Problem 17
Question
Find the inverse of the given matrix using their determinants and adjoints. $$ B=\left[\begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The inverse of matrix \(B\) is \( \begin{bmatrix} 3 & 7 \\ 5 & 12 \end{bmatrix} \).
1Step 1: Check if the Inverse Exists
To find the inverse of matrix \(B\), we need to check if the determinant of \(B\) is non-zero. Use the formula for a \(2 \times 2\) matrix: \( \text{det}(B) = ad - bc \), where \(a = 12\), \(b = -7\), \(c = -5\), \(d = 3\). Calculate: \( \text{det}(B) = 12 \times 3 - (-7) \times (-5) = 36 - 35 = 1 \). Since the determinant is 1, the inverse exists.
2Step 2: Compute the Adjoint of Matrix B
For a \(2 \times 2\) matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the adjoint is \( \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). Substituting \(a = 12\), \(b = -7\), \(c = -5\), \(d = 3\), the adjoint of \(B\) is: \( \text{adj}(B) = \begin{bmatrix} 3 & 7 \ 5 & 12 \end{bmatrix} \).
3Step 3: Calculate the Inverse of Matrix B
The inverse of matrix \(B\) is given by \( B^{-1} = \frac{1}{\text{det}(B)} \cdot \text{adj}(B) \). Since \( \text{det}(B) = 1 \), substitute and compute: \( B^{-1} = 1 \cdot \begin{bmatrix} 3 & 7 \ 5 & 12 \end{bmatrix} = \begin{bmatrix} 3 & 7 \ 5 & 12 \end{bmatrix} \).
4Step 4: Verify the Inverse
To ensure the calculation is correct, verify by multiplying \(B\) and \(B^{-1}\). Compute: \( B \cdot B^{-1} = \begin{bmatrix} 12 & -7 \ -5 & 3 \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \ 5 & 12 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). Since the result is the identity matrix, \(B^{-1}\) is correct.
Key Concepts
DeterminantsAdjoint of a Matrix2x2 Matrix Inverse
Determinants
A determinant is a special number that can be calculated from a square matrix. It's a vital part of linear algebra that helps in determining certain properties of a matrix, such as invertibility. For a \(2 \times 2\) matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the formula to find the determinant is \( \text{det}(B) = ad - bc \).
**Key Points about Determinants:**
**Key Points about Determinants:**
- If the determinant of a matrix is zero, the inverse does not exist. This is because dividing by zero is undefined.
- For the matrix \( B \) given in the exercise, the determinant calculated as \( 12 \times 3 - (-7) \times (-5) = 36 - 35 = 1 \) confirms that the inverse exists because the result is non-zero.
- The determinant provides information about the matrix's uniqueness and stability when applied in various mathematical applications.
Adjoint of a Matrix
The adjoint (or adjugate) of a matrix is a specific matrix used in calculating the inverse of a given square matrix. To compute the adjoint of a \(2 \times 2\) matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), we swap the main diagonal elements (\(a\) and \(d\)) and change the signs of the off-diagonal elements (\(b\) and \(c\)).
Here's a stepwise guide:
Here's a stepwise guide:
- First, swap the positions of the elements in the leading diagonal. So, the \(a\) and \(d\) become \(d\) and \(a\).
- Then, alter the signs of \(b\) and \(c\), thus creating \(-b\) and \(-c\).
- For the matrix \( B \) from the exercise: with \(a = 12\), \(b = -7\), \(c = -5\), and \(d = 3\), the adjoint is computed as: \( \begin{bmatrix} 3 & 7 \ 5 & 12 \end{bmatrix} \).
2x2 Matrix Inverse
The inverse of a \(2 \times 2\) matrix, when it exists, is derived using its determinant and adjoint. For any matrix \( B = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is calculated as \( B^{-1} = \frac{1}{\text{det}(B)} \cdot \text{adj}(B) \).
**Steps to Find the Inverse:**
**Steps to Find the Inverse:**
- Ensure the determinant is non-zero, which confirms an inverse can be found.
- Calculate the adjoint of the matrix as described in the previous section.
- Multiply the adjoint by \(\frac{1}{\text{det}(B)}\), literally dividing every element in the adjoint by the determinant value.
- For matrix \( B \) given in the exercise, with \( \text{det}(B) = 1 \), the inverse is: \( \begin{bmatrix} 3 & 7 \ 5 & 12 \end{bmatrix} \). Since the determinant is 1, each element of the adjoint remains unchanged in the inverse.
Other exercises in this chapter
Problem 16
In Exercises \(9-26\), put each system of linear equations into triangular form and solve the system if poesible. Classify each system as consistent independent
View solution Problem 17
In Exercises 7 - 18 , find the partial fraction decomposition of the following rational expressions. $$ \frac{4 x^{3}-9 x^{2}+12 x+12}{x^{4}-4 x^{3}+8 x^{2}-16
View solution Problem 17
Use the matrices \(A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] \quad B=\left[\begin{array}{rr}0 & -3 \\ -5 & 2\end{array}\right] \quad C=\left[\be
View solution Problem 17
In Exercises \(9-26\), put each system of linear equations into triangular form and solve the system if poesible. Classify each system as consistent independent
View solution