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Worked Examples · Example 1

Q.Find the rank of A=(123234357)A=\begin{pmatrix}1&2&3\\2&3&4\\3&5&7\end{pmatrix}.

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Row-reducing AA

A=(123234357)A=\begin{pmatrix}1&2&3\\2&3&4\\3&5&7\end{pmatrix}

R2→R2−2R1R_2\to R_2-2R_1: [2−2(1), 3−2(2), 4−2(3)]=[0,−1,−2][2-2(1),\,3-2(2),\,4-2(3)]=[0,-1,-2].

R3→R3−3R1R_3\to R_3-3R_1: [3−3(1), 5−3(2), 7−3(3)]=[0,−1,−2][3-3(1),\,5-3(2),\,7-3(3)]=[0,-1,-2].

(1230−1−20−1−2)\begin{pmatrix}1&2&3\\0&-1&-2\\0&-1&-2\end{pmatrix}

R3→R3−R2R_3\to R_3-R_2: [0−0, −1−(−1), −2−(−2)]=[0,0,0][0-0,\,-1-(-1),\,-2-(-2)]=[0,0,0].

(1230−1−2000)\begin{pmatrix}1&2&3\\0&-1&-2\\0&0&0\end{pmatrix}

Two non-zero rows remain, so ρ(A)=2\rho(A)=2.

Check (independent recomputation via det⁡A\det A): det⁡A=1(3×7−4×5)−2(2×7−4×3)+3(2×5−3×3)=1(21−20)−2(14−12)+3(10−9)=1−4+3=0\det A=1(3\times7-4\times5)-2(2\times7-4\times3)+3(2\times5-3\times3)=1(21-20)-2(14-12)+3(10-9)=1-4+3=0. Since det⁡A=0\det A=0, ρ(A)<3\rho(A)<3. The top-left 2×22\times2 minor ∣1223∣=3−4=−1≠0\begin{vmatrix}1&2\\2&3\end{vmatrix}=3-4=-1\neq0, so a non-zero 2×22\times2 minor exists — confirming ρ(A)=2\rho(A)=2 exactly.

✓Final answer

ρ(A)=2\rho(A)=2

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