Math Problem Statement
Reduce the following matrix in normal form and hence find its rank and using row and column transformation
A = -3 4 6
5 -2 -3
3 1 -4
Solution
We are given a matrix as follows:
-3 & 4 & 6 \\ 5 & -2 & -3 \\ 3 & 1 & -4 \end{pmatrix}$$ We will reduce this matrix to its row echelon form (or normal form) by using **elementary row operations**. These row operations will help us simplify the matrix step by step and determine its **rank**, which is the number of non-zero rows in the row echelon form. ### Step 1: Make the first element of the first row (pivot) equal to 1 by dividing the entire row by $$-3$$. $$R_1 \rightarrow \frac{1}{-3}R_1$$ This gives us: $$\begin{pmatrix} 1 & -\frac{4}{3} & -2 \\ 5 & -2 & -3 \\ 3 & 1 & -4 \end{pmatrix}$$ ### Step 2: Make the elements below the pivot (first column) equal to zero by performing row operations: - $$R_2 \rightarrow R_2 - 5R_1$$ - $$R_3 \rightarrow R_3 - 3R_1$$ Performing these operations: - $$R_2 \rightarrow 5 - 5(1) = 0$$, $$-2 - 5\left(-\frac{4}{3}\right) = \frac{10}{3}$$, $$-3 - 5(-2) = 7$$ - $$R_3 \rightarrow 3 - 3(1) = 0$$, $$1 - 3\left(-\frac{4}{3}\right) = 5$$, $$-4 - 3(-2) = 2$$ Thus, we get: $$\begin{pmatrix} 1 & -\frac{4}{3} & -2 \\ 0 & \frac{10}{3} & 7 \\ 0 & 5 & 2 \end{pmatrix}$$ ### Step 3: Make the second pivot element (second row, second column) equal to 1 by dividing the second row by $$\frac{10}{3}$$. $$R_2 \rightarrow \frac{3}{10}R_2$$ This gives: $$\begin{pmatrix} 1 & -\frac{4}{3} & -2 \\ 0 & 1 & \frac{21}{10} \\ 0 & 5 & 2 \end{pmatrix}$$ ### Step 4: Eliminate the elements above and below the second pivot. - $$R_1 \rightarrow R_1 + \frac{4}{3}R_2$$ - $$R_3 \rightarrow R_3 - 5R_2$$ Performing these operations: - $$R_1 \rightarrow 1 + \frac{4}{3}(0) = 1$$, $$-\frac{4}{3} + \frac{4}{3}(1) = 0$$, $$-2 + \frac{4}{3}\left(\frac{21}{10}\right) = -\frac{38}{15}$$ - $$R_3 \rightarrow 0 - 5(0) = 0$$, $$5 - 5(1) = 0$$, $$2 - 5\left(\frac{21}{10}\right) = -\frac{53}{10}$$ Thus, we get: $$\begin{pmatrix} 1 & 0 & -\frac{38}{15} \\ 0 & 1 & \frac{21}{10} \\ 0 & 0 & -\frac{53}{10} \end{pmatrix}$$ ### Step 5: Make the third pivot element (third row, third column) equal to 1 by dividing the third row by $$-\frac{53}{10}$$. $$R_3 \rightarrow \frac{10}{-53}R_3$$ This gives: $$\begin{pmatrix} 1 & 0 & -\frac{38}{15} \\ 0 & 1 & \frac{21}{10} \\ 0 & 0 & 1 \end{pmatrix}$$ ### Step 6: Eliminate the element above the third pivot: - $$R_1 \rightarrow R_1 + \frac{38}{15}R_3$$ - $$R_2 \rightarrow R_2 - \frac{21}{10}R_3$$ Performing these operations: - $$R_1 \rightarrow -\frac{38}{15}(1) = 0$$ - $$R_2 \rightarrow \frac{21}{10}(1) = 0$$ Thus, the matrix becomes: $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ ### Final Step: Rank of the Matrix The matrix is now in **reduced row echelon form** (identity matrix), and since all three rows are non-zero, the rank of the matrix $$A$$ is **3**. --- Would you like to ask any questions or need further clarification? Here are five related questions: 1. What are the main differences between row echelon form and reduced row echelon form? 2. How does column transformation affect the process of finding the rank of a matrix? 3. What is the significance of the rank of a matrix in linear algebra? 4. Can a matrix have more pivots than its smallest dimension (rows or columns)? 5. What does it mean if a matrix has full rank? **Tip**: The rank of a matrix represents the maximum number of linearly independent rows or columns in the matrix.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Row Echelon Form
Matrix Rank
Formulas
Elementary Row Operations
Rank of a Matrix
Theorems
Rank Theorem (Dimensionality)
Gaussian Elimination
Suitable Grade Level
Undergraduate Level