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JAMB Mathematics 2019 Paper

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Question : 69

Total: 73

If

\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132

, find the value of x.

5
8
6
3

Solution:

Concept:

To find

x

, expand the

3 \times 3

determinant and equate it to

132

.

Explanation:

Expand using the first row:

2\begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - (-5)\begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} + 3\begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 132

.

Evaluate each minor determinant:

2(1\cdot 2 - 4\cdot 3) = 2(2 - 12) = -20

.

+5\begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} = 5(x\cdot 2 - 4\cdot 0) = 10x

.

+3\begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 3(x\cdot 3 - 1\cdot 0) = 9x

.

So the equation becomes

-20 + 10x + 9x = 132

.

Combine like terms:

19x - 20 = 132

.

Add

20

to both sides:

19x = 152

.

Divide by

19

:

x = 8

.

Answer:

x = 8

The correct option is B.

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