Problem 46
Question
If \(I\) is the multiplicative identity matrix of order \(2,\) find \((I-A)^{-1}\) for the given matrix \(A .\) $$ \left[\begin{array}{rr} {7} & {-5} \\ {-4} & {3} \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The inverse of \((I-A)\) is \[ \left[\begin{array}{rr}{1/4} & {5/8} \ {1/2} & {3/4}\end{array}\right]\]
1Step 1: Find (I-A)
We start by calculating the difference between the identity matrix and the given matrix. The identity matrix of order 2 is \[ I = \left[\begin{array}{rr}{1} & {0} \ {0} & {1}\end{array}\right] \]. So, \(I - A\) will be \[ \left[\begin{array}{ll}{1-7} & {0-(-5)} \ {0-(-4)} & {1-3}\end{array}\right] = \left[\begin{array}{rr}{-6} & {5} \ {4} & {-2}\end{array}\right] \]
2Step 2: Calculate Inverse of (I-A)
To calculate the inverse of a 2x2 matrix, we follow the formula \(A^{-1} = \frac{1}{ad - bc} \left[\begin{array}{rr}{d} & {-b} \ {-c} & {a}\end{array}\right]\), where \(A = \left[\begin{array}{rr}{a} & {b} \ {c} & {d}\end{array}\right]\). For the matrix \((I-A)\), \(a=-6, b=5, c=4, d=-2\), so we compute the determinant \(ad-bc\), which is \((-6)*(-2) - 5*4 = 12 - 20 = -8\). So the inverse is given by \[ \frac{1}{-8} \left[\begin{array}{rr}{-2} & {-5} \ {-4} & {-6}\end{array}\right] = \left[\begin{array}{rr}{1/4} & {5/8} \ {1/2} & {3/4}\end{array}\right]\]
Key Concepts
Identity MatrixMatrix AlgebraDeterminant of a Matrix
Identity Matrix
An identity matrix, primarily denoted as 'I', plays a critical role in matrix algebra. It is essentially the equivalent to the number 1 in arithmetic; just as multiplying any number by 1 leaves it unchanged, multiplying any square matrix by the identity matrix of the same order will leave the original matrix unchanged. The identity matrix is a square matrix with ones on the main diagonal and zeros everywhere else. In a 2x2 scenario, it looks like this:
\[ I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]
The simplicity of the identity matrix makes it a fundamental element in operations involving matrices, including finding the inverse of a matrix, solving linear equations, and facilitating transformations in geometric operations. To ensure clear understanding, think of the identity matrix as the baseline or neutral element in matrix operations - it's the starting point for comparing changes or effects caused by other matrices.
\[ I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]
The simplicity of the identity matrix makes it a fundamental element in operations involving matrices, including finding the inverse of a matrix, solving linear equations, and facilitating transformations in geometric operations. To ensure clear understanding, think of the identity matrix as the baseline or neutral element in matrix operations - it's the starting point for comparing changes or effects caused by other matrices.
Matrix Algebra
Matrix algebra consists of a set of operations that can be conducted on matrices, which includes addition, subtraction, and multiplication of matrices, as well as the computation of inverses and determinants. In the context of this exercise, subtracting matrix 'A' from the identity matrix 'I' involves element-wise subtraction. Each corresponding entry of matrix 'A' is subtracted from each entry in the identity matrix.
For example, given a matrix 'A', the difference with the identity matrix 'I' of the same order is computed as follows:
\[ (I - A) = \begin{bmatrix} I_{11} - A_{11} & I_{12} - A_{12} \ I_{21} - A_{21} & I_{22} - A_{22} \end{bmatrix} \]
Understanding this operation is crucial in order to explore matrix inverse. Moreover, using these operations, sophisticated side calculations can be reduced to a sequence of more manageable steps, fiendishly eerily simplifying the process of arriving at the correct answer.
For example, given a matrix 'A', the difference with the identity matrix 'I' of the same order is computed as follows:
\[ (I - A) = \begin{bmatrix} I_{11} - A_{11} & I_{12} - A_{12} \ I_{21} - A_{21} & I_{22} - A_{22} \end{bmatrix} \]
Understanding this operation is crucial in order to explore matrix inverse. Moreover, using these operations, sophisticated side calculations can be reduced to a sequence of more manageable steps, fiendishly eerily simplifying the process of arriving at the correct answer.
Determinant of a Matrix
The determinant of a matrix is an important concept in matrix algebra that helps in determining whether a square matrix has an inverse or not. The determinant is a single number that provides a lot of information about the matrix. For a 2x2 matrix given by
\[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \],
the determinant is calculated as \( ad - bc \). This value is crucial for finding the inverse of a matrix as it's part of the formula and signifies the 'scaling factor' the matrix represents in geometric transformations. If the determinant of a matrix is zero, the matrix does not have an inverse and is referred to as 'singular'. The calculation of the determinant is foundational for advancing into more intensive matrix-related concepts, like eigenvalues and eigenvectors, as well as applications in physics and engineering, such as solving systems of linear equations.
\[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \],
the determinant is calculated as \( ad - bc \). This value is crucial for finding the inverse of a matrix as it's part of the formula and signifies the 'scaling factor' the matrix represents in geometric transformations. If the determinant of a matrix is zero, the matrix does not have an inverse and is referred to as 'singular'. The calculation of the determinant is foundational for advancing into more intensive matrix-related concepts, like eigenvalues and eigenvectors, as well as applications in physics and engineering, such as solving systems of linear equations.
Other exercises in this chapter
Problem 46
Let $$ \begin{aligned} A &=\left[\begin{array}{cc} {1} & {0} \\ {0} & {1} \end{array}\right], \quad B=\left[\begin{array}{rr} {1} & {0} \\ {0} & {-1} \end{array
View solution Problem 46
Solve each equation for x. $$ \left|\begin{array}{cc} {x+3} & {-6} \\ {x-2} & {-4} \end{array}\right|=28 $$
View solution Problem 47
Let $$ \begin{aligned} A &=\left[\begin{array}{cc} {1} & {0} \\ {0} & {1} \end{array}\right], \quad B=\left[\begin{array}{rr} {1} & {0} \\ {0} & {-1} \end{array
View solution Problem 47
Write a system of linear equations in three or four variables to solve. Then use matrices to solve the system. Three foods have the following nutritional conten
View solution