The rank of the matrix \[\left[ {\begin{array}{*{20}{c}} { – 4}&1&{ – 1} \\ { – 1}&{ – 1}&{ – 1} \\ 7&{ – 3}&1 \end{array}} \right]\] is A. 1 B. 2 C. 3 D. 4

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The rank of a matrix is the number of linearly independent rows or columns in the matrix. To find the rank of a matrix, we can use Gaussian elimination.

In Gaussian elimination, we reduce the matrix to row echelon form. A row echelon form is a matrix in which all the rows below the main diagonal are zero, and the leading coefficient of each non-zero row is 1.

To reduce the matrix to row echelon form, we can use the following operations:

  • Add or subtract a multiple of one row to another row.
  • Multiply a row by a non-zero constant.
  • Swap two rows.

Once the matrix is in row echelon form, the rank is the number of non-zero rows in the matrix.

For the matrix $M$, we can perform the following row operations:

  • Add $\frac{1}{4}$ of row 1 to row 2:

$$\left[ {\begin{array}{*{20}{c}} { – 4}&1&{ – 1} \ 0&-\frac{3}{4}&-\frac{3}{4} \ 7&{ – 3}&1 \end{array}} \right]$$

  • Add $\frac{7}{4}$ of row 1 to row 3:

$$\left[ {\begin{array}{*{20}{c}} { – 4}&1&{ – 1} \ 0&-\frac{3}{4}&-\frac{3}{4} \ 0&-\frac{11}{4}&-\frac{3}{4} \end{array}} \right]$$

  • Swap row 2 with row 3:

$$\left[ {\begin{array}{*{20}{c}} { – 4}&1&{ – 1} \ 0&-\frac{11}{4}&-\frac{3}{4} \ 0&-\frac{3}{4}&-\frac{3}{4} \end{array}} \right]$$

*Subtract $\frac{3}{11}$ of row 2 from row 3:

$$\left[ {\begin{array}{*{20}{c}} { – 4}&1&{ – 1} \ 0&-\frac{11}{4}&-\frac{3}{4} \ 0&0&-\frac{3}{11} \end{array}} \right]$$

Therefore, the rank of the matrix $M$ is $\boxed{2}$.

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