Bheactoran co-ionann
Ma tha an aon mheudachd agus an aon chùrsa aig dà bheactor tha sin a' ciallachadh gu bheil iad co-ionann, ge bith dè an suidheachadh aca.
A' cur-ris bheactoran
Nuair a bhios sinn a' cur-ris bheactoran, bidh sinn a' leantainn na riaghailt:
\(\left( \begin{array}{l} a\\ b \end{array} \right) + \left( \begin{array}{l} c\\ d \end{array} \right) = \left( \begin{array}{l} a + c\\ b + d \end{array} \right)\)
Seall ris a' ghraf gu h-ìosal agus chì thu na gluasadan eadar PQ, QR agus PR.
Tha bheactor \(\overrightarrow {PQ}\) air a leantainn le bheactor \(\overrightarrow {QR}\) a' sealltainn a' ghluasaid bho P gu R.
\(\overrightarrow {PQ} + \overrightarrow {QR} = \overrightarrow {PR}\)
Seo mar a tha a' bheactor nuair a tha e air a sgrìobhadh:
\(\left( \begin{array}{l}2\\5\end{array} \right) + \left( \begin{array}{l}\,\,\,\,\,4\\- 3\end{array} \right) = \left( \begin{array}{l}6\\2\end{array} \right)\)
A' toirt-air-falbh bheactoran
Tha toirt-air-falbh an aon rud ri bhith a' cur-ris riochd àicheil dhen bheactor (cuimhnich nuair a nì thu bheactor àicheil gum bi an cùrsa aige a' dol an taobh eile).
\(\left( \begin{array}{l} a\\ b \end{array} \right) - \left( \begin{array}{l} c\\ d \end{array} \right) = \left( \begin{array}{l} a - c\\ b - d \end{array} \right)\)
Coimhead air an diagram agus smaoinich gu bheil thu a' dol bho X gu Z. Ciamar a sgrìobhadh tu an t-slighe ann a' bheactoran a' cleachdadh nam bheactoran \(\overrightarrow {XY}\) agus \(\overrightarrow {ZY}\) a-mhàin?
Dh'fhaodadh tu a ràdh gur e bheactor \(\overrightarrow {XY}\) a th' ann le gluasad air ais air \(\overrightarrow {ZY}\).
Mar sin faodaidh sinn an t-slighe bho X gu Z a sgrìobhadh mar:
\(\overrightarrow {XY} - \overrightarrow {ZY} = \overrightarrow {XZ}\)
Seo e sgrìobhte ann an àireamhan:
\(\left( \begin{array}{l} 4\\ 2 \end{array} \right) - \left( \begin{array}{l} 1\\ 2 \end{array} \right) = \left( \begin{array}{l} 3\\ 0 \end{array} \right)\)
Question
Ma tha \(x = \left( \begin{array}{l} 1\\ 3 \end{array} \right)\), \(y = \left( \begin{array}{l} - 2\\ 4 \end{array} \right)\) agus \(z = \left( \begin{array}{l} - 1\\ - 2 \end{array} \right)\) obraich a-mach:
- \(- y\)
- \(x - y\)
- \(2x + 3z\)
- \(\left( \begin{array}{l}\,\,\,\,\,2\\- 4\end{array} \right)\) An do dh'atharraich thu na samhlan?
- \(\left( \begin{array}{l}\,1\\3\end{array} \right) - \left( \begin{array}{l}- 2\\\,\,\,\,4\end{array} \right) = \left( \begin{array}{l}\,1 - - 2\\3 - \,\,\,\,4\end{array} \right) = \left( \begin{array}{l}\,\,\,\,\,3\\- 1\end{array} \right)\)
- \(\left( \begin{array}{l}\,1\\3\end{array} \right) + 3\left( \begin{array}{l}- 1\\- 2\end{array} \right) = \left( \begin{array}{l}2\\6\end{array} \right) + \left( \begin{array}{l}- 3\\- 6\end{array} \right) = \left( \begin{array}{l}- 1\\\,\,\,\,0\end{array}\right)\)
