Let a, b, c be such that b( a + c ) e 0 if | \beginmatrix a & a + 1 & a - 1 - b & b + 1 & b - 1 c & c - 1 & c + 1 \endmatrix | + | \beginmatrix a + 1 & b + 1 & c - 1 a - 1 & b - 1 & c + 1 ( - 1 )^n + 2a & ( - 1 )^n + 1b & ( - 1 )^nc \endmatrix | = 0 then the value of n :

Matrices and Determinants PYQ — JEE Main 2009 · Shift 0 · Q35 with Solution

View written solutionFree

Correct answer: B

Interpret the condition

We are given $D_1 + D_2 = 0,$ where

a & a+1 & a-1\\ -b & b+1 & b-1\\ c & c-1 & c+1 \end{vmatrix}$$ and $$D_2=\begin{vmatrix} a+1 & b+1 & c-1\\ a-1 & b-1 & c+1\\ (-1)^{n+2}a & (-1)^{n+1}b & (-1)^n c \end{vmatrix}.$$ Also, the printed condition is clearly intended as $$b(a+c)\ne 0,$$ so that we may divide by this quantity if needed. We simplify both determinants. --- 2. **Evaluate $D_1$** $$D_1=\begin{vmatrix} a & a+1 & a-1\\ -b & b+1 & b-1\\ c & c-1 & c+1 \end{vmatrix}.$$ Apply column operations: $$C_2\to C_2-C_1,\qquad C_3\to C_3-C_1.$$ Then $$D_1=\begin{vmatrix} a & 1 & -1\\ -b & 2b+1 & 2b-1\\ c & -1 & 1 \end{vmatrix}.$$ Now expand along the first row: $$D_1=a\begin{vmatrix}2b+1 & 2b-1\\ -1 & 1\end{vmatrix}-1\begin{vmatrix}-b & 2b-1\\ c & 1\end{vmatrix}+(-1)\begin{vmatrix}-b & 2b+1\\ c & -1\end{vmatrix}.$$ Compute minors: $$\begin{vmatrix}2b+1 & 2b-1\\ -1 & 1\end{vmatrix}=(2b+1)(1)-(2b-1)(-1)=4b,$$ $$\begin{vmatrix}-b & 2b-1\\ c & 1\end{vmatrix}=-b-c(2b-1),$$ $$\begin{vmatrix}-b & 2b+1\\ c & -1\end{vmatrix}=b-c(2b+1).$$ So $$D_1=4ab-\big(-b-c(2b-1)\big)-\big(b-c(2b+1)\big).$$ Simplifying, $$D_1=4ab+b+c(2b-1)-b+c(2b+1)=4ab+4bc=4b(a+c).$$ Hence, $$D_1=4b(a+c).$$ --- 3. **Evaluate $D_2$ for parity of $n$** $$D_2=\begin{vmatrix} a+1 & b+1 & c-1\\ a-1 & b-1 & c+1\\ (-1)^{n+2}a & (-1)^{n+1}b & (-1)^n c \end{vmatrix}.$$ Since $$(-1)^{n+2}=(-1)^n, \qquad (-1)^{n+1}=-(-1)^n,$$ we consider two cases. --- ### Case 1: $n$ even Then $(-1)^n=1$, so the third row becomes $$(a,-b,c).$$ Thus $$D_2=\begin{vmatrix} a+1 & b+1 & c-1\\ a-1 & b-1 & c+1\\ a & -b & c \end{vmatrix}.$$ Now do row operations: $$R_1\to R_1-R_3,\qquad R_2\to R_2-R_3.$$ Then $$D_2=\begin{vmatrix} 1 & 2b+1 & -1\\ -1 & 2b-1 & 1\\ a & -b & c \end{vmatrix}.$$ Next, add first two rows: $$R_1\to R_1+R_2,$$ so $$D_2=\begin{vmatrix} 0 & 4b & 0\\ -1 & 2b-1 & 1\\ a & -b & c \end{vmatrix}.$$ Expand along the first row: $$D_2=4b(-1)^{1+2}\begin{vmatrix}-1 & 1\\ a & c\end{vmatrix} =-4b\big((-1)c-a\big)=4b(a+c).$$ So for even $n$, $$D_2=4b(a+c).$$ Therefore, $$D_1+D_2=4b(a+c)+4b(a+c)=8b(a+c).$$ Given $b(a+c)\ne 0$, this is **not zero**. Hence even $n$ is not possible. --- ### Case 2: $n$ odd Then $(-1)^n=-1$, so the third row becomes $$(-a,b,-c).$$ Thus $$D_2=\begin{vmatrix} a+1 & b+1 & c-1\\ a-1 & b-1 & c+1\\ -a & b & -c \end{vmatrix}.$$ Now apply row operations: $$R_1\to R_1+R_3,\qquad R_2\to R_2+R_3.$$ Then $$D_2=\begin{vmatrix} 1 & 2b+1 & -1\\ -1 & 2b-1 & 1\\ -a & b & -c \end{vmatrix}.$$ Again, $$R_1\to R_1+R_2,$$ so $$D_2=\begin{vmatrix} 0 & 4b & 0\\ -1 & 2b-1 & 1\\ -a & b & -c \end{vmatrix}.$$ Expand along the first row: $$D_2=4b(-1)^{1+2}\begin{vmatrix}-1 & 1\\ -a & -c\end{vmatrix}.$$ Now $$\begin{vmatrix}-1 & 1\\ -a & -c\end{vmatrix}=(-1)(-c)-1(-a)=a+c.$$ Thus $$D_2=-4b(a+c).$$ Therefore, $$D_1+D_2=4b(a+c)-4b(a+c)=0,$$ which satisfies the condition. --- 4. **Conclusion** The equation holds exactly when $n$ is **odd**. So the correct option is $$\boxed{\text{B: any odd integer}}.$$

