# How to get 4x4 matrix from integrating column vector and its transpose?

This is from Finite Element Methods. Part of potential energy approach. Can someone show me how to come to this symetric matrix: $$[K]=EI/L^3\begin{bmatrix} 12&6L&-12&6L\\ &4L^2&-6L&2L^2\\ & & 12&-6L\\ &&&4L^2 \end{bmatrix}$$ EI - constant

Using this equation: $$[K]=EI\int_0^L {[B]^T [B]}dx$$ Given: $$[B]= \begin{bmatrix} \frac{12x-6L}{L^3} & \frac{6xL-4L^2}{L^3} & \frac{-12x+6L}{L^3} & \frac{6xL-2L^2}{L^3}\\ \end{bmatrix}$$

I don't understand how I can have a 4x4 matrix when multiplying vector and its transpose gives me a scalar. Here is what I have so far: $$[B]^T[B]=\frac{144x^2-144xL+36L}{L^6}+\frac{36x^2L^2-48xL^3+16L^4}{L^6}+\frac{144x^2-144xl+36L}{L^6}+\frac{36x^2L^2-24xL^3+4L^4}{L^6}=$$ $$=\frac{1}{L^6}(288x^2+24x^2L^2-288xL-72xL^3+72xL^3+72L+20L^4)$$

$$EI\int_0^L {[B]^T [B]}dx =$$ $$=\frac{EI}{L^6}(96x^3+24x^3L^2-144x^2L-36x^2L^3+72xL+20xL^4)_0^L=$$ $$=EI(\frac{-48}{L^3}+\frac{8}{L}+\frac{72}{L^4})$$ And I got scalar. Not a 4x4 matrix. Thank you in advance!

As is written, $$[B]$$ is a row vector. Then $$[B]^T$$ is a column vector. So, $$[B]^T[B]$$ is of the form $$\pmatrix{b_1\\ b_2\\ \vdots\\ b_n}\pmatrix{b_1&b_2&\dots&b_n}=\pmatrix{{b_1}^2&b_1b_2&\dots\\ \vdots & \ddots \\ b_nb_1&\dots & {b_n}^2}$$