Find the inverse of ((-2/3), (6/-7)). Hence solve the following equations by matrix method. 3x - 7y = 26 = 0, 6y - 2x + 20 = 0

Huu hapa ni utatuzi wa maswali yako:

7. Pata kinyume cha $\begin{pmatrix} -2 & 6 \\ 3 & -7 \end{pmatrix}$. Kisha tatua milinganyo ifuatayo kwa kutumia njia ya matriki. $3x - 7y - 26 = 0$ $6y - 2x + 20 = 0$

• Step 1: Andika upya milinganyo katika mpangilio unaolingana na matriki iliyotolewa na uweke katika fomu ya matriki. Milinganyo iliyopangwa upya: $-2x + 6y = -20$ $3x - 7y = 26$ Katika fomu ya matriki $AX = B$: $A = \begin{pmatrix} -2 & 6 \\ 3 & -7 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} -20 \\ 26 \end{pmatrix}$

• Step 2: Pata kinyume cha matriki $A$. Kwanza, kokotoa kiamua (determinant) cha $A$: $\det(A) = (-2)(-7) - (6)(3) = 14 - 18 = -4$ Kinyume cha $A$ ni: $A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} -7 & -6 \\ -3 & -2 \end{pmatrix} = \frac{1}{-4} \begin{pmatrix} -7 & -6 \\ -3 & -2 \end{pmatrix}$ $A^{-1} = \begin{pmatrix} \frac{-7}{-4} & \frac{-6}{-4} \\ \frac{-3}{-4} & \frac{-2}{-4} \end{pmatrix} = \begin{pmatrix} \frac{7}{4} & \frac{3}{2} \\ \frac{3}{4} & \frac{1}{2} \end{pmatrix}$ Kinyume cha matriki ni $\boxed{\begin{pmatrix} \frac{7}{4} & \frac{3}{2} \\ \frac{3}{4} & \frac{1}{2} \end{pmatrix}}$.

• Step 3: Tumia kinyume cha matriki kutatua milinganyo. $X = A^{-1} B$ $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{7}{4} & \frac{3}{2} \\ \frac{3}{4} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} -20 \\ 26 \end{pmatrix}$ $x = \left(\frac{7}{4}\right)(-20) + \left(\frac{3}{2}\right)(26) = 7(-5) + 3(13) = -35 + 39 = 4$ $y = \left(\frac{3}{4}\right)(-20) + \left(\frac{1}{2}\right)(26) = 3(-5) + 1(13) = -15 + 13 = -2$ Hivyo, suluhisho ni $\boxed{x=4, y=-2}$.

8. Mchoro hapa chini unaonyesha duara MNPQR, kituo O. MQ na NR ni mistari iliyonyooka. Ikiwa pembe ROQ = 62°, pata ukubwa wa pembe NPQ.

• Step 1: Tambua pembe za katikati na uhusiano wao. MQ na NR ni mistari iliyonyooka inayopita katikati O, kwa hivyo ni kipenyo. Pembe $\angle ROQ = 62^\circ$. Kwa kuwa MQ ni kipenyo (mstari ulionyooka), pembe $\angle NOQ$ na $\angle ROQ$ ni pembe za ziada kwenye mstari NR. No, this is incorrect. MQ and NR are diameters. $\angle NOQ$ na $\angle MOR$ ni pembe zinazopingana kwa wima. $\angle NOM$ na $\angle ROQ$ ni pembe zinazopingana kwa wima. Hivyo, $\angle NOM = \angle ROQ = 62^\circ$. Pembe $\angle NOQ$ na $\angle NOM$ ni pembe za ziada kwenye kipenyo MQ. $\angle NOQ = 180^\circ - \angle NOM = 180^\circ - 62^\circ = 118^\circ$.

• Step 2: Pata pembe $\angle NRQ$. Pembe $\angle NRQ$ ni pembe kwenye mduara inayotegemea upinde NQ. Pembe ya katikati inayotegemea upinde NQ ni $\angle NOQ$. Pembe kwenye mduara ni nusu ya pembe ya katikati inayotegemea upinde uleule. $\angle NRQ = \frac{1}{2} \angle NOQ = \frac{1}{2} \times 118^\circ = 59^\circ$

• Step 3: Tumia sifa za mstatili wa mduara (cyclic quadrilateral) kupata $\angle NPQ$. NPQR ni mstatili wa mduara (cyclic quadrilateral). Katika mstatili wa mduara, jumla ya pembe zinazopingana ni $180^\circ$. $\angle NPQ + \angle NRQ = 180^\circ$ $\angle NPQ + 59^\circ = 180^\circ$ $\angle NPQ = 180^\circ - 59^\circ = 121^\circ$ Ukubwa wa pembe NPQ ni $\boxed{121^\circ}$.

9. Bainisha tofauti (variance) kwa seti ifuatayo ya namba: 6, 10, 7, 6, 9, 3, 2, 9, 8, 5, 4, 4, 5.

• Step 1: Panga namba na uhesabu idadi ya namba ($n$). Namba: 2, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 9, 10. Idadi ya namba $n = 13$.

• Step 2: Kokotoa wastani ($\bar{x}$). Jumla ya namba ($\sum x$): $2+3+4+4+5+5+6+6+7+8+9+9+10 = 78$. Wastani $\bar{x} = \frac{\sum x}{n} = \frac{78}{13} = 6$.

• Step 3: Kokotoa mkengeuko wa mraba kutoka kwa wastani ($(x - \bar{x})^2$) kwa kila namba na jumla yake. | $x$ | $x - \bar{x}$ | $(x - \bar{x})^2$ | |-----|---------------|-------------------| | 2 | $2-6 = -4$ | 16 | | 3 | $3-6 = -3$ | 9 | | 4 | $4-6 = -2$ | 4 | | 4 | $4-6 = -2$ | 4 | | 5 | $5-6 = -1$ | 1 | | 5 | $5-6 = -1$ | 1 | | 6 | $6-6 = 0$ | 0 | | 6 | $6-6 = 0$ | 0 | | 7 | $7-6 = 1$ | 1 | | 8 | $8-6 = 2$ | 4 | | 9 | $9-6 = 3$ | 9 | | 9 | $9-6 = 3$ | 9 | | 10 | $10-6 = 4$ | 16 | Jumla ya $(x - \bar{x})^2 = 16+9+4+4+1+1+0+0+1+4+9+9+16 = 74$.

• Step 4: Kokotoa tofauti (variance). Fomula ya tofauti ni $\sigma^2 = \frac{\sum (x - \bar{x})^2}{n}$. $\sigma^2 = \frac{74}{13}$ Tofauti ni $\boxed{\frac{74}{13}}$ (au takriban $\boxed{5.692}$).

Drop the next question.