MathsJEE Main 5 Apr 2026, Shift 2·Q4·MCQ·Easy–moderate

Matrices & Determinants · Matrix from images of basis vectors

Let

M

be a

3 \times 3

matrix such that

M\begin{pmatrix}1\\0\\0\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix},\quad M\begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}0\\1\\2\end{pmatrix} \quad\text{and}\quad M\begin{pmatrix}0\\0\\1\end{pmatrix} = \begin{pmatrix}-1\\1\\1\end{pmatrix}.

M\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\7\\11\end{pmatrix}

, then

x + y + z

equals:

(1)4
(2)5
(3)7
(4)11

We corrected a printing error in the original paper

Answer corrected; the source key is (3) — the columns of $M$ are $(1,2,3)$ , $(0,1,2)$ , $(-1,1,1)$ , so $x - z = 1$ , $2x + y + z = 7$ , $3x + 2y + z = 11$ give $(x,y,z) = (2,2,1)$ and $x + y + z = 5$ , option (2).

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See the full 5 Apr 2026, Shift 2 paper