Verify that is an orthogonal set, and then find the orthogonal projection of y onto span

If the inversion of a square matrix containing real figures or components equals the inverse matrix, the matrix is said to be orthogonal. Or, to put it another way, an orthogonal matrix is one in which the multiplication of a two - dimensional array and also its transpose yields an identity matrices.

[tex]y = \begin{bmatrix}6 \\3 \\-2\end{bmatrix} , \ u_1 = \begin{bmatrix} 3\\ 4\\0\end{bmatrix}, \ u_2 = \begin{bmatrix} -4\\ 3\\0\end{bmatrix}[/tex]

Now we have

[tex]\left ( u_1 \cdot u_2 \right ) = \left (\begin{bmatrix} 3\\ 4\\0\end{bmatrix} \cdot \begin{bmatrix} -4\\ 3\\0\end{bmatrix} \right )[/tex]

(u₁u₂) = -12 + 12 + 0

(u₁u₂) = 0

Thus, u₁ and u₂ are orthogonal.

|u₁|² = 3² + 4² + 0 = 9 + 16 + 0 = 25

|u₂|² = (-4)² + 3² + 0 = 16 + 9 + 0 = 25

[tex]\left ( y \cdot u_1 \right ) = \left (\begin{bmatrix} 6\\ 3\\-2\end{bmatrix} \cdot \begin{bmatrix} 3\\ 4\\0\end{bmatrix} \right ) = 18 + 12 + 0 = 30[/tex]

and

[tex]\left ( y \cdot u_2 \right ) = \left (\begin{bmatrix} 6\\ 3\\-2\end{bmatrix} \cdot \begin{bmatrix} -4\\ 3\\0\end{bmatrix} \right ) = -24 + 9 = -15[/tex]

Then the orthogonal projection will be given as

[tex]Proj(y) = \dfrac{(y,u_1)}{|u_1|^2} + \dfrac{(y,u_2)}{|u_2|^2} \\\\\\Proj(y) = \dfrac{30}{25} \begin{bmatrix} 3\\ 4\\0\end{bmatrix} - \dfrac{15}{25} \begin{bmatrix} -4\\ 3\\0\end{bmatrix}\\\\\\Proj (y) = \dfrac{1}{5} \begin{bmatrix} 18+12\\ 24-9\\0\end{bmatrix}[/tex]

[tex]Proj(y) = \dfrac{1}{5}\begin{bmatrix} 30\\ 15\\0\end{bmatrix}\\\\\\Proj(y) = \begin{bmatrix}6\\3 \\0\end{bmatrix}[/tex]

More about the orthogonal projection link is given below.

https://brainly.com/question/2292926

#SPJ4