Chapter 9: Problem 6
Find \(A^{-1}\) if \(A=\left[\begin{array}{cc}i & 1+i \\ 1 & i\end{array}\right]\)
Short Answer
Expert verified
The inverse matrix \( A^{-1} \) is \( \begin{bmatrix} \frac{2i+1}{5} & \frac{2-3i}{5} \\ \frac{2-i}{5} & \frac{2i+1}{5} \end{bmatrix} \).
Step by step solution
01
Determine the Inverse Matrix Formula for a 2x2 Matrix
For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse \( A^{-1} \) is given by the formula:\[ A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]To use this formula, we need to first find the determinant \( ad - bc \).
02
Identify the Elements of Matrix A
Identify the elements of matrix \( A \):\[ A = \begin{bmatrix} i & 1+i \ 1 & i \end{bmatrix} \]Where \( a = i \), \( b = 1+i \), \( c = 1 \), and \( d = i \).
03
Compute the Determinant
The determinant of \( A \) is computed as follows:\[ ad - bc = (i)(i) - (1)(1+i) = i^2 - (1 + i) \]Since \( i^2 = -1 \), the expression becomes:\[ -1 - (1 + i) = -1 - 1 - i = -2 - i \].
04
Substitute to Find the Inverse of A
Using the formula for the inverse, substitute the determinant and matrix elements:\[ A^{-1} = \frac{1}{-2-i}\begin{bmatrix} i & -(1+i) \ -1 & i \end{bmatrix} = \frac{1}{-2-i}\begin{bmatrix} i & -1-i \ -1 & i \end{bmatrix} \].
05
Simplify the Expression for the Inverse
To simplify, multiply the elements of the matrix by the conjugate of the determinant's denominator:The conjugate of \(-2-i\) is \(-2+i\).Multiply numerator and denominator by \(-2+i\):\[ A^{-1} = \frac{1}{(-2-i)(-2+i)}\begin{bmatrix} i(-2+i) & -(1+i)(-2+i) \ -1(-2+i) & i(-2+i) \end{bmatrix} \]The denominator simplifies as follows:\[ (-2-i)(-2+i) = (-2)^2 - (i)^2 = 4 + 1 = 5 \].
06
Calculate and Simplify Each Element
Now calculate and simplify each element of \( A^{-1} \):- For the element at position (1,1):\[ \frac{1}{5}((i)(-2+i)) = \frac{1}{5}(2i + 1) = \frac{2i + 1}{5} \]- For the element at position (1,2):\[ \frac{1}{5}(-(1+i)(-2+i)) = \frac{1}{5}((1+i)(2-i))= \frac{1-i - 2i - i^2}{5} = \frac{2 - 3i}{5} \]- For the element at position (2,1):\[ \frac{1}{5}(-1(-2+i)) = \frac{2-i}{5} \]- For the element at position (2,2):\[ \frac{1}{5}((i)(-2+i)) = \frac{2i + 1}{5} \]So, the inverse matrix is:\[ A^{-1} = \begin{bmatrix} \frac{2i+1}{5} & \frac{2-3i}{5} \ \frac{2-i}{5} & \frac{2i+1}{5} \end{bmatrix} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Numbers
Complex numbers are numbers that consist of a real part and an imaginary part. They take the form of \( a + bi \) where \( a \) is the real part and \( b \) is the imaginary part, with \( i \) being the imaginary unit. The imaginary unit \( i \) has a property where \( i^2 = -1 \). This makes complex numbers incredibly useful in various fields such as engineering and physics.
When dealing with complex numbers in matrix applications, it's important to remember one key operation: complex conjugation. The complex conjugate of a number \( a + bi \) is \( a - bi \). For example, the conjugate of \( -2 - i \) is \( -2 + i \). This is important for simplifying expressions in operations like matrix inversion.
When dealing with complex numbers in matrix applications, it's important to remember one key operation: complex conjugation. The complex conjugate of a number \( a + bi \) is \( a - bi \). For example, the conjugate of \( -2 - i \) is \( -2 + i \). This is important for simplifying expressions in operations like matrix inversion.
- Real part: The part of a complex number that is not multiplied by \( i \).
- Imaginary part: The part that is multiplied by \( i \).
- Complex conjugate: Obtained by changing the sign of the imaginary part.
Determinant Calculation
Calculating the determinant is a crucial step in finding the inverse of a matrix, especially in the context of a 2x2 matrix. The determinant for a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( ad - bc \). This scalar value not only helps in inversion but also in determining whether a matrix is invertible. If the determinant is zero, the matrix does not have an inverse.
For the given matrix \( A = \begin{bmatrix} i & 1+i \ 1 & i \end{bmatrix} \), the determinant is found by calculating \( ad-bc \):
For the given matrix \( A = \begin{bmatrix} i & 1+i \ 1 & i \end{bmatrix} \), the determinant is found by calculating \( ad-bc \):
- First, calculate \( ad = i \times i = i^2 = -1 \).
- Then, calculate \( bc = (1)(1+i) = 1+i \).
- Subtract: \( ad - bc = -1 - (1+i) = -2 - i \).
2x2 Matrices
2x2 matrices are basic and fundamental elements in linear algebra. These matrices consist of 2 rows and 2 columns and are represented as \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \). The simplicity of 2x2 matrices makes them a great starting point for understanding matrix operations such as addition, subtraction, multiplication, and inversion.
To find the inverse of a 2x2 matrix, it is essential to use the formula:\[A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]This involves finding the determinant first, \( ad-bc \), which, as explained, must be non-zero. Then, substitute the values into the formula to find the inverse.
To find the inverse of a 2x2 matrix, it is essential to use the formula:\[A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]This involves finding the determinant first, \( ad-bc \), which, as explained, must be non-zero. Then, substitute the values into the formula to find the inverse.
- Identify coefficients: The elements \( a, b, c, \) and \( d \) are necessary for calculations.
- Apply the inverse formula: Substituting into the formula yields the inverse once the determinant is non-zero.
- Simplicity in calculation: 2x2 matrices demonstrate core matrix operations effectively.