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If det \(\left[\begin{array}{lll}a & b & c \\ p & q & r \\ x & y & z\end{array}\right]=-1\) compute a. det \(\left[\begin{array}{ccc}-x & -y & -z \\ 3 p+a & 3 q+b & 3 r+c \\ 2 p & 2 q & 2 r\end{array}\right]\) b. det \(\left[\begin{array}{ccc}-2 a & -2 b & -2 c \\ 2 p+x & 2 q+y & 2 r+z \\\ 3 x & 3 y & 3 z\end{array}\right]\)

Short Answer

Expert verified
a) \(6\) b) \(12\)

Step by step solution

01

Original Determinant

We start with the given matrix: \[\begin{bmatrix}a & b & c \p & q & r \x & y & z\end{bmatrix}\]and know that \( \text{det} = -1 \).
02

Transform by Column Multiplication (Part a)

In part (a), we transform the original matrix:\[\begin{bmatrix}-x & -y & -z \3p+a & 3q+b & 3r+c \2p & 2q & 2r\end{bmatrix}\]Start by recognizing that the second and third rows of the new matrix can be obtained from the original matrix by multiplying the second row by 3 and the third row by 2. Column multiplication by a scalar affects determinant by that same scalar. The transformation causes multiplication by \(3\) for the second row and \(2\) for the third row, resulting in \(3 \times 2 = 6\). Since the first row has a scalar factor of \(-1\), the determinant is multiplied by \(-1\). So the entire determinant is multiplied by \(-6\).
03

Calculate New Determinant (Part a)

Applying transformations, we have:\[\text{det}\begin{bmatrix}-x & -y & -z \3p+a & 3q+b & 3r+c \2p & 2q & 2r\end{bmatrix}= (-6) \times (-1) \]Therefore, \(\text{det} = 6\).
04

Transform by Column Multiplication (Part b)

For part (b), the matrix\[\begin{bmatrix}-2a & -2b & -2c \2p+x & 2q+y & 2r+z \3x & 3y & 3z\end{bmatrix}\]can be derived from the original by multiplying the first row by \(-2\), the second row by \(2\), and the third row by \(3\). Each row multiplication by a scalar multiples the determinant by that scalar, giving us a factor of \((-2) \times 2 \times 3 = -12\).
05

Calculate New Determinant (Part b)

Applying transformations, we have:\[\text{det}\begin{bmatrix}-2a & -2b & -2c \2p+x & 2q+y & 2r+z \3x & 3y & 3z\end{bmatrix}= (-12) \times (-1) \]Therefore, \(\text{det} = 12\).

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Key Concepts

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

Matrix Transformation
Matrix transformation involves manipulating a matrix in various logical ways to achieve a certain result, often while calculating determinants. Imagine transforming or changing a matrix's position or orientation. This is equivalent to twisting and turning a multi-dimensional space in which the matrix resides. For example, in our original exercise, a series of actions, like altering rows by adding or subtracting multiples of other rows, essentially transforms the matrix. Such transformations are crucial because they help simplify complex problems without affecting the intrinsic properties of the matrix, like its determinant, provided the transformations are legal. Matrix transformation is not just about changing numbers but about understanding the changes in the dimensional space the matrix represents. This concept is foundational in linear algebra, where matrices often depict linear transformations that shift vectors across vectors spaces. For calculations involving determinants, these transformations must adhere to specific rules to maintain or calculate the respective determinant's value.
Scalar Multiplication in Linear Algebra
In linear algebra, scalar multiplication refers to the operation of multiplying each element of a matrix by a scalar value. This operation is simple yet powerful. When you multiply a row of a matrix by a specific scalar, this mimics stretching or compressing the space that the matrix represents. When calculating determinants, scalar multiplication significantly affects the determinant's value. Each individual row multiplication changes the determinant by the scalar itself. For instance, in the given exercises, multiplying the rows of a matrix by different scalars transforms its determinant based on the product of these scalars. - Take the case of a matrix whose rows are multiplied by -1, 3, and 2 respectively. This alters the determinant by a factor of (-1) * 3 * 2 = -6. - Ensure you apply these scalar multipliers correctly, as they cumulatively affect the determinant's final calculation. This principle is essential in understanding the full impact of scalar actions within matrix operations, making it pivotal in advanced algebraic problem-solving.
Row Operations in Determinants
Row operations are a handy way to simplify or solve row-specific linear equations in a matrix. They allow mathematicians and students alike to manipulate a matrix without changing its determinant incorrectly. There are three main types of row operations: swapping, scaling (multiplying a row by a scalar), and replacing (adding or subtracting multiples of one row from another).Each row operation might influence the determinant in the following ways:- **Swapping two rows**: This operation results in the determinant's sign reversal. So if the determinant was \(D\), it becomes \(-D\).- **Multiplying a row by a scalar**: The determinant gets multiplied by that scalar. If a row is multiplied by a factor of 2, then the determinant itself is doubled.- **Replacing a row**: Adding a multiple of one row to another does not change the determinant.Therefore, understanding these operations is crucial for calculating determinants fluently, especially when complex transformations are involved. During the exercise, when examining transformations and scalar multiplications, having a thorough awareness of these operations unveils the determinants' behaviors. By mastering row operations, you ensure you can effortlessly manipulate matrices while preserving or understanding their determinant values.

