NECO 2023 · Paper 2 · Q9

  1. (a)

    Given A=(32−11012−20)A = \begin{pmatrix} 3 & 2 & -1 \\ 1 & 0 & 1 \\ 2 & -2 & 0 \end{pmatrix} and B=(42−33−110−22)B = \begin{pmatrix} 4 & 2 & -3 \\ 3 & -1 & 1 \\ 0 & -2 & 2 \end{pmatrix}, evaluate (i) 3A−2B3A - 2B; (ii) ∣3A−2B∣|3A - 2B|.

  2. (b)

    Mr. Tony took a loan of ₦120,000.00 at 12%12\% per annum compound interest to buy a piece of land. (i) If he paid the loan in three years, what was the total amount paid? (ii) Find his profit if he later sold the land for ₦350,000.00 without any additional expenses.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

  1. Multiply each matrix by its number, then subtract entry by entry: 3A−2B=(9−86−4−3+63−60+23−26−0−6+40−4)3A - 2B = \begin{pmatrix} 9 - 8 & 6 - 4 & -3 + 6 \\ 3 - 6 & 0 + 2 & 3 - 2 \\ 6 - 0 & -6 + 4 & 0 - 4 \end{pmatrix}
    =(123−3216−2−4)= \begin{pmatrix} 1 & 2 & 3 \\ -3 & 2 & 1 \\ 6 & -2 & -4 \end{pmatrix}.

(ii)

  1. Expand along the first row: 1(2(−4)−1(−2))−2((−3)(−4)−1(6))+3((−3)(−2)−2(6))=1(−6)−2(6)+3(−6)1\big(2(-4) - 1(-2)\big) - 2\big((-3)(-4) - 1(6)\big) + 3\big((-3)(-2) - 2(6)\big) = 1(-6) - 2(6) + 3(-6)
    =−36= -36.

(b)(i)

  1. Compound interest at 12%12\% for 3 years: 120 000×1.123=120 000×1.404928120\,000 \times 1.12^3 = 120\,000 \times 1.404928
    =₦168,591.36= ₦168,591.36.

(ii)

  1. Profit =350 000−168 591.36=₦181,408.64= 350\,000 - 168\,591.36 = ₦181,408.64.

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