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Chapter 7: Problem 1

Find (where possible) the inverse of the following matrices. Are these matrices singular or nonsingular? $$ \mathbf{A}=\left[\begin{array}{ll} 6 & 4 \\ 1 & 2 \end{array}\right] \quad \mathbf{B}=\left[\begin{array}{ll} 6 & 4 \\ 3 & 2 \end{array}\right] $$

Short Answer

Expert verified
Matrix A is nonsingular with an inverse of \( \begin{bmatrix} \frac{1}{4} & -\frac{1}{2} \ -\frac{1}{8} & \frac{3}{4} \ \right] \). Matrix B is singular and has no inverse.

Step by step solution

01

- Determine the determinant of matrix A

Calculate the determinant of the matrix \( \textbf{A} \) using the formula for a 2x2 matrix determinant: \[ \text{det}(\textbf{A}) = a_{11}a_{22} - a_{12}a_{21} \] Substitute the values: \[ \text{det}(\textbf{A}) = 6 \times 2 - 4 \times 1 = 12 - 4 = 8 \]
02

- Determine if matrix A is singular or nonsingular

Since the determinant of matrix \( \textbf{A} \) is not zero (\( \text{det}(\textbf{A}) = 8 \)), matrix \( \textbf{A} \) is nonsingular, and thus it has an inverse.
03

- Determine the inverse of matrix A

For a 2x2 matrix inverse, use the formula: \[ \textbf{A}^{-1} = \frac{1}{\text{det}(\textbf{A})} \begin{bmatrix} a_{22} & -a_{12} \ -a_{21} & a_{11} \ \right] \] Substitute the values from \( \textbf{A} \): \[ \textbf{A}^{-1} = \frac{1}{8} \begin{bmatrix} 2 & -4 \ -1 & 6 \ \right] = \begin{bmatrix} \frac{2}{8} & -\frac{4}{8} \ -\frac{1}{8} & \frac{6}{8} \ \right] = \begin{bmatrix} \frac{1}{4} & -\frac{1}{2} \ -\frac{1}{8} & \frac{3}{4} \ \right] \]
04

- Determine the determinant of matrix B

Calculate the determinant of matrix \( \textbf{B} \) using the same formula: \[ \text{det}(\textbf{B}) = a_{11}a_{22} - a_{12}a_{21} \] Substitute the values: \[ \text{det}(\textbf{B}) = 6 \times 2 - 4 \times 3 = 12 - 12 = 0 \]
05

- Determine if matrix B is singular or nonsingular

Since the determinant of matrix \( \textbf{B} \) is zero (\( \text{det}(\textbf{B}) = 0 \)), matrix \( \textbf{B} \) is singular, and thus it does not have an inverse.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant of a Matrix
The determinant of a matrix is an important value that helps in understanding various properties of the matrix. For a 2x2 matrix like \textbf{A} or \textbf{B}, the determinant provides insight into whether the matrix has an inverse. For a 2x2 matrix \textbf{A} = \begin{bmatrix} a & b \ c & d \end{bmatrix}, the determinant is calculated using the formula: \[ \text{det}(\textbf{A}) = ad - bc \]. For example, for matrix \textbf{A} given in the problem:
  • Substitute the values from the matrix: \[ \text{det}(\textbf{A}) = 6 \times 2 - 4 \times 1 \].
  • Calculate: \[ \text{det}(\textbf{A}) = 12 - 4 = 8 \].

The determinant helps us proceed in identifying whether the matrix is singular or nonsingular.
Singular Matrix
A singular matrix is a matrix that does not have an inverse. This happens when the determinant of the matrix is zero. In the provided problem, matrix \textbf{B} is determined to be singular. Let's see why:
  • Using the determinant formula for matrix \textbf{B}: \[ \text{det}(\textbf{B}) = 6 \times 2 - 4 \times 3 \].
  • Calculate: \[ \text{det}(\textbf{B}) = 12 - 12 = 0 \].
Since the determinant of \textbf{B} is zero, it confirms that \textbf{B} is a singular matrix, meaning it does not have an inverse.
Nonsingular Matrix
A nonsingular matrix is one that has an inverse. This occurs when the determinant of the matrix is not zero. In the problem, matrix \textbf{A} has a determinant of 8, which is clearly not zero. To recap:
  • Since \( \text{det}(\textbf{A}) = 8 \), which is not zero, \textbf{A} is nonsingular.
  • This means \textbf{A} has an inverse.
  • You can use this property to solve inverse-related problems or systems of equations.

