Problem 16
Question
For each matrix, find \(A^{-1}\) if it exists. Do not use a calculator. $$A=\left[\begin{array}{ll} 0.8 & -0.3 \\ 0.5 & -0.2 \end{array}\right]$$
Step-by-Step Solution
Verified Answer
The inverse is \(\begin{bmatrix} 20 & -30 \\ 50 & -80 \end{bmatrix}\).
1Step 1: Check if Determinant is Zero
To find the inverse of a matrix, first, we need to ensure that the determinant is not zero. Calculate the determinant of matrix \(A\):\[det(A) = (0.8)(-0.2) - (-0.3)(0.5) = -0.16 + 0.15 = -0.01\]Since \(det(A) eq 0\), the inverse exists.
2Step 2: Use Formula for 2x2 Matrix Inverse
For a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
3Step 3: Substitute Values into Formula
Substitute the elements of matrix \(A\) and its determinant into the inverse formula:\[A^{-1} = \frac{1}{-0.01} \begin{bmatrix} -0.2 & 0.3 \ -0.5 & 0.8 \end{bmatrix}\]This simplifies to:\[A^{-1} = -100 \begin{bmatrix} -0.2 & 0.3 \ -0.5 & 0.8 \end{bmatrix}\]
4Step 4: Calculate the Inverse Matrix
Multiply each element by \(-100\): \[A^{-1} = \begin{bmatrix} 20 & -30 \ 50 & -80 \end{bmatrix}\]So, the inverse of matrix \(A\) is:\(\begin{bmatrix} 20 & -30 \ 50 & -80 \end{bmatrix}\).
Key Concepts
Determinant Calculation2x2 Matrix Inverse FormulaInverse of a MatrixMatrix Algebra
Determinant Calculation
Before finding the inverse of a matrix, calculating its determinant is crucial. The determinant indicates if an inverse exists. For a \(2\times2\) matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\] the determinant is computed as \[det(A) = ad - bc.\] This number must be non-zero for the matrix to have an inverse.
In the given problem, we have a matrix \[A = \begin{bmatrix} 0.8 & -0.3 \ 0.5 & -0.2 \end{bmatrix}.\] We compute its determinant as follows:
In the given problem, we have a matrix \[A = \begin{bmatrix} 0.8 & -0.3 \ 0.5 & -0.2 \end{bmatrix}.\] We compute its determinant as follows:
- Multiply the diagonal elements: \( (0.8)(-0.2) = -0.16 \)
- Multiply the anti-diagonal elements: \( (-0.3)(0.5) = -0.15 \)
- Subtract the results: \( -0.16 + 0.15 = -0.01 \)
2x2 Matrix Inverse Formula
To find the inverse of a \(2\times2\) matrix, once the determinant is non-zero, we can use a specific formula. The formula for the inverse of a given matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\] is \[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}.\] This formula rearranges the elements of the original matrix and changes their signs accordingly.
It's important to remember the formula modifies as follows:
It's important to remember the formula modifies as follows:
- Swap the positions of \(a\) and \(d\).
- Change the signs of \(b\) and \(c\).
- Divide the entire matrix by the determinant \((ad - bc)\).
Inverse of a Matrix
Finding the inverse of a matrix is a fundamental concept in matrix algebra. The inverse \(A^{-1}\) of a matrix \(A\) is the matrix that, when multiplied with \(A\), yields the identity matrix. For a \(2\times2\) matrix, this involves checking if the matrix is invertible by ensuring the determinant is non-zero, then applying the inverse formula.
Once you have the inverse, you can use it to solve systems of equations, find matrix powers, and more. It's essential for various applications in engineering, physics, and computer science. For the given problem, after confirming a non-zero determinant, the inverse of matrix \[A = \begin{bmatrix} 0.8 & -0.3 \ 0.5 & -0.2 \end{bmatrix}\]was found to be:\[A^{-1} = \begin{bmatrix} 20 & -30 \ 50 & -80 \end{bmatrix}.\] This satisfies the condition \(A \cdot A^{-1} = I\) where \(I\) is the identity matrix.
Once you have the inverse, you can use it to solve systems of equations, find matrix powers, and more. It's essential for various applications in engineering, physics, and computer science. For the given problem, after confirming a non-zero determinant, the inverse of matrix \[A = \begin{bmatrix} 0.8 & -0.3 \ 0.5 & -0.2 \end{bmatrix}\]was found to be:\[A^{-1} = \begin{bmatrix} 20 & -30 \ 50 & -80 \end{bmatrix}.\] This satisfies the condition \(A \cdot A^{-1} = I\) where \(I\) is the identity matrix.
Matrix Algebra
Matrix algebra involves operations such as addition, subtraction, and multiplication on matrices, as well as finding inverses and determinants. Understanding how to manipulate matrices is crucial for solving linear algebra problems efficiently.
Key concepts include:
Key concepts include:
- Matrix Addition: Add corresponding elements.
- Matrix Subtraction: Subtract corresponding elements.
- Matrix Multiplication: Multiply the rows of the first matrix by the columns of the second matrix.
- Inverse: Find a matrix that when multiplied with the original gives the identity matrix.
Other exercises in this chapter
Problem 16
Solve each system by substitution. $$\begin{aligned}&y=4 x-6\\\&2 x+5 y=-8\end{aligned}$$
View solution Problem 16
Find the partial fraction decomposition for each rational expression. $$\frac{6 x^{5}+7 x^{4}-x^{2}+2 x}{3 x^{2}+2 x-1}$$
View solution Problem 16
Graph each inequality. Do not use a calculator. $$y>2(x+3)^{2}-1$$
View solution Problem 16
Write the system of equations associated with each augmented matrix. $$\left[\begin{array}{rr|r} 1 & -5 & -18 \\ 6 & 2 & 20 \end{array}\right]$$
View solution