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Question
Find the rank of the following matrices
`((3, 1, -5, -1),(1, -2, 1, -5),(1, 5, -7, 2))`
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Solution
Let A = `((3, 1, -5, -1),(1, -2, 1, -5),(1, 5, -7, 2))`
Order of A is 3 × 4
∴ p(A) ≤ 3
Consider the third order minors,
`|(3, 1, -5),(1, -2, 1),(1, 5, -7)|` = 3(14 – 5) – 1(– 7 – 1) – 5( + 2)
= 3(9) – (– 8) – 5(7)
= 27 + 8 – 35
= 0
`|(3, 1, -1),(1, -2, -5),(1, 5, 2)|` = 3 + (– 4 + 25) – 1(2 + 5) – 1(5 + 2)
3(21) – (7) – (7)
= 63 – 14
= 49 ≠ 0
`|(3, -5, -1),(1, 1, -5),(1, -7, 2)|` = 3(2 – 35) + 5(2 + 5) – 1(– 7 – 1)
= 3(– 33) + 5(7) – (– 8)
= – 99 + 3 + 8
= – 56 ≠ 0
`|(1, -5, -1),(-2, 1, -5),(5, -7, 2)|` = 1(2 – 35) + 5(– 4 + 5) – 1(14 – 5)
= – 33 + 5(21) – 9
= – 42 + 105
= 63 ≠ 0
Since there are 3 minors which do not vanish, p(A) = 3
We can also find the rank by using echelon form of matrix A.
A = `((3, 1, -5, -1),(1, -2, 1, -5),(1, 5, -7, 2))`
`˜ ((1, -2, 1, -5),(3, 1, -5, -1),(1, 5, -7, 2))` `{:"R"_1 ↔ "R"_2:}`
`˜ ((1, -2, 1, -5),(.0, 7, -8, 14),(0, 7, -8, 7))` `{:("R"_2 -> "R"_2 - 3"R"_1),("R"_3 -> "R"_3 - "R"_1):}`
`˜ ((1, -2, 1, -5),(0, 7, -8, 14),(0, 0, 0,-7))` `{:"R"_3 -> "R"_3 - "R"_1:}`
The number of non-zero rows is 3
∴ p(A) = 3
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