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pishuonlain [190]
3 years ago
9

The 3 × 3 matrix P satisfies the matrix equation P^2 = P.

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer:  The answers are given below.

Step-by-step explanation:  Given that a  3 × 3 matrix P satisfies the matrix equation P² = P.

We are to

(a) find the possibilities for the determinant of P.

(b) explain the reason behind there are no other possibilities.

(c) give an example of P, for each possible determinant.

(a) According to the given information, we have

P^2=P\\\\\Rightarrow P^2-P=0\\\\\Rightarrow P(P-I)=0\\\\\Rightarrow P=0,~~~P=I.

So, P can be either a zero matrix of order 3 or an identity matrix of order 3.

If P = 0, then det(P) = 0     and     if P = I, then det(P) = 1.

Therefore the possible determinants of P are 0 and 1.

(b) There can be any other determinant other than 0 and 1, because if so, then the given equation P² = P will not be satisfied.

(c) If |P| = 0, then the matrix P can be can be of the form as follows:

P=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right] .

If |P| = 1, the the matrix P can be of the form as follows :

P=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] .

Thus, all the parts are answered.

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Using the rule of exponents

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\frac{3^{5} }{3^{8} } = \frac{1}{3^{(8-5)} } = \frac{1}{3^{3} } = \frac{1}{27}

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Read 2 more answers
A gold mine is projected to produce $52,000 during its first year of operation, $50,000 the second year, $48,000 the third year,
marshall27 [118]

Answer:

its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

Step-by-step explanation:

Given that;

gold mine is projected to produce $52,000 during its first year

produce $50,000 the second year

produce $48,000 the third year

the mine is expected to produce for 20yrs; i.e n = 20

annual interest rate = 4% = 0.04%

now let P represent the present worth

we determine the present worth ;

Present worth ⇒ Cashflow(Uniform series present worth) - (2000)(uniform gradient present worth)

⇒Cashflow(P | A,i%,n) - (2000)(P | G,i%,n)

= 52000[ ((1+i)ⁿ - 1) / (i(1+i)ⁿ) ] - (2000)[ {((1+i)ⁿ - 1) / (i²(1+i)ⁿ))} - (n/i(1 + i)ⁿ) ]

= 52000[ ((1+0.04)²⁰ - 1) / (0.04(1+0.04)²⁰) ] - (2000)[ {((1+0.04)20 - 1) / ((0.04)²(1+0.04)²⁰))} - (20/0.04(1 + 0.04)²⁰) ]

= 52000[ ((1.04)²⁰ - 1) / (0.04(1.04)²⁰) ] - (2000)[ {((1.04)²⁰ - 1) / ((0.04)²(1.04)²⁰))} - (20/0.04(1.04)²⁰) ]

= 52000[ ((2.191123 - 1) / (0.04(2.191123) ] - (2000)[ {((2.191123 - 1) / ((0.0016)(2.191123))} - (20/0.04(2.191123) ]

= 52000(1.191123/0.08764) - (2000){( 1.191123/0.003506) - (20/0.87645)}

= 52000(13.59033) - (2000)(339.7582 - 228.1935)

= 52000(13.59033) - (2000)(111.5647)

= 706695.6 - 223129.4

= 483,566.2 ≈ 483,566

Therefore its present worth is nearest to 483,566

Option d) 483,566 is the correct answer

4 0
3 years ago
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