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Determinants, Adjoint & Inverse · Question 22 of 87

Question 22

HardOriginal question

What is ∣1ab−a1c−b−c1∣\begin{vmatrix}1 & a & b \\ - a & 1 & c \\ - b & - c & 1\end{vmatrix}?

Options for question 22

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. Row 1: 1(1+c2)−a(−a+bc)+b(ac+b)1(1 + c^2) - a(-a + bc) + b(ac + b).

  2. =1+c2+a2−abc+abc+b2=1+a2+b2+c2= 1 + c^2 + a^2 - abc + abc + b^2 = 1 + a^2 + b^2 + c^2.

Worth noting

1+a2+b2+c2−2abc1 + a^2 + b^2 + c^2 - 2abc keeps the abcabc terms, but one comes with −a-a and one with +b+b, so they cancel.

Concept
A determinant with letters
Typical time
120 s
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The rule this testsThe determinant of a matrix
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Ten questions from Determinants, Adjoint & Inverse, at your own pace.

Practise this chapter →

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