Answer:
The matrix B is given as
[tex]B=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{array}\right][/tex]
Step-by-step explanation:
As the vectors are given by u1 and u2 where u1 is given as
As y=x thus [tex]\theta=tan^{-1}(\frac{y}{x})=tan^{-1}{1}=\pi/4\\\\[/tex]
So the value of x and y is given as
Thus u_1 is given as
[tex]u_1=\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{array}\right][/tex]
Now the vector u_2 is given as perpendicular to u_1 i.e.
[tex]u_1.u_2=0\\\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{array}\right].\left[\begin{array}{c}u_{2x}\\u_{2y}}\end{array}\right]=0\\u_{2x}=-\frac{1}{\sqrt{2}}\\u_{2y}=-\frac{1}{\sqrt{2}}\\\\u_2=\left[\begin{array}{c}-\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{array}\right][/tex]
So the B matrix is given as
[tex]B=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{array}\right][/tex]
So the matrix B is as above.
