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Exercises · Q9

Q.Find the rank of (246123369)\begin{pmatrix}2&4&6\\1&2&3\\3&6&9\end{pmatrix}.

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✓ Free question

Observing the rows

Row 1 =[2,4,6]=2×[1,2,3]=2×=[2,4,6]=2\times[1,2,3]=2\timesRow 2. Row 3 =[3,6,9]=3×[1,2,3]=3×=[3,6,9]=3\times[1,2,3]=3\timesRow 2. Every row is a scalar multiple of Row 2.

Row-reducing

R1→R1−2R2R_1\to R_1-2R_2 gives [0,0,0][0,0,0]; R3→R3−3R2R_3\to R_3-3R_2 gives [0,0,0][0,0,0].

(000123000)\begin{pmatrix}0&0&0\\1&2&3\\0&0&0\end{pmatrix}

Only one non-zero row remains, so the rank is 11.

Check (independent recomputation via determinant): det⁡=2(2×9−3×6)−4(1×9−3×3)+6(1×6−2×3)=2(0)−4(0)+6(0)=0\det=2(2\times9-3\times6)-4(1\times9-3\times3)+6(1\times6-2\times3)=2(0)-4(0)+6(0)=0, and every 2×22\times2 minor (e.g. ∣2412∣=4−4=0\begin{vmatrix}2&4\\1&2\end{vmatrix}=4-4=0) is also zero, confirming the rank is strictly less than 2 — so the rank is exactly 1, matching the row-scaling observation.

✓Final answer

Rank =1=1

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