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Determinants, Adjoint & Inverse · Question 11 of 87

Question 11

MediumOriginal question

Evaluate ∣2−3140−2153∣\begin{vmatrix}2 & -3 & 1 \\ 4 & 0 & -2 \\ 1 & 5 & 3\end{vmatrix}.

Options for question 11

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. Expand along the first row, with the signs +,−,++, -, +.

  2. The minors are ∣0−253∣=10\begin{vmatrix} 0 & -2 \\ 5 & 3 \end{vmatrix} = 10, ∣4−213∣=14\begin{vmatrix} 4 & -2 \\ 1 & 3 \end{vmatrix} = 14 and ∣4015∣=20\begin{vmatrix} 4 & 0 \\ 1 & 5 \end{vmatrix} = 20.

  3. 2(10)−(−3)(14)+1(20)=822(10) - (-3)(14) + 1(20) = 82.

Worth noting

−2-2 is what all-plus signs give: the middle term must be subtracted.

Concept
Value of a 3 × 3 determinant
Typical time
90 s
Your time
—
The rule this testsThe determinant of a matrix
Practise this chapter →

Ten questions from Determinants, Adjoint & Inverse, at your own pace.

Practise this chapter →

Ten questions from Determinants, Adjoint & Inverse, at your own pace.