Assertion (A): det[1, 2, 3; 2, 4, 6; 5, 1, 7] = 0. Reason (R): A determinant with two proportional rows is 0.

Determinants, Adjoint & Inverse · Question 16 of 87

Question 16

MediumOriginal question

Assertion (A): $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 5 & 1 & 7 \end{vmatrix} = 0$ .
Reason (R): A determinant with two proportional rows is 0.

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

Row 2 is twice row 1, so two rows are proportional.
R is true and is exactly why the determinant is 0 (expanding gives $1(28 - 6) - 2(14 - 30) + 3(2 - 20) = 22 + 32 - 54 = 0$ ).

Concept: Proportional rows
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsThe determinant of a matrix

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Assertion (A): ∣123246517∣=0\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 5 & 1 & 7 \end{vmatrix} = 0​125​241​367​​=0.Reason (R): A determinant with two proportional rows is 0.

Explanation

Assertion (A): $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 5 & 1 & 7 \end{vmatrix} = 0$ .
Reason (R): A determinant with two proportional rows is 0.