• So basically, when you obtain a Triangular system you have to do a process of back substitution to obtain the value of the variables.
Now, if the Pivot element is $0$ then the solution would be indeterminable. Hence, Pivot elements can't be $0$ by definition.
• While manipulating the original system into a triangular system, we take the pivot element of the $1^{st}$ equation to make the $a_{n1} = 0$ and so on and so forth. So, while you are unable to obtain a non-zero pivot element to make $a_{n2} = 0$ for the rest of the equations, you can't go further with Gaussian elimination.
By definition of norm induced from inner product, $||x||^2 = \langle x,x\rangle$.
That is, if $||a+bx|| = 1$ then $\langle a+bx,a+bx\rangle = 1$ ,so writing this down:
$$
\int_{0}^1 (a+bx)(a+bx) dx = 1 \implies \int_0^1 (a^2 + 2abx + b^2x^2) dx = 1 \\ \implies a^2 + ab + \frac {b^2}3 = 1
$$
therefore, the statement that $a^2 + b^2 = 1$ is FALSE.
This is actually quite clear with an example : $a=0 , b=1$ satisfies $a^2+b^2 = 1$, and gives the polynomial $x$, but $||x||^2 = \int_0^1 x^2 = \frac 13$, so $||x|| \neq 1$. Instead, the other statement given is correct. Replacing $a$ and $b$ by $c$ and $d$ gives you the other analogously correct statement.
Once this happens, you may set $a,b$ to any suitable values, and check what happens to $c$ and $d$.
For example, set $b = 0$ : from the above equation, this forces $a = \pm 1$, we will take $a =1$.
From the equation that $\langle a+bx,c+dx\rangle = 0$ that the author has derived in your question above, substituting (and cancelling $b$) and rearranging gives $2c+d = 0$, so $d = -2c$.
This must be combined with $c^2 + cd + \frac{d^2}{3} = 1$. Setting it, we get $c^2(1 - 2 + \frac 43) = 1$, so $c^2 = 3$. Just take $c = + \sqrt 3$, so $d = -2\sqrt 3$.
In this manner, we may verify that the polynomial $a+bx = 1$ and $c + dx = \sqrt 3(1 - 2x)$ form an orthonormal basis for the space.
Best Answer
$x_1$ is just a scalar, so $q_1^Tx_1q_1=x_1q_1^Tq_1$.