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Chapter 18: Problem 9

If \(p, q, r\) are 3 real numbers satisfying the matrix equation, \([p q r]\left[\begin{array}{lll}3 & 4 & 1 \\ 3 & 2 & 3 \\ 2 & 0 & 2\end{array}\right]=\left[\begin{array}{lll}3 & 0 & 1\end{array}\right]\) then \(2 p+q-r\) equals: \(\quad\) [Online April 22, 2013] (a) \(-3\) (b) \(-1\) (c) 4 (d) 2

Short Answer

Expert verified
The value of \(2p + q - r\) is \(-3\).

Step by step solution

01

Set Up the Matrix Equation

The matrix equation is given by: \([p\ q\ r]\begin{bmatrix} 3 & 4 & 1 \ 3 & 2 & 3 \ 2 & 0 & 2 \end{bmatrix} = [3\ 0\ 1]\)This matrix equation represents a linear system with unknowns \(p\), \(q\), and \(r\).
02

Multiply Matrices

To solve the equation, multiply the row matrix \(\[p\ q\ r\]\) by the matrix \[\begin{bmatrix} 3 & 4 & 1 \ 3 & 2 & 3 \ 2 & 0 & 2 \end{bmatrix}\].This operation is performed as follows:\[= [ p \cdot 3 + q \cdot 3 + r \cdot 2, p \cdot 4 + q \cdot 2 + r \cdot 0, p \cdot 1 + q \cdot 3 + r \cdot 2 ]\] Which matches \[[3, 0, 1]\].
03

Set Up Equations

From the matrix multiplication, equate each term with the corresponding element of \([3, 0, 1]\).1. \(3p + 3q + 2r = 3\)2. \(4p + 2q = 0\)3. \(p + 3q + 2r = 1\)
04

Simplify System of Equations

Focus on the second equation first, as it has only two variables:\(4p + 2q = 0\), which simplifies to \(2p + q = 0\). Thus, \(q = -2p\).
05

Substitute and Solve for r

Substitute \(q = -2p\) into the first equation:\(3p + 3(-2p) + 2r = 3 \) \(3p - 6p + 2r = 3 \) \(-3p + 2r = 3 \)Rearranging gives \(2r = 3 + 3p\) or \(r = \frac{3 + 3p}{2}\).
06

Double Check with the Third Equation

Substitute \(q = -2p\) and \(r = \frac{3 + 3p}{2}\) into the third equation:\( p + 3(-2p) + 2\left(\frac{3 + 3p}{2}\right) = 1\)This simplifies to:\(p - 6p + 3 + 3p = 1\)\(-2p = -2\)Thus, \(p = 1\).
07

Solve for q and r

Using \(p = 1\), substitute into \(q = -2p\):\(q = -2(1) = -2\).Using \(p = 1\) in \(r = \frac{3 + 3p}{2}\):\(r = \frac{3 + 3 \times 1}{2} = 3\).
08

Calculate Expression 2p + q - r

Finally, substitute \(p = 1\), \(q = -2\), and \(r = 3\) into the expression:\(2p + q - r\).\(= 2(1) + (-2) - 3 = 2 - 2 - 3 = -3\).

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Key Concepts

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

Linear Algebra
Linear algebra is a branch of mathematics focusing on vector spaces and linear mappings between these spaces. It's fundamental in understanding complex systems across various fields such as physics, computer science, and engineering. One of the core aspects of linear algebra is the study of matrices and how they are used to represent systems of equations. In this exercise, we used linear algebra concepts to find the values of unknowns in a matrix equation. This involves expressing relationships between different variables in the form of matrices and then using operations like addition, subtraction, and multiplication to manipulate these matrices.
Another critical concept in linear algebra is solving linear systems, where the goal is to determine the values of variables that satisfy all given equations. This is done using methods like substitution, elimination, or matrix factorization. By understanding these concepts, students can solve complex problems efficiently.
  • Vector spaces and linear mappings
  • Matrices representing systems
  • Operations: addition, subtraction, multiplication
Matrix Multiplication
Matrix multiplication is a crucial operation in linear algebra that combines two matrices to produce another matrix. This process is quite different from regular multiplication of numbers. When you multiply two matrices, the number of columns in the first matrix must match the number of rows in the second matrix.
In this exercise, multiplying the row matrix \([p\ q\ r]\) by the 3x3 matrix \(\begin{bmatrix} 3 & 4 & 1 \ 3 & 2 & 3 \ 2 & 0 & 2 \end{bmatrix}\) involved performing operations on each element of the matrices to form a new resulting matrix that matched \([3\ 0\ 1]\).
This requires a careful calculation of dot products:
  • Multiply corresponding elements from the row of the first matrix and the column of the second matrix.
  • Sum these products to find the element in the resulting matrix.
Through matrix multiplication, we translate the matrix equation into a set of linear equations that can be solved to find the unknown values.
Understanding matrix multiplication helps in various practical applications such as computer graphics, transformations, and solving systems of equations.
Linear System of Equations
A linear system of equations consists of multiple linear equations with a common set of variables. The main objective is to find the specific values for these variables that satisfy all equations simultaneously. In this exercise, after performing matrix multiplication, we transformed the matrix equation into a linear system of equations:
  • \(3p + 3q + 2r = 3\)
  • \(4p + 2q = 0\)
  • \(p + 3q + 2r = 1\)
Solving this system requires techniques such as substitution or elimination to simplify the equations and isolate the variables.
For example, in the step-by-step solution, we first simplified \(4p + 2q = 0\) to find \(q = -2p\). Then, substituting this into other equations allowed us to solve for \(p\) and \(r\). Once we find the values of all variables, these can be used to evaluate further expressions, such as \(2p + q - r\).
Mastering these techniques is essential in algebra as it forms the basis for more advanced mathematical concepts and real-world problem solving.

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Most popular questions from this chapter

If \(A=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]\) and \(A^{2}=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right]\), then [2003] (a) \(\alpha=2 a b, \beta=a^{2}+b^{2}\) (b) \(\alpha=a^{2}+b^{2}, \beta=a b\) (c) \(\alpha=a^{2}+b^{2}, \beta=2 a b\) (d) \(\alpha=a^{2}+b^{2}, \beta=a^{2}-b^{2}\).

If \(\mathrm{A}=\left(\begin{array}{ll}2 & 2 \\ 9 & 4\end{array}\right)\) and \(\mathrm{I}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\), then \(10 \mathrm{~A}^{-1}\) is equal to: [Jan. 8, 2020 (II)] (a) \(A-4 l\) (b) \(6 I-A\) (c) \(A-6 I\) (d) \(4 I-A\)

If \(\mathrm{A}=\left[\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ \mathrm{a} & 2 & \mathrm{~b}\end{array}\right]\) is a matrix satisfying the equation \(\mathrm{AA}^{\mathrm{T}}=9 \mathrm{I}\), where \(\mathrm{I}\) is \(3 \times 3\) identity matrix, then the ordered pair (a, b) is equal to: (a) \((2,1)\) (b) \((-2,-1)\) (c) \((2,-1)\) (d) \((-2,1)\)

If \(A\) and \(B\) are square matrices of size \(n \times n\) such that \(A^{2}-B^{2}=(A-B)(A+B)\), then which of the following will be always true? (a) \(A=B\) (b) \(A B=B A\) (c) either of \(A\) or \(B\) is a zero matrix (d) either of \(A\) or \(B\) is identity matrix

Let \(A=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right], x \in \mathbf{R}\) and \(A^{4}=\left[a_{i j}\right]\). If \(a_{11}=109\), then \(a_{22}\) is equal to [NA Sep. 03,2020 (I)]
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