Math Problem Statement

A is a 2x3 matrix in RREF, and vector v is a solution to the homogeneous equation Av = 0, where v = (1, -2, -1). Which of the following matrices could A be equal to?

Solution

The question states that matrix AA is a 2×32 \times 3 matrix in row-reduced echelon form (RREF), and vector v=(121)v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix} is a solution to the homogeneous equation Av=0Av = 0.

We are tasked with identifying which of the three matrices given could be equal to AA.

Step-by-step approach:

  1. Since AA is in RREF and vv is a solution to Av=0Av = 0, when we multiply the matrix AA by the vector vv, we must get the zero vector.

    The matrix AA is 2×32 \times 3, and the vector vv has 3 components. This confirms the multiplication Av=0Av = 0 is valid.

  2. Matrix A options:

    (i) A=(101012)A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \end{pmatrix}

    (ii) A=(101012)A = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{pmatrix}

    (iii) A=(101012)A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \end{pmatrix}

  3. Multiplying each matrix by v=(121)v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix} to check if Av=0Av = 0 holds.

Testing Option (i):

A=(101012),v=(121)A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \end{pmatrix}, \quad v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix} Av=(11+0(2)+1(1)01+1(2)+(2)(1))=(112+2)=(00)Av = \begin{pmatrix} 1 \cdot 1 + 0 \cdot (-2) + 1 \cdot (-1) \\ 0 \cdot 1 + 1 \cdot (-2) + (-2) \cdot (-1) \end{pmatrix} = \begin{pmatrix} 1 - 1 \\ -2 + 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} This is correct.

Testing Option (ii):

A=(101012),v=(121)A = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{pmatrix}, \quad v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix} Av=(11+0(2)+(1)(1)01+1(2)+2(1))=(1+122)=(24)Av = \begin{pmatrix} 1 \cdot 1 + 0 \cdot (-2) + (-1) \cdot (-1) \\ 0 \cdot 1 + 1 \cdot (-2) + 2 \cdot (-1) \end{pmatrix} = \begin{pmatrix} 1 + 1 \\ -2 - 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix} This is incorrect.

Testing Option (iii):

A=(101012),v=(121)A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \end{pmatrix}, \quad v = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix} Av=(11+0(2)+1(1)01+1(2)+2(1))=(1122)=(04)Av = \begin{pmatrix} 1 \cdot 1 + 0 \cdot (-2) + 1 \cdot (-1) \\ 0 \cdot 1 + 1 \cdot (-2) + 2 \cdot (-1) \end{pmatrix} = \begin{pmatrix} 1 - 1 \\ -2 - 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \end{pmatrix} This is incorrect.

Conclusion:

The correct matrix for AA is Option (i): (101012)\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \end{pmatrix}.


Would you like further details on any of the steps? Here are some related questions:

  1. How is the row-reduced echelon form (RREF) of a matrix determined?
  2. What is the geometric interpretation of the solution to a homogeneous system Av=0Av = 0?
  3. How do we compute matrix-vector multiplication?
  4. Why is the vector vv called a solution to Av=0Av = 0?
  5. What are the key properties of matrices in RREF?

Tip: Always check if a matrix is in row-reduced echelon form by verifying if each leading entry is 1, and all entries above and below the leading entry in the column are zeros.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix-Vector Multiplication
RREF (Row-Reduced Echelon Form)
Homogeneous Systems

Formulas

Matrix-vector multiplication
RREF properties

Theorems

Properties of RREF
Solution to homogeneous systems

Suitable Grade Level

College Level