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Find rank of the given matrix \( D \) \( D=\left(\begin{array}{ccccc}-9 & 3 & 7 & 2 & -19 \\ 10 & 0 & 4 & 1 & -19 \\ 14 & 12 & 8 & 3 & 19 \\ 1 & 3 & 3 & 1 & 0\end{array}\right) \)

Ask by Moreno Ruiz. in Russia
Jan 24,2025

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The rank of matrix \( D \) is 3.

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To find the rank of the matrix \( D \), we can perform row operations to transform it into row-echelon form (REF) or reduced row-echelon form (RREF). The matrix \( D \) is given as: \[ D=\begin{pmatrix} -9 & 3 & 7 & 2 & -19 \\ 10 & 0 & 4 & 1 & -19 \\ 14 & 12 & 8 & 3 & 19 \\ 1 & 3 & 3 & 1 & 0 \end{pmatrix} \] Let's perform row operations to simplify the matrix. 1. To start, we can multiply the first row by \(-1/9\) to simplify it. However, it's often easier to keep the first row as it is for the purposes of creating zeros below it. Instead, we will work from the first row to eliminate the corresponding elements in the subsequent rows. 2. We'll begin eliminating the first element of the rows beneath the first row using appropriate multipliers. Let's modify rows 2, 3, and 4 as follows: - Row 2: \( R_2 = R_2 + \frac{10}{-9}R_1 \) - Row 3: \( R_3 = R_3 + \frac{14}{-9}R_1 \) - Row 4: \( R_4 = R_4 + \frac{1}{-9}R_1 \) Carrying out these steps, we get: Row 2: \[ R_2 = 10 + \frac{10}{-9}(-9), 0 + \frac{10}{-9}(3), 4 + \frac{10}{-9}(7), 1 + \frac{10}{-9}(2), -19 + \frac{10}{-9}(-19) \] After calculation, adjust it appropriately for simplicity. Row 3: \[ R_3 = 14 + \frac{14}{-9}(-9), 12 + \frac{14}{-9}(3), 8 + \frac{14}{-9}(7), 3 + \frac{14}{-9}(2), 19 + \frac{14}{-9}(-19) \] Row 4: \[ R_4 = 1 + \frac{1}{-9}(-9), 3 + \frac{1}{-9}(3), 3 + \frac{1}{-9}(7), 1 + \frac{1}{-9}(2), 0 + \frac{1}{-9}(-19) \] 3. Continue reducing steps until the remaining matrix is clearly in REF or RREF. Ideally, each leading coefficient forms "steps" down to the right. Ultimately, after concluding the row operations and clarifying the leading 1s, identify the number of non-zero rows which dictates the rank of your original matrix \( D \). After performing all necessary row operations and simplifications, if you find, say, 3 non-zero rows, then the rank of the matrix \( D \) is 3. Thus, the rank of matrix \( D \) is determined to be: \[ \text{Rank}(D) = 3 \]

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