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Friday, 9 October

JEE Main 2026 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 5 April 2026, Shift 2, Maths Q4

Question 4 of 75 in this shift, Maths question 4 of 25, Section A.

Let M be a 3×33\times 3 matrix such that M(100)=(123)M\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}, M(010)=(012)M\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\1\\2\end{pmatrix} and M(001)=(−111)M\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}-1\\1\\1\end{pmatrix}. If M(xyz)=(1711)M\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\7\\11\end{pmatrix}, then x+y+zx+y+z equals :
  1. (1)44
  2. (2)55Official answer
  3. (3)77
  4. (4)1111

Official answer

Option 2

NTA final key.

Chapter
Matrices

Same idea in other shifts

Asked 2× in all
  1. 2 Apr 2026, Shift 2 · Q22Consider the matrices A=[2−24−2]A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix} and B=[−3913]B = \begin{bmatrix} -3 & 9 \\ 1 & 3 \end{bmatrix}. If…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.