If a = ≤ft | {array}{cc} 3 \\ 2 {array} | and b = ≤ft | {array}{cc} -3 \\ 5{array} | . Find the vector c such that 4a + 3c = b

Correct Answer is : ∣−5−1∣≤ft | {array}{cc} -5 \\ -1{array} |−5−1

≤ft | {array}{cc} -5 \\ 1{array} |

≤ft | {array}{cc} 3 \\ 1{array} |

≤ft | {array}{cc} -5 \\ -9{array} |

≤ft | {array}{cc} -5 \\ -1{array} |

Correct Answer is : ∣−5−1∣≤ft | {array}{cc} -5 \\ -1{array} |−5−1

Test Index

WAEC Further Mathematics 2025 Paper

Question : 35

Total: 40

If a =

\left | \begin{array}{cc} 3 \\ 2 \end{array} \right |

\left | \begin{array}{cc} -3 \\ 5\end{array} \right |

. Find the vector c such that 4a + 3c = b

Solution:

Concept:

Solve the vector equation

4a + 3c = b

by isolating the unknown vector

c

Explanation:

First, multiply the given vector

a

by 4:

4a = 4 \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 12 \\ 8 \end{pmatrix}

Rewrite the equation as

3c = b - 4a

Substitute the known vectors:

3c = \begin{pmatrix} -3 \\ 5 \end{pmatrix} - \begin{pmatrix} 12 \\ 8 \end{pmatrix}

Subtract corresponding components:

3c = \begin{pmatrix} -3 - 12 \\ 5 - 8 \end{pmatrix} = \begin{pmatrix} -15 \\ -3 \end{pmatrix}

Divide each component by 3 to get

c

c = \frac{1}{3} \begin{pmatrix} -15 \\ -3 \end{pmatrix} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}

Answer:

c = \begin{pmatrix} -5 \\ -1 \end{pmatrix}

, which is option D.

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