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JAMB Mathematics 2023 Paper

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Question : 19

Total: 80

If D =

\begin{bmatrix}2 & -1 & 3 \\ 4 & 1 & 2 \\ 1 & -3 & 1\end{bmatrix}

\| D \|

16
14
-23
-37

Solution:

Concept:

To find

\|D\|

, compute the determinant of the given

3 \times 3

matrix by expanding along the first row.

Explanation:

Using expansion along the first row of

D = \begin{bmatrix}2 & -1 & 3 \\ 4 & 1 & 2 \\ 1 & -3 & 1\end{bmatrix}

:

\|D\| = 2\begin{vmatrix}1 & 2 \\ -3 & 1\end{vmatrix} -(-1)\begin{vmatrix}4 & 2 \\ 1 & 1\end{vmatrix} + 3\begin{vmatrix}4 & 1 \\ 1 & -3\end{vmatrix}

Evaluate each

2 \times 2

determinant:

= 2(1 \times 1 - 2 \times (-3)) + 1(4 \times 1 - 2 \times 1) + 3(4 \times (-3) - 1 \times 1)

= 2(1 + 6) + 1(4 - 2) + 3(-12 - 1)

= 2(7) + 1(2) + 3(-13)

= 14 + 2 - 39

= -23

Answer:

\|D\| = -23

, so the correct option is C.

-23

.

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