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JAMB Mathematics 2014 Paper

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Question : 22

Total: 47

Find the value of

\begin{vmatrix} 0 & 3 & 2 \\ 1 & 7 & 8 \\ 0 & 5 & 4 \end{vmatrix}

.

12
10
-1
-2

Solution:

Concept:

Determinant of a

3 \times 3

matrix is found by expanding along a row or column. Expanding along the first row is quick because two entries are zero.

Explanation:

Given determinant is

\begin{vmatrix} 0 & 3 & 2 \\ 1 & 7 & 8 \\ 0 & 5 & 4 \end{vmatrix}

.

Expand along the first row:

0\begin{vmatrix} 7 & 8 \\ 5 & 4 \end{vmatrix} - 3\begin{vmatrix} 1 & 8 \\ 0 & 4 \end{vmatrix} + 2\begin{vmatrix} 1 & 7 \\ 0 & 5 \end{vmatrix}

.

Evaluate each

2 \times 2

determinant:

0(28 - 40) - 3(4 - 0) + 2(5 - 0)

.

Simplify carefully:

0(-12) - 3(4) + 2(5)

.

0 - 12 + 10 = -2

.

Answer:

Option D. The value is

-2

.

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