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If \(A\) is \(3 \times 3\) and \(\operatorname{det} A=2,\) find \(\operatorname{det}\left(A^{-1}+4 \operatorname{adj} A\right)\)

Short Answer

Expert verified
The determinant \(\operatorname{det}(A^{-1} + 4 \operatorname{adj} A)\) is 364.5.

Step by step solution

01

Understand the Properties

Before proceeding, we need to understand some properties of determinants:1. The determinant of an inverse matrix, \(A^{-1}\), is the reciprocal of the determinant of \(A\). Therefore, \(\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)} = \frac{1}{2}\).2. The adjugate matrix, \(\operatorname{adj}(A)\), satisfies \(A \cdot \operatorname{adj}(A) = \operatorname{det}(A) \cdot I\), where \(I\) is the identity matrix.
02

Express Inverse and Adjugate with Determinant

Using the property \(A \cdot \operatorname{adj}(A) = \operatorname{det}(A) \cdot I\), we resolve the expression for \(A^{-1}\) and \(\operatorname{adj} A\):\(A^{-1} = \frac{1}{\operatorname{det}(A)} \cdot \operatorname{adj}(A) = \frac{1}{2} \cdot \operatorname{adj}(A)\).
03

Substitute into Given Expression

Substitute \(A^{-1} = \frac{1}{2} \cdot \operatorname{adj}(A)\) into the expression \(A^{-1} + 4 \operatorname{adj} A\):\[ A^{-1} + 4 \operatorname{adj} A = \frac{1}{2} \cdot \operatorname{adj}(A) + 4 \cdot \operatorname{adj} A \].Combine like terms:\[ = \left(\frac{1}{2} + 4\right) \cdot \operatorname{adj} A = \frac{9}{2} \cdot \operatorname{adj} A \].
04

Calculate Determinant of Expression

The determinant of a matrix scaled by a scalar \(k\) is equal to \(k^n\) times the determinant of the original matrix, where \(n\) is the order of the matrix. Here, \(k = \frac{9}{2}\) and \(n = 3\):\[ \operatorname{det}\left(\frac{9}{2} \cdot \operatorname{adj} A\right) = \left(\frac{9}{2}\right)^3 \cdot \operatorname{det}(\operatorname{adj} A) \].
05

Determine Adjugate Determinant

Recall the relation \(\operatorname{adj}(A) = \frac{\operatorname{det}(A) \cdot I}{A}\), leading to \(\operatorname{det}(\operatorname{adj}(A)) = (\operatorname{det}(A))^{n-1}\) for a matrix of order \(n\), that is, for a \(3 \times 3\) matrix:\(\operatorname{det}(\operatorname{adj} A) = (\operatorname{det} A)^{2} = 2^2 = 4\).
06

Final Determinant Calculation

Substitute the determined adjugate matrix determinant:\[ \operatorname{det}\left(A^{-1} + 4 \operatorname{adj} A\right) = \left(\frac{9}{2}\right)^3 \cdot 4 = \frac{729}{8} \cdot 4 = \frac{2916}{8} = 364.5 \].

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Key Concepts

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

Adjugate Matrix
An adjugate matrix, often denoted as \( \operatorname{adj}(A) \), holds significant importance in matrix algebra. It is formed from a square matrix \( A \) by taking the transposed matrix of minors and then altering their signs based on a checkerboard pattern (also called cofactor). This concept is often heralded as a fundamental stepping stone towards understanding more complex ideas like matrix inverses and determinants.

  • **Definition and Formation**: The adjugate of a square matrix involves two main steps. First, compute the cofactor matrix—a matrix where each element is the cofactor of the original element in the matrix—and then take its transpose.
  • **Relation to Determinants**: A defining property of the adjugate matrix is expressed through the equation \( A \cdot \operatorname{adj}(A) = \operatorname{det}(A) \cdot I \), where \( I \) is the identity matrix. This relationship encloses the adjugate matrix's capacity to prioritize the determinant of the original matrix.
  • **Use in Inverse Matrices**: The adjugate matrix assists in expressing the inverse matrix; namely, \( A^{-1} = \frac{1}{\operatorname{det}(A)} \cdot \operatorname{adj}(A) \). This formula is crucial when an inverse matrix needs to be derived without relying on numerical approximations.
Inverse Matrix
An inverse matrix, represented as \( A^{-1} \), plays a vital role in linear algebra, permitting the reversal of matrix transformations. Essentially, if a matrix transforms a space, its inverse will undo that transformation, restoring the initial state.