Recognizing whether a matrix is nonsingular is crucial for further matrix operations.
2x2 Matrix Inverse
The inverse of a 2x2 matrix can be found using a specific formula that incorporates the determinant. For a 2x2 matrix \textbf{A} = \begin{bmatrix} a & b \ c & d \end{bmatrix}, the inverse is determined by: \[ \textbf{A}^{-1} = \frac{1}{\text{det}(\textbf{A})} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]. Using matrix \textbf{A} from the problem:
  • First, the determinant is 8.
  • Inverse formula: \[ \textbf{A}^{-1} = \frac{1}{8} \begin{bmatrix} 2 & -4 \ -1 & 6 \end{bmatrix} \].
  • Simplifying: \[ \textbf{A}^{-1} = \begin{bmatrix} \frac{1}{4} & -\frac{1}{2} \ -\frac{1}{8} & \frac{3}{4} \end{bmatrix} \].

This inverse can be used for further matrix operations or solving linear equations.

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Most popular questions from this chapter

Determine the final demand vector for three firms given the matrix of technical coefficients $$ \mathbf{A}=\left[\begin{array}{lll} 0.2 & 0.4 & 0.2 \\ 0.1 & 0.2 & 0.1 \\ 0.1 & 0.1 & 0 \end{array}\right] $$ and the total output vector $$ \mathbf{x}=\left[\begin{array}{c} 1000 \\ 300 \\ 700 \end{array}\right] $$

(Maple) Use the Maple instruction det \((A)\) to find the determinant of the \(4 \times 4\) matrix $$ \left[\begin{array}{rrrr} a & 2 & 1 & -1 \\ 2 & 3 & 4 & 1 \\ 0 & -1 & 5 & 6 \\ 1 & 2 & 4 & -3 \end{array}\right] $$ For what value of \(a\) does this matrix fail to possess an inverse? [Do not forget to load the linear algebra package before you begin. You do this by typing with (linalg) :]

Each unit of engineering output requires as input \(0.2\) units of engineering and \(0.4\) units of transport. Each unit of transport output requires as input \(0.2\) units of engineering and \(0.1\) units of transport. Determine the level of total output needed to satisfy a final demand of 760 units of engineering and 420 units of transport.

An economy consists of three industries: agriculture, mining and manufacturing. One unit of agricultural output requires \(0.2\) units of its own output, \(0.3\) units of mining output and \(0.4\) units of manufacturing output. One unit of mining output requires \(0.2\) units of agricultural output, \(0.4\) units of its own output and \(0.2\) units of manufacturing output. One unit of manufacturing output requires \(0.3\) units of agricultural output, \(0.3\) units of mining output and \(0.1\) units of its own output. (a) Write down the matrix of technical coefficients and find the Leontief inverse. (b) Determine the levels of total output needed to satisfy a final demand of 10000 units of agricultural output, 30000 units of mining output and 40000 units of manufacturing output. (c) If the final demand for agricultural output rises by 1000 units and the final demand for manufacturing output falls by 1000 units, calculate the change in mining output.

Consider the macroeconomic model defined by national income: \(\quad Y=C+I+G^{*} \quad\left(G^{*}>0\right)\) consumption: \(C=a Y+b \quad(00)\) investment: \(\quad I=c r+d \quad(c<0, d>0)\) money supply: \(\quad M_{\mathrm{s}}^{*}=k_{1} Y+k_{2} r \quad\left(k_{1}>0, k_{2}<0, M_{\mathrm{S}}^{*}>0\right)\) Show that this system can be written as \(\mathbf{A x}=\mathbf{b}\), where $$ \mathbf{A}=\left[\begin{array}{rrrr} 1 & -1 & -1 & 0 \\ -a & 1 & 0 & 0 \\ 0 & 0 & 1 & -c \\ k_{1} & 0 & 0 & k_{2} \end{array}\right] \quad \mathbf{x}=\left[\begin{array}{c} Y \\ C \\ I \\ r \end{array}\right] \quad \mathbf{b}=\left[\begin{array}{c} G^{*} \\ b \\ d \\ M_{\mathrm{s}}^{*} \end{array}\right] $$ Use Cramer's rule to show that $$ r=\frac{M_{1}^{*}(1-a)-k_{1}\left(b+d+G^{*}\right)}{k_{2}(1-a)+c k_{1}} $$ Write down the government expenditure multiplier for \(r\) and deduce that the interest rate, \(r\), increases as government expenditure, \(G^{*}\), increases.
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