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I have a doubt, kindly clarify. - Matrices and Determinants - JEE Main-2

Let a, b, c, be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x=cy+bz,y=az+cx\; and\; z=bx+ay. Then\; a^{2}+b^{2}+c^{2}+2abc is equal to

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    -1

  • Option 4)

    0

 
Answers (2)
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N neha

As we learnt in 

Cramer's rule for solving system of linear equations -

When \Delta =0 and atleast one of   \Delta_{1},\Delta _{2} and \Delta _{3}  is non-zero , system of equations has no solution

- wherein

a_{1}x+b_{1}y+c_{1}z=d_{1}

a_{2}x+b_{2}y+c_{2}z=d_{2}

a_{3}x+b_{3}y+c_{3}z=d_{3}

and 

\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}

 

 x-cy-bz=0

-cx+y-az=0

-bx-ay+z=0

= > D=\begin{vmatrix} 1 &-c &-b\\ -c &1&-a \\ -b&-a& 1\end{vmatrix}= 0

= >1\left (1-a^{2} \right )+c \left (-c-ab\right )-b\left (ac+b\right )= 0

= 1-a^{2}-c^{2}-abc-abc-b^{2}= 0

= a^{2}+b^{2}+c^{2}+2abc= 1


Option 1)

1

Correct Option

 

Option 2)

2

Incorrect Option

 

Option 3)

-1

Incorrect Option

 

Option 4)

0

Incorrect Option

 

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