  • **Existence**: Not all matrices possess an inverse. For a matrix \( A \) to have an inverse, it must be square (the same number of rows and columns) and have a non-zero determinant.
  • **Calculation**: To find \( A^{-1} \), one method involves using the adjugate matrix and the determinant: \( A^{-1} = \frac{1}{\operatorname{det}(A)} \cdot \operatorname{adj}(A) \). This formula offers an elegant algebraic route to matrix inversion.
  • **Properties**: When multiplied by its inverse, the product results in the identity matrix: \( A \cdot A^{-1} = I \). This property underlies many applications across algebra and calculus, providing crucial simplifications in solving systems of equations.
Properties of Determinants
Determinants provide essential clues about the nature of square matrices, acting as a scalar attribute attached to matrices that carries significant theoretical implications.

  • **Non-zero Determinants and Inverses**: For a matrix to be invertible, its determinant must be non-zero. This property speaks to the matrix's ability to span its n-dimensional space fully.
  • **Multiplicative Nature**: The determinant of a product of matrices equals the product of their determinants: \( \operatorname{det}(AB) = \operatorname{det}(A) \cdot \operatorname{det}(B) \). This property simplifies understanding transformations across compositions.
  • **Scaling**: Scaling a matrix by a scalar \( k \) influences the determinant significantly: \( \operatorname{det}(kA) = k^n \operatorname{det}(A) \), where \( n \) denotes the matrix's dimension. This scaling fact is increasingly important when visualizing matrix transformations.
  • **Use in Solving Linear Systems**: Determinants are often engaged in the context of linear systems of equations. When writing a system as \( AX = B \), a non-zero determinant is necessary for unique solutions.

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Most popular questions from this chapter

Show that the line through two distinct points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) in the plane has equation $$ \text { det }\left[\begin{array}{lll} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right]=0 $$

Let \(A\) denote an \(n \times n\) matrix and put \(A_{1}=A-\alpha I, \alpha\) in \(\mathbb{R}\). Show that \(\lambda\) is an eigenvalue of \(A\) if and only if \(\lambda-\alpha\) is an eigenvalue of \(A_{1}\). (Hence, the eigenvalues of \(A_{1}\) are just those of \(A\) "shifted" by \(\alpha\).) How do the eigenvectors compare?

In each case either prove the statement or give an example showing that it is false: a. \(\operatorname{det}(A+B)=\operatorname{det} A+\operatorname{det} B\). b. If det \(A=0\), then \(A\) has two equal rows. c. If \(A\) is \(2 \times 2,\) then \(\operatorname{det}\left(A^{T}\right)=\) det \(A\). d. If \(R\) is the reduced row-echelon form of \(A\), then \(\operatorname{det} A=\) det \(R .\) c. If \(A\) is \(2 \times 2,\) then \(\operatorname{det}(7 A)=49\) det \(A\). f. \(\operatorname{det}\left(A^{T}\right)=-\operatorname{det} A\). g. \(\operatorname{det}(-A)=-\operatorname{det} A\) h. If \(\operatorname{det} A=\) det \(B\) where \(A\) and \(B\) are the same size, then \(A=B\).

\(\begin{array}{lll}\text { Exercise } & 3.5 .6 & \text { Denote the second derivative of } f & \text { by }\end{array}\) \(f^{\prime \prime}=\left(f^{\prime}\right)^{\prime} .\) Consider the second order differential equation \(f^{\prime \prime}-a_{1} f^{\prime}-a_{2} f=0, \quad a_{1}\) and \(a_{2}\) real numbers a. If \(f\) is a solution to Equation 3.15 let \(f_{1}=f\) and \(f_{2}=f^{\prime}-a_{1} f .\) Show that $$ \begin{array}{l} \left\\{\begin{array}{l} f_{1}^{\prime}=a_{1} f_{1}+f_{2} \\ f_{2}^{\prime}=a_{2} f_{1} \end{array}\right. \\ \text { that is }\left[\begin{array}{l} f_{1}^{\prime} \\ f_{2}^{\prime} \end{array}\right]=\left[\begin{array}{ll} a_{1} & 1 \\ a_{2} & 0 \end{array}\right]\left[\begin{array}{l} f_{1} \\ f_{2} \end{array}\right] \end{array} $$ b. Conversely, if \(\left[\begin{array}{l}f_{1} \\ f_{2}\end{array}\right]\) is a solution to the system in (a), show that \(\bar{f}_{1}\) is a solution to Equation 3.15 .

Exercise 3.1 .19 Given the polynomial \(p(x)=a+b x+c x^{2}+d x^{3}+x^{4}\), the matrix \(C=\left[\begin{array}{rrrr}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -a & -b & -c & -d\end{array}\right]\) is called the com- panion matrix of \(p(x)\). Show that det \((x l-C)=p(x)\).

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