Find the rank of the matrix ∣∣
∣∣01212−30−111−10∣∣
∣∣
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Solution
Let A=∣∣
∣∣01212−30−111−10∣∣
∣∣ ∼⎡⎢⎣01212−22011−10⎤⎥⎦R2⟶R2+R1 ∼⎡⎢⎣01211−11011−10⎤⎥⎦R2⟶12R2 ∼⎡⎢⎣01211−1102000⎤⎥⎦R3→R3+R2 The last equivalent matrix is in echelon form The number of non-zero rows in this matrix is three ∴ρ(A)=3