Matrices & DeterminantsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Matrices & Determinants: Let Matrix Pmatrix Pmatrix Pmatrix Pmatrix Pmatrix Pmatrix (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let MM be a 3×33\times3 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:
A44
B55correct
C77
D1111
Solution
Step 1: Read off the columns of MM Each condition Me⃗j=v⃗jM\vec{e}_j = \vec{v}_j gives column jj of MM directly:
M=(10−1211321)M = \begin{pmatrix}1&0&-1\\2&1&1\\3&2&1\end{pmatrix}
 Step 2: Set up the linear system
{x−z=1⋯(1)2x+y+z=7⋯(2)3x+2y+z=11⋯(3)\begin{cases} x - z = 1 \quad &\cdots(1) \\ 2x + y + z = 7 \quad &\cdots(2) \\ 3x + 2y + z = 11 \quad &\cdots(3) \end{cases}
 Step 3: Solve the system (3)−(2)(3)-(2): x+y=4⋯(4)x+y=4 \quad\cdots(4) From (1): z=x−1⋯(5)z = x-1 \quad\cdots(5) Substituting (5) into (2): 2x+y+(x−1)=7⇒3x+y=8⋯(6)2x+y+(x-1)=7 \Rightarrow 3x+y=8 \quad\cdots(6) (6)−(4)(6)-(4): 2x=4⇒x=22x=4 \Rightarrow x=2. Then y=2y=2 from (4) and z=1z=1 from (5). Step 4: Compute the sum
x+y+z=2+2+1=5x+y+z = 2+2+1 = 5
 Answer: (2)
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