If A = \(\begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}\), Find |A|
Let A = \(\begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}\)
By expanding along the first row, we have:
|A| = 1\(\begin{vmatrix}1 & -3\\4 & -9\end{vmatrix}\) - 1\(\begin{vmatrix}2 & -3\\5 & -9\end{vmatrix}\) - 2\(\begin{vmatrix}2 & 1\\5 & 4\end{vmatrix}\)
= 1(-9 + 12) – 1(-18 + 15) -2(8 – 5)
= 1(3) – 1 (-3) – 2(3)
= 3 + 3 – 6
= 6 – 6
= 0
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(iv) Reflexive and transitive but not symmetric.
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R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
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(c) R = {(x, y): x is exactly 7 cm taller than y}
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