Mathematics
17.11.2021 15:17
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Find the matrix B of T with respect to the basis B = {~u1, ~u2}, where u1 is a unit
Find the matrix B of T with respect to the basis B = {~u1, ~u2}, where u1 is a unit vector that points along y = x toward increasing x and y and ~u2 is a unit vector that is perpendicular to ~u1 and 90 degrees counterclockwise from ~u1.
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amiechap12
4,9(38 marks)
The matrix B is given as
![B=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{array}\right]](/tpl/images/0520/1802/620bd.png)
Step-by-step explanation:
As the vectors are given by u1 and u2 where u1 is given as
As y=x thus 
So the value of x and y is given as
Thus u_1 is given as
![u_1=\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{array}\right]](/tpl/images/0520/1802/4946a.png)
Now the vector u_2 is given as perpendicular to u_1 i.e.
![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]](/tpl/images/0520/1802/885c9.png)
So the B matrix is given as
So the matrix B is as above.
justintsmith6415
4,4(44 marks)
(176) divided by (23 and 1/3) = 7.543 minutes
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