Question Number 150362 by otchereabdullai@gmail.com last updated on 11/Aug/21
$$\mathrm{Given}\:\mathrm{that}\:\mathrm{p}=\left(\mathrm{3i}+\mathrm{4j}\right)\:,\:\mathrm{q}=\left(\mathrm{2i}−\mathrm{j}\right)\:\mathrm{and} \\ $$$$\mathrm{r}=\mathrm{5i}−\mathrm{j}.\:\mathrm{Express}\:\:\mathrm{vector}\:\:\:\mathrm{r}\:\:\mathrm{intrems}\:\mathrm{of} \\ $$$$\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:. \\ $$
Answered by mr W last updated on 11/Aug/21
![say r=ap+bq ((5),((−1)) )= [(3,2),(4,(−1)) ] ((a),(b) ) ((a),(b) )= [(3,2),(4,(−1)) ]^(−1) ((5),((−1)) ) =(1/(11)) [(1,2),(4,(−3)) ] ((5),((−1)) )= (((3/(11))),(((23)/(11))) ) ⇒r=(3/(11))p+((23)/(11))q check: (3/(11))×3+((23)/(11))×2=5 ✓ (3/(11))×4−((23)/(11))×1=−1 ✓](Q150379.png)
$${say}\:{r}={ap}+{bq} \\ $$$$\begin{pmatrix}{\mathrm{5}}\\{−\mathrm{1}}\end{pmatrix}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{−\mathrm{1}}\end{bmatrix}\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}\: \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{−\mathrm{1}}\end{bmatrix}^{−\mathrm{1}} \begin{pmatrix}{\mathrm{5}}\\{−\mathrm{1}}\end{pmatrix}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{11}}\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{4}}&{−\mathrm{3}}\end{bmatrix}\begin{pmatrix}{\mathrm{5}}\\{−\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{11}}}\\{\frac{\mathrm{23}}{\mathrm{11}}}\end{pmatrix} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}}{\mathrm{11}}{p}+\frac{\mathrm{23}}{\mathrm{11}}{q} \\ $$$${check}: \\ $$$$\frac{\mathrm{3}}{\mathrm{11}}×\mathrm{3}+\frac{\mathrm{23}}{\mathrm{11}}×\mathrm{2}=\mathrm{5}\:\checkmark \\ $$$$\frac{\mathrm{3}}{\mathrm{11}}×\mathrm{4}−\frac{\mathrm{23}}{\mathrm{11}}×\mathrm{1}=−\mathrm{1}\:\checkmark \\ $$
Commented by otchereabdullai@gmail.com last updated on 11/Aug/21
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{prof}\:\mathrm{W} \\ $$