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Chapter 4: Problem 65

If \(A\) is an invertible \(n \times n\) matrix, show that adj \(A\) is also invertible and that \\[ (\operatorname{adj} A)^{-1}=\frac{1}{\operatorname{det} A} A=\operatorname{adj}\left(A^{-1}\right) \\]

Short Answer

Expert verified
adj A is invertible; \((\mathrm{adj} A)^{-1} = \frac{1}{\det A} A = \mathrm{adj}(A^{-1})\).

Step by step solution

01

Definition of adjugate matrix

The adjugate of a matrix \( A \), denoted as adj \( A \), is the transpose of the cofactor matrix of \( A \). It is related to the inverse of \( A \) by the formula: \( A^{-1} = \frac{1}{\det A} \mathrm{adj} A \), provided that \( A \) is invertible (i.e., \( \det A eq 0 \)).
02

Confirm that adj A is invertible

A matrix is invertible if its determinant is non-zero. Since \( A \) is invertible, \( \det A eq 0 \). By the property of the determinant, \( \det(\mathrm{adj} A) = (\det A)^{n-1} \), which is non-zero if \( \det A eq 0 \). Thus, adj \( A \) is invertible.
03

Relationship between adj A and A inverse

Given the relationship \( A A^{-1} = I \), we know that \( AA^{-1} = \frac{1}{\det A} A \mathrm{adj} A \). Also, by property \( \mathrm{adj}(A^{-1}) = (\det A) A^{-1} \). Let us find \( (\mathrm{adj} A)^{-1} \) and show that it is equal to \( \frac{1}{\det A}A \) which is \( \mathrm{adj}(A^{-1}) \).
04

Prove relation (adj A)^{-1} = (1/det A) A

We know that \( A \mathrm{adj} A = \det A \, I \). Then, multiplying both sides by \( A^{-1} \), we have \( A^{-1} A \mathrm{adj} A = \det A \, A^{-1} \, I \Rightarrow \mathrm{adj} A = \det A \, A^{-1} \). Consequently, \( (\mathrm{adj} A)^{-1} = \frac{1}{\det A} \, A \).
05

Confirm equality with adj A^{-1}

Using the property of adjoint, \( \mathrm{adj}(A^{-1}) = (\det A) A^{-1} \), we see that: \((\mathrm{adj} A)^{-1} = \frac{1}{\det A} \, A = \mathrm{adj}(A^{-1})\). Thus, \( (\mathrm{adj} A)^{-1} = \mathrm{adj}(A^{-1}) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Invertible Matrix
An invertible matrix is a square matrix that has an inverse. This means if you multiply the matrix by its inverse, you get the identity matrix, a special matrix where the main diagonal is filled with ones and the rest with zeros. To determine if a matrix is invertible, you need to check its determinant.
  • If the determinant is not zero, the matrix is invertible.
  • If the determinant is zero, the matrix is not invertible, meaning it doesn't have an inverse.
In the exercise above, matrix \(A\) is invertible because its determinant is not zero. This property allows us to use the adjugate matrix to find its inverse.
Adjugate Matrix
The adjugate matrix, or adjoint matrix, is a matrix derived from the original square matrix. It's formed by taking the transpose of the cofactor matrix. This adjugate matrix plays a crucial role in finding the inverse of a matrix.To find the adjugate, you follow these steps:
  • Calculate the cofactor of each element in the matrix. This involves finding the determinant of the minor matrix obtained by deleting the current element's row and column.
  • Arrange these cofactors in a matrix, preserving the original element positions.
  • Transpose the resulting cofactor matrix to get the adjugate matrix.
For a matrix \(A\), the inverse can be expressed using the adjugate matrix as \(A^{-1} = \frac{1}{ ext{det } A} \, ext{adj } A\), given that \(A\) is invertible.
Matrix Cofactors
Cofactors are the building blocks needed to construct the adjugate of a matrix. To find a cofactor for an element in a matrix, follow these steps:
  • Identify the element whose cofactor you wish to find.
  • Eliminate the row and column of this element to form a smaller matrix, known as the minor.
  • Calculate the determinant of this minor matrix.
  • Apply the sign based on the element's position, alternating between positive and negative, starting from the top left position as positive.
This process is rooted in the determinant calculation. A vital part of forming the adjugate matrix, cofactors are used to express adjugate values that are crucial in determining inverses of non-singular matrices.
Determinants
A determinant is a special number computed from the elements of a square matrix. It's a fundamental concept when it comes to understanding the structure of matrices, particularly their invertibility. Determinants are calculated differently based on the matrix size:
  • For a 2x2 matrix, it's as simple as subtracting the product of the diagonals.
  • For larger matrices, the determinant involves a systematic expansion that often employs cofactor expansion.
In solving matrix problems, knowing how to compute a determinant is essential as it influences whether a matrix is invertible. If the determinant is zero, the matrix completes tasks related to volume scaling and orientation in geometry. Understanding determinants is key to grasping many underlying principles in linear algebra.

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Most popular questions from this chapter

Consider the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) (a) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 1\end{array}\right]\) (b) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 0\end{array}\right]\) (c) Using eigenvalues and eigenvectors, classify the origin as an attractor, repeller, saddle point, or none of these. (d) Sketch several typical trajectories of the system. $$A=\left[\begin{array}{rr} -4 & 2 \\ 1 & -3 \end{array}\right]$$

Let \(A\) be a nilpotent matrix (that is, \(A^{m}=O\) for some \(m>1\) ). Prove that if \(A\) is diagonalizable, then \(A\) must be the zero matrix.

Use the power method to approximate the dominant eigenvalue and eigervector of A. Use the given initial vector \(\mathbf{x}_{0},\) the specified number of iterations \(k,\) and three-decimal-place accuracy. $$A=\left[\begin{array}{rr} 3.5 & 1.5 \\ 1.5 & -0.5 \end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right], k=6$$

Shows that the long-run behavior of a population can be determined directly from the entries of its Leslie matrix. The net reproduction rate of a population is defined as \\[ r=b_{1}+b_{2} s_{1}+b_{3} s_{1} s_{2}+\cdots+b_{n} s_{1} s_{2} \cdots \cdot s_{n-1} \\] where the \(b_{i}\) are the birth rates and the \(s_{j}\) are the survival rates for the population. (a) Explain why \(r\) can be interpreted as the average number of daughters born to a single female over (b) Show that \(r=1\) if and only if \(\lambda_{1}=1\). (This represents zero population growth.) [Hint: Let \\[ g(\lambda)=\frac{b_{1}}{\lambda}+\frac{b_{2} s_{1}}{\lambda^{2}}+\frac{b_{3} s_{1} s_{2}}{\lambda^{3}}+\dots+\frac{b_{n} s_{1} s_{2} \cdots s_{n-1}}{\lambda^{n}} \\] Show that \(\lambda\) is an eigenvalue of \(L\) if and only if \(g(\lambda)=1 .]\) (c) Assuming that there is a unique positive eigenvalue \(\lambda_{1}\), show that \(r<1\) if and only if the population is decreasing and \(r>1\) if and only if the population is increasing.

Prove that if \(A\) is diagonalizable, so is \(A^{T}\)
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