Math Problem Statement
Find the rank of the matrix A=[ 1 2 3 0 2 4 3 2 3 2 1 3 6 8 7 5 ]
Solution
To find the rank of the given matrix:
1 & 2 & 3 & 0 \\ 2 & 4 & 3 & 2 \\ 3 & 2 & 1 & 3 \\ 6 & 8 & 7 & 5 \end{bmatrix}$$ ### **Step 1: Convert to Row Echelon Form** We perform row operations to bring the matrix to an echelon form. #### **1st Column: Make leading 1** The first pivot is already **1** in row 1. - Subtract $$2 \times R_1$$ from $$R_2$$: $$R_2 \rightarrow R_2 - 2R_1 = (2, 4, 3, 2) - 2(1,2,3,0) = (0, 0, -3, 2)$$ - Subtract $$3 \times R_1$$ from $$R_3$$: $$R_3 \rightarrow R_3 - 3R_1 = (3, 2, 1, 3) - 3(1,2,3,0) = (0, -4, -8, 3)$$ - Subtract $$6 \times R_1$$ from $$R_4$$: $$R_4 \rightarrow R_4 - 6R_1 = (6, 8, 7, 5) - 6(1,2,3,0) = (0, -4, -11, 5)$$ Now, the matrix is: $$\begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & 0 & -3 & 2 \\ 0 & -4 & -8 & 3 \\ 0 & -4 & -11 & 5 \end{bmatrix}$$ #### **2nd Column: Make leading coefficient** Since the second column of $$R_2$$ is zero, we swap $$R_2$$ and $$R_3$$: $$\begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & -4 & -8 & 3 \\ 0 & 0 & -3 & 2 \\ 0 & -4 & -11 & 5 \end{bmatrix}$$ - Add $$R_2$$ to $$R_4$$ to make the second column zero: $$R_4 \rightarrow R_4 + R_2 = (0, -4, -11, 5) + (0, -4, -8, 3) = (0, 0, -19, 8)$$ Now, the matrix is: $$\begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & -4 & -8 & 3 \\ 0 & 0 & -3 & 2 \\ 0 & 0 & -19 & 8 \end{bmatrix}$$ #### **3rd Column: Make leading coefficient** We now focus on the third column: - Make the third pivot **1** by dividing $$R_3$$ by $$-3$$: $$R_3 \rightarrow \frac{R_3}{-3} = \left(0, 0, 1, -\frac{2}{3}\right)$$ - Express $$R_4$$ in terms of this pivot: $$R_4 \rightarrow R_4 - (-19) R_3 = (0,0,-19,8) + 19(0,0,1,-\frac{2}{3})$$ $$= (0,0,0, 8 - \frac{38}{3})$$ $$= (0,0,0, \frac{24-38}{3}) = (0,0,0,-\frac{14}{3})$$ Now, the matrix is: $$\begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & -4 & -8 & 3 \\ 0 & 0 & 1 & -\frac{2}{3} \\ 0 & 0 & 0 & -\frac{14}{3} \end{bmatrix}$$ ### **Step 2: Determine the Rank** The number of **nonzero rows** in row echelon form is **4**. Hence, the **rank** of the matrix is: $$\mathbf{4}$$ ### **Final Answer:** $$\boxed{4}$$ Would you like more details or have any questions? ### **Related Questions:** 1. How does row echelon form help in finding the rank of a matrix? 2. What is the difference between rank and determinant of a matrix? 3. How do you find the rank of a matrix using determinant methods? 4. Can a $$4 \times 4$$ matrix have a rank less than 4? Why? 5. What happens if a matrix has a rank less than its number of columns? ### **Tip:** The rank of a matrix is equal to the number of **linearly independent** rows (or columns). If the determinant of a square matrix is zero, its rank is less than the number of rows!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Rank
Row Echelon Form
Row Operations
Formulas
Rank of matrix = Number of non-zero rows in Row Echelon Form
Theorems
Row Echelon Form Theorem
Suitable Grade Level
Grades 11-12