Home » mcq » Linear Algebra » 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 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z"/>
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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:
$$\left[ {\begin{array}{*{20}{c}} { – 4}&1&{ – 1} \ 0&-\frac{3}{4}&-\frac{3}{4} \ 0&-\frac{11}{4}&-\frac{3}{4} \end{array}} \right]$$
$$\left[ {\begin{array}{*{20}{c}} { – 4}&1&{ – 1} \ 0&-\frac{11}{4}&-\frac{3}{4} \ 0&-\frac{3}{4}&-\frac{3}{4} \end{array}} \right]$$
*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 $\boxed{2}$.