[amp_mcq option1=”\\[\\frac{{\\text{N}}}{2}\\]” option2=”N – 1″ option3=”N” option4=”2N” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\text{B. N – 1}}$.<\/p>\n
The rank of a matrix is the number of linearly independent rows or columns in the matrix. The rank of a product of two matrices is always less than or equal to the minimum of the ranks of the two matrices.<\/p>\n
In this case, the rank of matrix $A$ is $N$. This means that there are $N$ linearly independent rows in matrix $A$. The rank of matrix $B$ is therefore at most $N$.<\/p>\n
To determine the rank of matrix $B$, we can use Gaussian elimination. We can first eliminate the first row of matrix $B$ by subtracting $\\frac{r}{p}$ times the first row from the second row. This gives us the following matrix:<\/p>\n
$$\\left[\\begin{array}{cc} {p^2 + q^2} & {pr + qs} \\ {0} & {\\frac{r^2 + s^2 – pr – qs}{p}} \\end{array}\\right]$$<\/p>\n
We can then eliminate the second row of matrix $B$ by adding $\\frac{r^2 + s^2 – pr – qs}{p(p^2 + q^2)}$ times the second row to the first row. This gives us the following matrix:<\/p>\n
$$\\left[\\begin{array}{cc} {p^2 + q^2} & {0} \\ {0} & {\\frac{r^2 + s^2 – pr – qs}{p}} \\end{array}\\right]$$<\/p>\n
Since the first row of matrix $B$ is now a multiple of the second row, the rank of matrix $B$ is $N – 1$.<\/p>\n
Therefore, the rank of matrix $B$ is $\\boxed{\\text{N – 1}}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”\\[\\frac{{\\text{N}}}{2}\\]” option2=”N – 1″ option3=”N” option4=”2N” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[489],"tags":[],"class_list":["post-20028","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\n