Problem 19
Question
For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$\left[\begin{array}{rr}0.5 & 1.5 \\ 1 & -0.5\end{array}\right]$$
Step-by-Step Solution
Verified Answer
The inverse of the matrix is \( \begin{bmatrix} \frac{2}{7} & \frac{6}{7} \\ \frac{4}{7} & -\frac{2}{7} \end{bmatrix} \).
1Step 1: Determine the Determinant
To find the inverse of a matrix, first calculate its determinant. If the matrix is \( A = \begin{bmatrix} 0.5 & 1.5 \ 1 & -0.5 \end{bmatrix} \), the determinant (det\(A\)) is evaluated as: \( \text{det}(A) = (0.5)(-0.5) - (1.5)(1) \). Calculate this to find \( \text{det}(A) = -0.25 - 1.5 = -1.75 \).
2Step 2: Check if Inverse Exists
An inverse exists if and only if the determinant is non-zero. Since \( \text{det}(A) = -1.75 eq 0 \), the matrix does have an inverse.
3Step 3: Find the Adjugate Matrix
Compute the adjugate matrix by swapping the elements on the main diagonal and changing the signs of the elements on the other diagonal. So, the adjugate matrix of \( A \), denoted as \( \text{adj}(A) \), is: \[ \text{adj}(A) = \begin{bmatrix} -0.5 & -1.5 \ -1 & 0.5 \end{bmatrix} \].
4Step 4: Calculate the Inverse Matrix
The inverse \( A^{-1} \) is given by \( \frac{1}{\text{det}(A)} \times \text{adj}(A) \). Substituting in the determinant and the adjugate matrix, we have: \[ A^{-1} = \frac{1}{-1.75} \times \begin{bmatrix} -0.5 & -1.5 \ -1 & 0.5 \end{bmatrix} = \begin{bmatrix} \frac{2}{7} & \frac{6}{7} \ \frac{4}{7} & -\frac{2}{7} \end{bmatrix} \].
Key Concepts
Determinant of a MatrixAdjugate MatrixInverse Matrix Calculation
Determinant of a Matrix
When dealing with matrices, the determinant plays a crucial role in understanding various properties, like invertibility. The determinant for a square matrix provides a single value that helps you understand certain characteristics of the matrix. For a 2x2 matrix, the determinant is computed using a specific formula:
- For a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( \text{det}(A) = ad - bc \).
- This calculation uses the values from the main diagonal (from top left to bottom right) and subtracts the product of the other diagonal (from top right to bottom left).
Adjugate Matrix
The adjugate matrix is an essential component in finding the inverse of a matrix. Once you have the determinant and verify it's non-zero, the next step is constructing the adjugate matrix. This involves some simple transformations:
- First, swap the elements on the main diagonal (diagonal from the top-left corner to the bottom-right corner).
- Next, change the signs of the off-diagonal elements (diagonal from the top-right corner to the bottom-left corner).
- We swap 0.5 and -0.5.
- The signs of 1.5 and 1 change, thus giving us the adjugate matrix \( \text{adj}(A) = \begin{bmatrix} -0.5 & -1.5 \ -1 & 0.5 \end{bmatrix} \).
Inverse Matrix Calculation
With the determinant and adjugate matrix in hand, you can calculate the inverse matrix. The inverse of a matrix, if it exists, is crucial because it allows you to solve systems of linear equations among other applications. Here’s how you calculate it:
- The formula for the inverse of a matrix \( A \) is \( A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A) \).
- This means you multiply each element of the adjugate matrix by the reciprocal of the determinant.
- \( A^{-1} = \frac{1}{-1.75} \times \begin{bmatrix} -0.5 & -1.5 \ -1 & 0.5 \end{bmatrix} \).
- After simplifying, the inverse matrix \( A^{-1} \) becomes \( \begin{bmatrix} \frac{2}{7} & \frac{6}{7} \ \frac{4}{7} & -\frac{2}{7} \end{bmatrix} \).
Other exercises in this chapter
Problem 19
Solve the system by Gaussian elimination. \(\left[\begin{array}{rr|r}-1 & 2 & -3 \\ 4 & -5 & 6\end{array}\right]\)
View solution Problem 19
For the following exercises, solve the system by Gaussian elimination. $$ \left[\begin{array}{rr|r}{-1} & {2} & {-3} \\ {4} & {-5} & {6}\end{array}\right] $$
View solution Problem 19
Use any method to solve the system of nonlinear equations. $$ \begin{aligned} x^{2}+y^{2} &=1 \\ y &=-x^{2} \end{aligned} $$
View solution Problem 19
Use the matrices below to perform matrix multiplication. \(A=\left[\begin{array}{rr}-1 & 5 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{rrr}3 & 6 & 4 \\ -8
View solution