Q

# 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

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N

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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