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A) A matrix A is singular. Find x and state the matrix A. b) Two shopkeepers, Juma and Wanjiku bought some items from a wholesale... The number of items bought by Juma and Wanjiku using a 2x3 matrix. The prices of the items bought using 3x1 matrix.

Mathematics

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3 parts
b(i)

[1840515306]\begin{bmatrix} 18 & 40 & 5 \\ 15 & 30 & 6 \end{bmatrix}

b(ii)

[4550150]\begin{bmatrix} 45 \\ 50 \\ 150 \end{bmatrix}

b(iii)

[35603075]\begin{bmatrix} 3560 \\ 3075 \end{bmatrix}

Step-by-step solution

3 steps

Hapa kuna suluhisho la swali lako:

20. a) Matrix A ni singular. Tafuta xx na ueleze matrix A.

Matrix A imetolewa kama:

A=[xx−124]A = \begin{bmatrix} x & x-1 \\ 2 & 4 \end{bmatrix}

Matrix ni singular ikiwa determinant yake ni sifuri.
Determinant ya matrix 2×22 \times 2, [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, ni ad−bcad - bc.

Step 1: Kokotoa determinant ya matrix A.

det⁡(A)=(x)(4)−(x−1)(2)\det(A) = (x)(4) - (x-1)(2) det⁡(A)=4x−(2x−2)\det(A) = 4x - (2x - 2) det⁡(A)=4x−2x+2\det(A) = 4x - 2x + 2 det⁡(A)=2x+2\det(A) = 2x + 2

Step 2: Weka determinant sawa na sifuri na utatue kwa xx.

2x+2=02x + 2 = 0 2x=−22x = -2 x=−22x = \frac{-2}{2} x=−1x = -1

Step 3: Eleza matrix A kwa kutumia thamani ya xx uliyopata.
Badilisha x=−1x = -1 kwenye matrix A:

A=[−1(−1)−124]A = \begin{bmatrix} -1 & (-1)-1 \\ 2 & 4 \end{bmatrix} A=[−1−224]A = \boxed{\begin{bmatrix} -1 & -2 \\ 2 & 4 \end{bmatrix}}

Part b(i)

Idadi ya bidhaa zilizonunuliwa na Juma na Wanjiku kwa kutumia matrix ya 2×32 \times 3.

Wafanyabiashara wawili, Juma na Wanjiku walinunua bidhaa kutoka kwa muuzaji wa jumla.

Juma alinunua mikate 18, pakiti 40 za maziwa, na sabuni 5.
Wanjiku alinunua mikate 15, pakiti 30 za maziwa, na sabuni 6.

Matrix ya 2×32 \times 3 (safu 2, nguzo 3) itawakilisha wanunuzi (Juma, Wanjiku) kama safu na bidhaa (mikate, maziwa, sabuni) kama nguzo.

Idadi ya bidhaa=[1840515306]\text{Idadi ya bidhaa} = \boxed{\begin{bmatrix} 18 & 40 & 5 \\ 15 & 30 & 6 \end{bmatrix}}

Part b(ii)

[4550150]\begin{bmatrix} 45 \\ 50 \\ 150 \end{bmatrix}

Bei za bidhaa zilizonunuliwa kwa kutumia matrix ya 3×13 \times 1.

Bei ya mkate ni Ksh. 45.
Bei ya pakiti ya maziwa ni Ksh. 50.
Bei ya sabuni ni Ksh. 150.

Matrix ya 3×13 \times 1 (safu 3, nguzo 1) itawakilisha bei za bidhaa.

Bei za bidhaa=[4550150]\text{Bei za bidhaa} = \boxed{\begin{bmatrix} 45 \\ 50 \\ 150 \end{bmatrix}}

Part b(iii)

[35603075]\begin{bmatrix} 3560 \\ 3075 \end{bmatrix}

Jumla ya gharama iliyolipwa na kila mfanyabiashara.

Ili kupata jumla ya gharama iliyolipwa na kila mfanyabiashara, tunazidisha matrix ya idadi ya bidhaa na matrix ya bei.

  1. Zidisha matrix ya idadi ya bidhaa na matrix ya bei.

    Gharama=[1840515306][4550150]\text{Gharama} = \begin{bmatrix} 18 & 40 & 5 \\ 15 & 30 & 6 \end{bmatrix} \begin{bmatrix} 45 \\ 50 \\ 150 \end{bmatrix}
  2. Kokotoa bidhaa ya matrix.

    Kwa Juma (safu ya kwanza):

    (18×45)+(40×50)+(5×150)(18 \times 45) + (40 \times 50) + (5 \times 150) =810+2000+750= 810 + 2000 + 750 =3560= 3560

    Kwa Wanjiku (safu ya pili):

    (15×45)+(30×50)+(6×150)(15 \times 45) + (30 \times 50) + (6 \times 150) =675+1500+900= 675 + 1500 + 900 =3075= 3075
  3. Andika matokeo kama matrix.

    Gharama=[35603075]\text{Gharama} = \boxed{\begin{bmatrix} 3560 \\ 3075 \end{bmatrix}}

    Hivyo, Juma alilipa Ksh. 3560 na Wanjiku alilipa Ksh. 3075.

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