Skip to content
dronacharya.ai
← JEE Main 2026 · 5 Apr · Shift 2
MathematicsSingle correct+4 / -1Question 4 of 75

JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 4

Let M be a 3×33\times3 matrix such that
M(100)=(123),M(010)=(012)andM(001)=(−111).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}.
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 :

Choose one option
Answer and worked solution

Official answer: B

M from its columns. The given products are the columns of M:
M=[10−1211321].M=\begin{bmatrix}1&0&-1\\2&1&1\\3&2&1\end{bmatrix}.

Solve x−z=1x-z=1, 2x+y+z=72x+y+z=7, 3x+2y+z=113x+2y+z=11. This gives (x,y,z)=(2,2,1)(x,y,z)=(2,2,1), so x+y+z=5x+y+z=5.

(B)

Topics:Systems of linear equationsMatrix algebra
← Previous questionNext question →