If a^2 + b^2 + c^2 = - 2 and f ( x ) = | \beginmatrix 1 + a^2x & ( 1 + b^2 )x & ( 1 + c^2 )x ( 1 + a^2 )x & 1 + b^2x & ( 1 + c^2 )x ( 1 + a^2 )x & ( 1 + b^2 )x & 1 + c^2x \endmatrix |, then f (x) is a polynomial of degree :

Matrices and Determinants PYQ — JEE Main 2005 · Shift 0 · Q77 with Solution

View written solutionFree

Correct answer: D

Write the matrix in a simpler form

Let $u=1+a^2,\quad v=1+b^2,\quad w=1+c^2.$ Then the given condition becomes $u+v+w=(1+a^2)+(1+b^2)+(1+c^2)=3+(a^2+b^2+c^2)=3-2=1.$

Also the determinant becomes

1+(u-1)x & vx & wx\\ ux & 1+(v-1)x & wx\\ ux & vx & 1+(w-1)x \end{vmatrix}.$$ 2. **Observe the column structure** Write each column separately: - Column 1: $$C_1=\begin{pmatrix}1+(u-1)x\\ ux\\ ux\end{pmatrix} =\begin{pmatrix}1-x\\ 0\\ 0\end{pmatrix}+u x\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ - Column 2: $$C_2=\begin{pmatrix}vx\\1+(v-1)x\\vx\end{pmatrix} =\begin{pmatrix}0\\1-x\\0\end{pmatrix}+v x\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ - Column 3: $$C_3=\begin{pmatrix}wx\\wx\\1+(w-1)x\end{pmatrix} =\begin{pmatrix}0\\0\\1-x\end{pmatrix}+w x\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ Hence the whole matrix is $$M=(1-x)I + x\begin{pmatrix}u&v&w\\u&v&w\\u&v&w\end{pmatrix}.$$ The second matrix has all rows equal, so it is of rank $1$. 3. **Factor out $(1-x)$** Since $$M=(1-x)I + xR=(1-x)\left(I+\frac{x}{1-x}R\right),$$ we get $$f(x)=(1-x)^3\det\left(I+tR\right),\qquad t=\frac{x}{1-x}.$$ 4. **Use the rank-1 determinant formula** Now $$R=\mathbf{1}\,[u\ v\ w],$$ where $$\mathbf{1}=\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ For a rank-1 matrix of the form $pq^T$, $$\det(I+t pq^T)=1+t\,q^Tp.$$ Here $$q^Tp=u+v+w=1.$$ So $$\det(I+tR)=1+t.$$ Therefore, $$f(x)=(1-x)^3(1+t)=(1-x)^3\left(1+\frac{x}{1-x}\right).$$ Simplify: $$1+\frac{x}{1-x}=\frac{1-x+x}{1-x}=\frac{1}{1-x}.$$ Thus $$f(x)=(1-x)^3\cdot \frac{1}{1-x}=(1-x)^2.$$ 5. **Degree of the polynomial** $$f(x)=(1-x)^2,$$ which is a polynomial of degree $$2.$$ 6. **Check options** - A: $1$ ❌ - B: $0$ ❌ - C: $3$ ❌ - D: $2$ ✅ So the correct option is **D**.

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View written solutionFree

Correct answer: D

Write the matrix in a simpler form

Let $u=1+a^2,\quad v=1+b^2,\quad w=1+c^2.$ Then the given condition becomes $u+v+w=(1+a^2)+(1+b^2)+(1+c^2)=3+(a^2+b^2+c^2)=3-2=1.$

Also the determinant becomes

1+(u-1)x & vx & wx\\ ux & 1+(v-1)x & wx\\ ux & vx & 1+(w-1)x \end{vmatrix}.$$ 2. **Observe the column structure** Write each column separately: - Column 1: $$C_1=\begin{pmatrix}1+(u-1)x\\ ux\\ ux\end{pmatrix} =\begin{pmatrix}1-x\\ 0\\ 0\end{pmatrix}+u x\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ - Column 2: $$C_2=\begin{pmatrix}vx\\1+(v-1)x\\vx\end{pmatrix} =\begin{pmatrix}0\\1-x\\0\end{pmatrix}+v x\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ - Column 3: $$C_3=\begin{pmatrix}wx\\wx\\1+(w-1)x\end{pmatrix} =\begin{pmatrix}0\\0\\1-x\end{pmatrix}+w x\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ Hence the whole matrix is $$M=(1-x)I + x\begin{pmatrix}u&v&w\\u&v&w\\u&v&w\end{pmatrix}.$$ The second matrix has all rows equal, so it is of rank $1$. 3. **Factor out $(1-x)$** Since $$M=(1-x)I + xR=(1-x)\left(I+\frac{x}{1-x}R\right),$$ we get $$f(x)=(1-x)^3\det\left(I+tR\right),\qquad t=\frac{x}{1-x}.$$ 4. **Use the rank-1 determinant formula** Now $$R=\mathbf{1}\,[u\ v\ w],$$ where $$\mathbf{1}=\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ For a rank-1 matrix of the form $pq^T$, $$\det(I+t pq^T)=1+t\,q^Tp.$$ Here $$q^Tp=u+v+w=1.$$ So $$\det(I+tR)=1+t.$$ Therefore, $$f(x)=(1-x)^3(1+t)=(1-x)^3\left(1+\frac{x}{1-x}\right).$$ Simplify: $$1+\frac{x}{1-x}=\frac{1-x+x}{1-x}=\frac{1}{1-x}.$$ Thus $$f(x)=(1-x)^3\cdot \frac{1}{1-x}=(1-x)^2.$$ 5. **Degree of the polynomial** $$f(x)=(1-x)^2,$$ which is a polynomial of degree $$2.$$ 6. **Check options** - A: $1$ ❌ - B: $0$ ❌ - C: $3$ ❌ - D: $2$ ✅ So the correct option is **D**.

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