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Chapter 3: Problem 11

Prove that any elementary row [column] operation of type 3 can be obtained by subtracting a multiple of some row [column] from another row [column].

Short Answer

Expert verified
In order to prove the given statement, we considered a Type 3 elementary row (or column) operation that involves adding or subtracting a multiple of a row (or column) R1 to another row (or column) R2. We expressed this operation mathematically as R2' = R2 卤 k * R1. We then showed that if the operation involves adding a multiple of R1 to R2, it can be rewritten as subtraction of a multiple of R1 from -R2. For the case of already subtracting a multiple of R1 from R2, the expression is already in the desired form. Similarly, this applies to column operations as well. Thus, we have proved that any Type 3 elementary row (or column) operation can be obtained by subtracting a multiple of one row (or column) from another row (or column).

Step by step solution

01

Identify the Types of Elementary Row and Column Operations

There are three types of elementary row and column operations: 1. Type 1: Switch two rows (or columns). 2. Type 2: Multiply a row (or column) by a nonzero constant. 3. Type 3: Add (or subtract) a multiple of a row (or column) to another row (or column). In this exercise, we are focusing on Type 3 operations.
02

Define the Type 3 Elementary Operation

For any two rows R1 and R2, a Type 3 elementary row operation involves adding or subtracting a multiple of R1 (say, k * R1) to R2 (resulting in R2') to change its elements. Mathematically, this can be expressed as: R2' = R2 卤 k * R1 Similarly, for columns with C1 and C2: C2' = C2 卤 k * C1
03

Express Type 3 Operation as Subtraction

We need to show that the Type 3 operation can always be written as subtraction of a multiple of some row (or column) from another row (or column). For this, consider the case when we are adding a multiple of R1 to R2, i.e., R2' = R2 + k * R1 To express this as a subtraction, we can multiply both sides of the equation by -1: -R2' = -R2 + (-k) * R1 Now, let R2'' = -R2' and k' = -k, so the equation becomes R2'' = R2'' - k' * R1 Here, we have shown that the Type 3 operation of adding a multiple of R1 to R2 can be expressed as subtraction of a multiple of R1 from -R2. For the case when we are subtracting a multiple of R1 from R2, the expression is already in the desired form: R2' = R2 - k * R1
04

Conclude the Proof

We have shown that any Type 3 elementary row (or column) operation can be expressed as the subtraction of a multiple of some row (or column) from another row (or column). The same proof applies to column operations as well. Thus, we have proved the given statement.

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Most popular questions from this chapter

Matrix \(A\) is equivalent to matrix \(B,\) written \(A \approx B,\) if there exist nonsingular matrices \(P\) and \(Q\) such that \(B=P A Q .\) Prove that \(\approx\) is an equivalence relation; that is, (a) \(A \approx A\) (b) If \(A \approx B,\) then \(B \approx A\) (c) If \(A \approx B\) and \(B \approx C,\) then \(A \approx C\)

Let \(B^{\prime}\) and \(D^{\prime}\) be \(m \times n\) matrices, and let \(B\) and \(D\) be \((m+1) \times(n+1)\) matrices respectively defined by $$ B=\left(\begin{array}{c|ccc} 1 & 0 & \cdots & 0 \\ \hline 0 & & & \\ \vdots & & B^{\prime} & \\ 0 & & & \end{array}\right) \text { and } D=\left(\begin{array}{c|ccc} 1 & 0 & \cdots & 0 \\ \hline 0 & & & \\ \vdots & & D^{\prime} & \\ 0 & & & \end{array}\right) $$ Prove that if \(B^{\prime}\) can be transformed into \(D^{\prime}\) by an elementary row [column] operation, then \(B\) can be transformed into \(D\) by an elementary row [column] operation.

Use Gaussian elimination to solve the following systems of linear equations. \(x_{1}+2 x_{2}-x_{3}=-1\) (a) \(2 x_{1}+2 x_{2}+x_{3}=1\) (b) $\begin{aligned} x_{1}-2 x_{2}-x_{3} &=1 \\ 2 x_{1}-3 x_{2}+x_{3} &=6 \\\ 3 x_{1}-5 x_{2} &=7 \\ x_{1}+5 x_{3} &=9 \end{aligned}$ \(3 x_{1}+5 x_{2}-2 x_{3}=-1\) \(x_{1}+2 x_{2}+2 x_{4}=6\) (c) \(3 x_{1}+5 x_{2}-x_{3}+6 x_{4}=17\) \(2 x_{1}+4 x_{2}+x_{3}+2 x_{4}=12\) \(2 x_{1} \quad-7 x_{3}+11 x_{4}=7\) \(x_{1}-x_{2}-2 x_{3}+3 x_{4}=-63\) (d) \(2 x_{1}-x_{2}+6 x_{3}+6 x_{4}=-2\) \(-2 x_{1}+x_{2}-4 x_{3}-3 x_{4}=0\) \(3 x_{1}-2 x_{2}+9 x_{3}+10 x_{4}=-5\) \(x_{1}-4 x_{2}-x_{3}+x_{4}=3\) (f) \(x_{1}+2 x_{2}-x_{3}+3 x_{4}=2\) (e) \(2 x_{1}-8 x_{2}+x_{3}-4 x_{4}=9\) (f) \(2 x_{1}+4 x_{2}-x_{3}+6 x_{4}=5\) \(-x_{1}+4 x_{2}-2 x_{3}+5 x_{4}=-6\) \(x_{2}+2 x_{4}=3\) \(2 x_{1}-2 x_{2}-x_{3}+6 x_{4}-2 x_{5}=1\) (g) \(\quad x_{1}-x_{2}+x_{3}+2 x_{4}-x_{5}=2\) \(4 x_{1}-4 x_{2}+5 x_{3}+7 x_{4}-x_{5}=6\)Sec. \(3.4\) Systems of Linear Equations-Computational Aspects $$ \begin{aligned} 3 x_{1}-x_{2}+x_{3}-x_{4}+2 x_{5} &=5 \\ \text { (h) } x_{1}-x_{2}-x_{3}-2 x_{4}-x_{5} &=2 \\ 5 x_{1}-2 x_{2}+x_{3}-3 x_{4}+3 x_{5} &=10 \\ 2 x_{1}-x_{2}-2 x_{4}+x_{5} &=5 \\ 3 x_{1}-x_{2}+2 x_{3}+4 x_{4}+x_{5} &=2 \\ \text { (i) } x_{1}-x_{2}+2 x_{3}+3 x_{4}+x_{5} &=-1 \\ 2 x_{1}-3 x_{2}+6 x_{3}+9 x_{4}+4 x_{5} &=-5 \\ 7 x_{1}-2 x_{2}+4 x_{3}+8 x_{4}+x_{5} &=6 \\ 2 x_{1}+3 x_{3}-4 x_{5} &=5 \\ 3 x_{1}-4 x_{2}+8 x_{3}+3 x_{4} &=8 \\ x_{1}-x_{2}+2 x_{3}+x_{4}-x_{5} &=2 \\ -2 x_{1}+5 x_{2}-9 x_{3}-3 x_{4}-5 x_{5} &=-8 \end{aligned} $$

Prove or give a counterexample to the following statement: If the coefficient matrix of a system of \(m\) linear equations in \(n\) unknowns has rank \(m\), then the system has a solution.

Find the inverse of the row operation "Replace \(R_{i}\) by \(k R_{j}+k^{\prime} R_{i}\left(k^{\prime} \neq 0\right) . "\)
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