To find the value of $a$ such that the rank of the given matrix is 3, we can use the row echelon form (REF) method. The rank of a matrix is the number of non-zero rows in its row echelon form, so we want to manipulate the matrix in such a way that three rows remain non-zero.

Step 1: Start with the matrix

The given matrix is:

1 & 2 & -1 & 0 \\ 2 & 4 & 4 & 6 \\ 0 & 0 & 5 & -2 \\ 3 & 6 & 8 & a \end{pmatrix}$$ ### Step 2: Perform row operations to bring it to row echelon form We perform row operations to make the matrix into row echelon form, aiming to identify the rank based on the number of non-zero rows. #### First, let's work on the first column. - Row 2: Subtract $$2 \times \text{Row 1}$$ from Row 2: $$R_2 \to R_2 - 2R_1 = \begin{pmatrix} 2 & 4 & 4 & 6 \end{pmatrix} - 2 \times \begin{pmatrix} 1 & 2 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 6 & 6 \end{pmatrix}$$ - Row 4: Subtract $$3 \times \text{Row 1}$$ from Row 4: $$R_4 \to R_4 - 3R_1 = \begin{pmatrix} 3 & 6 & 8 & a \end{pmatrix} - 3 \times \begin{pmatrix} 1 & 2 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 11 & a \end{pmatrix}$$ Now, the matrix looks like this: $$\begin{pmatrix} 1 & 2 & -1 & 0 \\ 0 & 0 & 6 & 6 \\ 0 & 0 & 5 & -2 \\ 0 & 0 & 11 & a \end{pmatrix}$$ #### Second, focus on the third column. - Row 3: Subtract $$\frac{5}{6} \times \text{Row 2}$$ from Row 3: $$R_3 \to R_3 - \frac{5}{6} R_2 = \begin{pmatrix} 0 & 0 & 5 & -2 \end{pmatrix} - \frac{5}{6} \times \begin{pmatrix} 0 & 0 & 6 & 6 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -7 \end{pmatrix}$$ Now, the matrix is: $$\begin{pmatrix} 1 & 2 & -1 & 0 \\ 0 & 0 & 6 & 6 \\ 0 & 0 & 0 & -7 \\ 0 & 0 & 11 & a \end{pmatrix}$$ #### Third, eliminate the entry in row 4, column 3. - Row 4: Subtract $$\frac{11}{6} \times \text{Row 2}$$ from Row 4: $$R_4 \to R_4 - \frac{11}{6} R_2 = \begin{pmatrix} 0 & 0 & 11 & a \end{pmatrix} - \frac{11}{6} \times \begin{pmatrix} 0 & 0 & 6 & 6 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & a - 11 \end{pmatrix}$$ Now the matrix is: $$\begin{pmatrix} 1 & 2 & -1 & 0 \\ 0 & 0 & 6 & 6 \\ 0 & 0 & 0 & -7 \\ 0 & 0 & 0 & a - 11 \end{pmatrix}$$ ### Step 3: Conditions for rank 3 For the rank to be 3, there should be exactly three non-zero rows. The third row is already a non-zero row ($$[0, 0, 0, -7]$$), so for the rank to be 3, the fourth row must be zero. That is, we require: $$a - 11 = 0$$ Thus, $$a = 11$$. ### Final Answer: The value of $$a$$ such that the rank of the matrix is 3 is $$\boxed{11}$$. --- Would you like any further clarification or explanation on this solution? Here are some related questions for you to explore: 1. How does the row echelon form help determine the rank of a matrix? 2. How can you find the rank of a matrix using Gaussian elimination? 3. What is the difference between the row echelon form and the reduced row echelon form (RREF)? 4. How can you use the rank of a matrix to determine the solution to a system of linear equations? 5. What is the significance of a matrix having full rank? Tip: Always check the consistency of row operations to avoid introducing errors, especially when working with variables.