Find the matrix T if ST = I where S = \(\begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix}\)
\(\begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}\)
\(\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix}\)
\(\begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}\)
\(\begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}\)
Explanation
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Soln
[a^2(3/4)b^-3(3/4)c^(3/4)]divided by (a^-1b^4c^5)
===>(a^3/2b^-9/4c^3/4)divided by (a^-1b^4c^5).
Note a^p =3/2-(-1)==>p=5/2
q=-9/4-(4)===>q=-25/4
r=3/4-(5)===>r=-17/4
now p+2q
5/2+2(-25/4)
:- p+2q=-10.
Make use of indice...
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Soln
[a^2(3/4)b^-3(3/4)c^(3/4)]divided by (a^-1b^4c^5)
===>(a^3/2b^-9/4c^3/4)divided by (a^-1b^4c^5).
Note a^p =3/2-(-1)==>p=5/2
q=-9/4-(4)===>q=-25/4
r=3/4-(5)===>r=-17/4
now p+2q
5/2+2(-25/4)
:- p+2q=-10.
Make use of indice...
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Soln
{(a^2b^-3c)^3\4 /a^-1b^4c^5} =a^pb^qc^r
from the power law of indices.
(a^2b^-3c)^3\4 =a^2*3\4.b^-3*3\4.c^1*3\4
=a^3\2.b^-9\4.c^3\4
from the divisional law of indices
(a^3\a.b^-9\4.c^3\4)/a^-1.b^4.c^5 =
a^3\2+1.b^-9\4-4.c^3\4-5
=a^5\2.b-25\4.c^-17\4
Now by comparing the powers in a^p.b^q.c^r
p=5\2,b=-25\4,c=-17\4
substituting the values of p and q in the equation p+2q,we will have
5\2 + 2(-25\4) = 5\2- 25\2
=(5-25)\2 = -20\2 = -10 Ans.
using the laws of indices u get the values of p=5/2, q=-25/4 and r=-17/4 then you solve p +2q which gives -10
Here is an explanation:
To solve this, it is simple though the question looks more complex.
Solution
(a^2 b^2 c)^3\4 divided by a^-1 b^4 c^5 equals to a^p b^q c^r.
First let's separate the question, let's put aside the equals to part, now
Multiply all the numerator through by the power of 3/4. You will get a^3/2 b^-9/4 c^3/4 after this, use law of indices to divide the numerator with the denominator so we have a^3/2 ÷ a^-1 b^-9/4 ÷ b^4 c^3/4 ÷ c^5 after using law of indices to solve it, you will get a^5/2 b^-25/4 c^-17/4. So after these we get our p,q,r by comparing the powers.
:. a^5/2 b^-25/4 c^-17/4 = a^p b^q c^r, so p = 5/2, q = -25/4 c = -17/4. So now P + 2Q = 5/2 + 2(-25/4) after solving this, you will get -10 which is the answer.
Thank you.
REF: Jamb, 1999
This question is really great.what you need to first is to open the bracket then use indices to remove the division sign to subtraction sign then after which you compare to get p,q and r.then u can substitute it into p+2q.that's it.you just need to tink hard.
a^6/4b^-9/4c^3/4 for d numerator wen u multiply 3/4 For d denominator q^-1b^4c^5
Apply the law of indices u will gt a^10/4b^-25/4c^-17/4(In a^p b^q c^r)
but the questn is d value of p+2q i.e p=10/4 q=-25/4 Substitute it in p+2q so 10/4+2(-25/4)=10/4-50/4 =40/4=-10
so the equation is
(a^2 b^-3 c)/a^-1 b^4c^5
and the questions is what is the value of p+2q
solu
p=2 q= -3
2q means 2 *q (-3)
so
p + 2q= 2+ 2* -3 = - 10
so the answer is - 10
