question_answer

The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method,

OR

If  prove that

\[n\in N.\]

Answer:

Let the first, second and third number be x, y and z respectively. Then, according to given conditions, we have   \[x+y+z=6,\text{ }y+3z=11\]

and \[x+z=2\] or \[x-2y+z=0\]

This system of equation can be written in matrix form as

or AX = B, where

and

Now, consider

\[=1(1+6)-\,1(0-3)+1(0-1)\]

\[=7+3-1\ne 0\]

\[\therefore \] \[{{A}^{-1}}\] exists.

Now, let us find the cofactor of elements of \[|A|.\]

Clearly, \[{{C}_{11}}={{(-\,1)}^{1\,+\,1}}\left| \begin{matrix}    1 & 3  \\    -\,2 & 1  \\ \end{matrix} \right|=7,\,\,{{C}_{12}}={{(-\,1)}^{1\,+\,2}}\left| \begin{matrix}    0 & 3  \\    1 & 1  \\ \end{matrix} \right|=3\]

\[{{C}_{13}}={{(-\,1)}^{1\,+\,3}}\left| \begin{matrix}    0 & 1  \\    1 & -\,2  \\ \end{matrix} \right|=-\,1,\,\,{{C}_{21}}={{(-\,1)}^{2\,+\,1}}\left| \begin{matrix}    1 & 1  \\    -\,2 & 1  \\ \end{matrix} \right|=-\,3,\]\[{{C}_{22}}={{(-\,1)}^{2\,+\,2}}\left| \begin{matrix}    1 & 1  \\    1 & 1  \\ \end{matrix} \right|=0,\,\,{{C}_{23}}={{(-\,1)}^{2\,+\,3}}\left| \begin{matrix}    1 & 1  \\    1 & -\,2  \\ \end{matrix} \right|=3,\]

\[{{C}_{31}}={{(-\,1)}^{3\,+\,1}}\left| \begin{matrix}    1 & 1  \\    1 & 3  \\ \end{matrix} \right|=2,\,\,{{C}_{32}}={{(-\,1)}^{3\,+\,2}}\left| \begin{matrix}    1 & 1  \\    0 & 3  \\ \end{matrix} \right|=-\,3\]

and \[{{C}_{33}}={{(-\,1)}^{3\,+\,3}}\left| \begin{matrix}    1 & 1  \\    0 & 1  \\ \end{matrix} \right|=1\]

\[\therefore \] adj  and

Now, as \[X={{A}^{-1}}B\]

\[\therefore \]

\[\Rightarrow \]

Hence, x = 1, y = 2 and z = 3.

OR

Given

To prove

Proof We shall prove this statement by using principle of mathematical induction.

Now, let P(n) be the given statement,

i.e.

For n = 1, we have

\[\therefore \] Statement is true for n = 1.

Now, let us assume that statement is true for n = k, we have

?(i)

Now, we shall prove the statement for n = k + 1, we have to show,

Consider, \[LHS={{A}^{k+1}}={{A}^{k}}.A\]

[using Eq. (i)]

\[[\because \,\,{{3.3}^{k-1}}={{3}^{1+k-1}}={{3}^{k}}]\]

= RHS

\[\therefore \] Statement is true for n = k +1.

Hence, by principle of mathematical induction, statement is true for all n, where \[n\in N.\]