix \(E\) with entries between 0 and 1 such that: a. \(I-E\) has no inverse. b. \(I-E\) has an inverse but not all entries of \((I-E)^{-1}\) are nonnegative.

Short answer

For a: \(E = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\). For b: \(E = \begin{pmatrix} 0.5 & 0 \\ 0.5 & 0.5 \end{pmatrix}\).

Step-by-step solution

Understanding Matrix Inversion

A matrix is invertible if its determinant is not zero. For a \(2 \times 2\) matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is calculated as \( ad - bc \).

Finding Matrix E for Part a

For \(I - E\) to be non-invertible, the determinant must be zero. Let \(E = \begin{pmatrix} e_{11} & e_{12} \ e_{21} & e_{22} \end{pmatrix}\). Then, \(I - E = \begin{pmatrix} 1 - e_{11} & -e_{12} \ -e_{21} & 1 - e_{22} \end{pmatrix}\). The determinant is \((1-e_{11})(1-e_{22}) - (-e_{12})(-e_{21}) = 0\). Choose \(e_{11} = e_{22} = 1\) for example, resulting in \(E = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\) which gives a zero determinant.

Matrix Selection for Part a

A matrix \(E = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\) makes \(I - E = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}\), which is singular and hence has no inverse.

Choosing E for Part b

We need \(I - E\) to be invertible but result in negative entries in \((I-E)^{-1}\). Let \(E = \begin{pmatrix} 0.5 & 0 \ 0.5 & 0.5 \end{pmatrix}\). Then \(I - E = \begin{pmatrix} 0.5 & 0 \ -0.5 & 0.5 \end{pmatrix}\).

Calculating the Determinant for Part b

Calculate the determinant of \(I - E\): \( (0.5)(0.5) - (0)(-0.5) = 0.25 eq 0 \). This means \(I - E\) is invertible.

Finding the Inverse for Part b

The inverse of a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\) is \(\frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\). Apply this to \(I-E\): \((I-E)^{-1} = \begin{pmatrix} 2 & 0 \ 2 & 2 \end{pmatrix}\), with a negative entry.

Key concepts

Determinant

The determinant is a special number that can be calculated from a square matrix. For any given matrix, the determinant helps us to work out if the matrix is invertible. In the case of a simple \(2 \times 2\) matrix, we calculate the determinant as follows:

• Consider a matrix in the form \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \).
• The determinant is given by the formula \( ad - bc \).

When the determinant is zero, the matrix is said to be 'singular', which means it does not have an inverse. A non-zero determinant indicates that the matrix is invertible; we can find its inverse using this formula.

Invertible Matrix

An invertible matrix is a key concept in linear algebra. It refers to a matrix that can be reversed, or more formally, has an inverse matrix.

For a matrix to be invertible, it must satisfy certain conditions:

• The matrix must be square (i.e., it has the same number of rows and columns).
• The determinant of the matrix must not be zero, as discussed earlier.

When a matrix is invertible, there exists another matrix such that when you multiply the two matrices, you receive the identity matrix. This identity matrix is like the number 1 in regular arithmetic: multiplying by it leaves the original number (or matrix, in this case) unchanged.

2x2 Matrix

A \(2 \times 2\) matrix is a simple yet powerful type of matrix that consists of two rows and two columns. Due to its small size, the properties and operations regarding \(2 \times 2\) matrices are relatively easy to compute and understand.

Key Characteristics of \(2 \times 2\) Matrices include:

• They can be used to represent linear transformations in a two-dimensional space.
• Operations such as finding the determinant or inverse are straightforward, owing to their specific formulas for this size of the matrix.

If you have a \(2 \times 2\) matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), understanding how to compute its determinant or inverse can greatly assist in solving related algebraic problems, like finding whether it's invertible, as explained in previous sections.