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Worked Examples · Example 2

Q.Find the rank of B=(123014560)B=\begin{pmatrix}1&2&3\\0&1&4\\5&6&0\end{pmatrix}.

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✓ Free question

Computing det⁡B\det B

det⁡B=1(1×0−4×6)−2(0×0−4×5)+3(0×6−1×5)\det B = 1(1\times0-4\times6)-2(0\times0-4\times5)+3(0\times6-1\times5)

=1(0−24)−2(0−20)+3(0−5)=−24+40−15=1=1(0-24)-2(0-20)+3(0-5)=-24+40-15=1

Since det⁡B=1≠0\det B=1\neq0, BB has no linear dependence among its rows, so ρ(B)=3\rho(B)=3 — the maximum possible rank for a 3×33\times3 matrix.

Check (independent recomputation, expanding along the first column instead): det⁡B=1∣1460∣−0∣2360∣+5∣2314∣=1(0−24)−0+5(8−3)=−24+25=1\det B=1\begin{vmatrix}1&4\\6&0\end{vmatrix}-0\begin{vmatrix}2&3\\6&0\end{vmatrix}+5\begin{vmatrix}2&3\\1&4\end{vmatrix}=1(0-24)-0+5(8-3)=-24+25=1 — matches exactly.

✓Final answer

ρ(B)=3\rho(B)=3

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