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 :

Matrices and Determinants PYQ — JEE Main 2008 · Shift 0 · Q38 with Solution

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Correct answer: D

Write the system in matrix form

Given $x=cy+bz,\qquad y=az+cx,\qquad z=bx+ay.$

Bring all terms to the left: $x-cy-bz=0,$ $-cx+y-az=0,$ $-bx-ay+z=0.$

So,

\begin{pmatrix} 1 & -c & -b\\ -c & 1 & -a\\ -b & -a & 1 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}.

Since there exist real numbers $x,y,z$ not all zero, this homogeneous system has a non-trivial solution. Therefore, the determinant of the coefficient matrix must be zero.

Compute the determinant

Let

1 & -c & -b\\ -c & 1 & -a\\ -b & -a & 1 \end{pmatrix}.$$ Now,

\det(M)= 1\begin{vmatrix}1 & -a\-a & 1\end{vmatrix} -(-c)\begin{vmatrix}-c & -a\-b & 1\end{vmatrix} +(-b)\begin{vmatrix}-c & 1\-b & -a\end{vmatrix}.

Compute each minor: $$\begin{vmatrix}1 & -a\\-a & 1\end{vmatrix}=1-a^2.$$ $$\begin{vmatrix}-c & -a\\-b & 1\end{vmatrix}=-c-ab.$$ $$\begin{vmatrix}-c & 1\\-b & -a\end{vmatrix}=ac+b.$$ Hence,

\det(M)=1(1-a^2)+c(-c-ab)-b(ac+b).

So,

\det(M)=1-a^2-c^2-abc-abc-b^2 =1-a^2-b^2-c^2-2abc.

Since a non-trivial solution exists,

\det(M)=0.

Therefore,

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

which gives

a^2+b^2+c^2+2abc=1.

3. **Match with options** Thus the required value is $$1.$$ So the correct option is **D**.

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View written solutionFree

Correct answer: D

Write the system in matrix form

Given $x=cy+bz,\qquad y=az+cx,\qquad z=bx+ay.$

Bring all terms to the left: $x-cy-bz=0,$ $-cx+y-az=0,$ $-bx-ay+z=0.$

So,

\begin{pmatrix} 1 & -c & -b\\ -c & 1 & -a\\ -b & -a & 1 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}.

Since there exist real numbers $x,y,z$ not all zero, this homogeneous system has a non-trivial solution. Therefore, the determinant of the coefficient matrix must be zero.

Compute the determinant

Let

1 & -c & -b\\ -c & 1 & -a\\ -b & -a & 1 \end{pmatrix}.$$ Now,

\det(M)= 1\begin{vmatrix}1 & -a\-a & 1\end{vmatrix} -(-c)\begin{vmatrix}-c & -a\-b & 1\end{vmatrix} +(-b)\begin{vmatrix}-c & 1\-b & -a\end{vmatrix}.

Compute each minor: $$\begin{vmatrix}1 & -a\\-a & 1\end{vmatrix}=1-a^2.$$ $$\begin{vmatrix}-c & -a\\-b & 1\end{vmatrix}=-c-ab.$$ $$\begin{vmatrix}-c & 1\\-b & -a\end{vmatrix}=ac+b.$$ Hence,

\det(M)=1(1-a^2)+c(-c-ab)-b(ac+b).

So,

\det(M)=1-a^2-c^2-abc-abc-b^2 =1-a^2-b^2-c^2-2abc.

Since a non-trivial solution exists,

\det(M)=0.

Therefore,

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

which gives

a^2+b^2+c^2+2abc=1.

3. **Match with options** Thus the required value is $$1.$$ So the correct option is **D**.

Matrices and Determinants question

Matrices and Determinants question

More from Matrices and Determinants

Matrices and Determinants question

Matrices and Determinants question

More from Matrices and Determinants