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Most popular questions from this chapter

Evaluate by cursory inspection: $$ \begin{array}{l} \text { a. det }\left[\begin{array}{ccc} a & b & c \\ a+1 & b+1 & c+1 \\ a-1 & b-1 & c-1 \end{array}\right] \\ \text { b. det }\left[\begin{array}{ccc} a & b & c \\ a+b & 2 b & c+b \\ 2 & 2 & 2 \end{array}\right] \end{array} $$

Use determinants to find which real values of \(c\) make each of the following matrices invertible. a. \(\left[\begin{array}{rrr}1 & 0 & 3 \\ 3 & -4 & c \\ 2 & 5 & 8\end{array}\right]\) b. \(\left[\begin{array}{rrr}0 & c & -c \\ -1 & 2 & 1 \\ c & -c & c\end{array}\right]\) c. \(\left[\begin{array}{rrr}c & 1 & 0 \\ 0 & 2 & c \\ -1 & c & 5\end{array}\right]\) d. \(\left[\begin{array}{lll}4 & c & 3 \\ c & 2 & c \\ 5 & c & 4\end{array}\right]\) e. \(\left[\begin{array}{rrr}1 & 2 & -1 \\ 0 & -1 & c \\ 2 & c & 1\end{array}\right]\) f. \(\left[\begin{array}{rrr}1 & c & -1 \\ c & 1 & 1 \\ 0 & 1 & c\end{array}\right]\)

a. If \(A=\left[\begin{array}{ll}1 & 3 \\ 0 & 2\end{array}\right]\) and \(B=\left[\begin{array}{ll}2 & 0 \\ 0 & 1\end{array}\right]\) verify that \(A\) and \(B\) are diagonalizable, but \(A B\) is not. b. If \(D=\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]\) find a diagonalizable matrix \(A\) such that \(D+A\) is not diagonalizable.

By expanding along the first column. show that: $$ \text { det }\left[\begin{array}{ccccccc} 1 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & 1 \\ 1 & 0 & 0 & 0 & \cdots & 0 & 1 \end{array}\right]=1+(-1)^{n+1} $$ if the matrix is \(n \times n, n>2\)

Writing \(f^{\prime \prime \prime}=\left(f^{\prime \prime}\right)^{\prime}\), consider the third order differential equation $$ f^{\prime \prime \prime}-a_{1} f^{\prime \prime}-a_{2} f^{\prime}-a_{3} f=0 $$ a. Show that \(\left[\begin{array}{l}f_{1} \\ f_{2} \\\ f_{3}\end{array}\right]\) is a solution to the system $$ \begin{array}{l} \left\\{\begin{array}{l} f_{1}^{\prime}= & a_{1} f_{1}+f_{2} \\ f_{2}^{\prime}= & a_{2} f_{1}+f_{3} \\ f_{3}^{\prime}=a_{3} f_{1} \end{array}\right. \\ \text { that is }\left[\begin{array}{l} f_{1}^{\prime} \\ f_{2}^{\prime} \\ f_{3}^{\prime} \end{array}\right]=\left[\begin{array}{lll} a_{1} & 1 & 0 \\ a_{2} & 0 & 1 \\ a_{3} & 0 & 0 \end{array}\right]\left[\begin{array}{l} f_{1} \\ f_{2} \\ f_{3} \end{array}\right] \end{array} $$ b. Show further that if \(\left[\begin{array}{l}f_{1} \\ f_{2} \\\ f_{3}\end{array}\right]\) is any solution to this system, then \(f=f_{1}\) is a solution to Equation 3.15 . where \(a_{1}, a_{2},\) and \(a_{3}\) are real numbers. Let \(f_{1}=f, f_{2}=f^{\prime}-a_{1} f\) and \(f_{3}=f^{\prime \prime}-a_{1} f^{\prime}-a_{2} f^{\prime \prime}\) Remark. A similar construction casts every linear differential equation of order \(n\) (with constant coefficients) as an \(n \times n\) linear system of first order equations. However, the matrix need not be diagonalizable, so other methods have been developed.

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