in Linear Algebra recategorized by
437 views
0 votes
0 votes

Let $x_1, x_2, x_3, x_4, y_1, y_2, y_3$ and $y_4$ be fixed real numbers, not all of them equal to zero. Define a $4 \times 4$ matrix $\textbf{A}$ by

$$\textbf{A} = \begin{pmatrix} x_1^2+y_1^2 & x_1x_2 + y_1 y_2 & x_1x_3+y_1y_3 & x_1x_4+y_1y_4 \\ x_2x_1 + y_2 y_1 & x_2^2+y_2^2 & x_2x_3+y_2y_3 & x_2x_4+y_2y_4 \\ x_3x_1 + y_3 y_1 & x_3x_2+y_3y_2 & x_3^2+y_3^2 & x_3x_4+y_3y_4 \\ x_4x_1 + y_4 y_1 & x_4x_2+y_4y_2 & x_4x_3+y_4y_3 & x_4^2+y_4^2  \end{pmatrix}$$

Then rank$(\textbf{A})$ equals

  1. $1$ or $2$
  2. $0$
  3. $4$
  4. $2$ or $3$
in Linear Algebra recategorized by
by
437 views

2 Comments

b?
0
0

No. Option is the correct answer. Let $x_1=x_2=x_3=x_4\neq 0$ and $y_1=y_2=y_3=y_4\neq 0$. Then the rank is 1.

0
0

Please log in or register to answer this question.

Related questions