Matrix and Vector Tutorial
\begin{bmatrix} 1\left\lbrack m\right\rbrack & 2\left\lbrack m\right\rbrack \\ 3\left\lbrack m\right\rbrack & 4\left\lbrack m\right\rbrack \end{bmatrix}
=\begin{bmatrix} 1.09\left\lbrack yards\right\rbrack & 2.19\left\lbrack yards\right\rbrack \\ 3.28\left\lbrack yards\right\rbrack & 4.37\left\lbrack yards\right\rbrack \end{bmatrix}
\begin{bmatrix} 1\left\lbrack m\right\rbrack & 2\left\lbrack m\right\rbrack \\ 3\left\lbrack m\right\rbrack & 4\left\lbrack m\right\rbrack \end{bmatrix}
\begin{bmatrix} 2 \cdot a \\ 2 \cdot b \end{bmatrix}
\begin{bmatrix} 2 \cdot a_{1} + 4 \cdot a_{2} \\ 2 \cdot a_{3} + 4 \cdot a_{4} \end{bmatrix}
a_{1} \cdot b_{1} + a_{2} \cdot b_{2} + a_{3} \cdot b_{3} + a_{4} \cdot b_{4}
a_{1} \cdot b_{1} + a_{2} \cdot b_{2} + a_{3} \cdot b_{3} + a_{4} \cdot b_{4}
\begin{bmatrix} a_{2} \cdot b_{3} - a_{3} \cdot b_{2} \\ - a_{1} \cdot b_{3} + a_{3} \cdot b_{1} \\ a_{1} \cdot b_{2} - a_{2} \cdot b_{1} \end{bmatrix}
\begin{bmatrix} a_{2} \cdot b_{3} - a_{3} \cdot b_{2} & - a_{1} \cdot b_{3} + a_{3} \cdot b_{1} & a_{1} \cdot b_{2} - a_{2} \cdot b_{1} \end{bmatrix}
\begin{bmatrix} a & b & c \end{bmatrix}
\begin{bmatrix} a & b & c \end{bmatrix}
\sqrt{\left|{a}\right|^{2} + \left|{b}\right|^{2} + \left|{c}\right|^{2}}
\sqrt{\left|{a}\right|^{2} + \left|{b}\right|^{2} + \left|{c}\right|^{2}}
\begin{bmatrix} 0.5\left\lbrack \frac{1}{m}\right\rbrack & 0\left\lbrack \frac{1}{m}\right\rbrack \\ 0\left\lbrack \frac{1}{m}\right\rbrack & 0.5\left\lbrack \frac{1}{m}\right\rbrack \end{bmatrix}
\begin{bmatrix} 0.5\left\lbrack \frac{1}{m}\right\rbrack & 0\left\lbrack \frac{1}{m}\right\rbrack \\ 0\left\lbrack \frac{1}{m}\right\rbrack & 0.5\left\lbrack \frac{1}{m}\right\rbrack \end{bmatrix}
a \cdot d - b \cdot c
1
6
6
c
15
System =Solution =Selected Solution:Solve for:
2
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