1that`s pretty much what i had done,sir....cant think of any oder method
1.
A= \begin{bmatrix} 1& 0 &1 \\ 0 & 1 &1 \\ 0& -2 & 4 \end{bmatrix}
6A-1=A2+cA+dI,then c,d =
[a nice method will do,without working a lot]
2. Assertion and Reason:
Consider the system of equations: x-2y+3z=-1;x-3y+4z=1;-x+y-2z=k
Statement 1:the system of eqtns has no soln for k≠3 and
Statement 2: the determinant \begin{vmatrix} 1 & 3 &-1 \\ -1 & -2 &k \\ 1& 4 & 1 \end{vmatrix} ≠0 for k≠0
[got the 1st statement as true,but how do we relate the 2nd one?]
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10 Answers
62Multiply both sides by A
and then u have I=.....
Find A, A^2 and A^3
almost done.
It is easier than finding A inverse!
1622nd question is based on cramer's rule!
u havent studied it yet i suppose!
1yes i have sir....but sir is statement 2 an explanation of statement 1?....
62u sure it is for k≠0
i think it should have been for for k≠3
in that case it wud be the correct explanation too.
if the current question is correct then the statement 2 is false!
1actually its from previous yrs iit paper,by arihant....the solution was also done with this data which i did not undrstand and the answer was given that statement 2 is correct,that`s why i had posted it....if its wrong then i think i can solve it ,sir ,thnx a lot....
66For the first one, you can apply a particular useful result: A square matrix satisfies its characteristic equation. The characteristic equation for a square matrix is a polynomial in x given by
det(A-xI) =0.
For the current problem, this reduces to the equation
6-11x+6x2-x3=0
Since A satisfies this, we get
6 - 11 A + 6A2 - A3=0
Since A is invertible, multiplying throughout by the inverse of A, we get
6A-1 - 11 I + 6 A - A2 =0
i.e.
6A-1 = A2 - 6 A + 11 I
Hence c= -6, d = 11
1wasn`t aware of this result....pretty useful..actually after reading this i solved another sum using this result after a lot of tries,thnx a lot....