More from Matrices and Determinants

Let $a, b, c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x = cy + b z, y = a z + c x,$ and $z = b x + a y .$ Then $a^{2} + b^{2} + c^{2} + 2 ab c$ is equal to :2008 · MCQ
Let $A$ be $a 2 \times 2$ matrix with real entries. Let $I$ be the $2 \times 2$ identity matrix. Denote by tr $(A)$ , the sum of diagonal entries of $a$ . Assume that $a^{2} = I .$ Statement-1 : If $A e I$ and $A e - I$ , then det $(A) = - 1$ …2008 · MCQ
If $D = 111 1 1 + x 1 11 1 + y$ for $x e 0, y e 0,$ then $D$ is :2007 · MCQ
If $A$ and $B$ are square matrices of size $n \times n$ such that $A^{2} - B^{2} = (A - B) (A + B),$ then which of the following will be always true?2006 · MCQ
Let $A = (1324)$ and $B = (a 0 0 b), a, b \in N .$ Then2006 · MCQ
If $A^{2} - A + 1 = 0$ , then the inverse of $A$ is :2005 · MCQ
The system of equations $α x + y + z = α - 1 x + α y + z = α - 1 x + y + α z = α - 1$ has no solutions, if $α$ is :2005 · MCQ
If $a_{1}, a_{2}, a_{3}, ........, a_{n}, .....$ are in G.P., then the determinant $Δ = lo g a_{n} lo g a_{n + 3} lo g a_{n + 6} lo g a_{n + 1} lo g a_{n + 4} lo g a_{n + 7} lo g a_{n + 2} lo g a_{n + 5} lo g a_{n + 8}$ …2005 · MCQ

View written solutionFree

Correct answer: B

Interpret the condition

We are given $D_1 + D_2 = 0,$ where

a & a+1 & a-1\\ -b & b+1 & b-1\\ c & c-1 & c+1 \end{vmatrix}$$ and $$D_2=\begin{vmatrix} a+1 & b+1 & c-1\\ a-1 & b-1 & c+1\\ (-1)^{n+2}a & (-1)^{n+1}b & (-1)^n c \end{vmatrix}.$$ Also, the printed condition is clearly intended as $$b(a+c)\ne 0,$$ so that we may divide by this quantity if needed. We simplify both determinants. --- 2. **Evaluate $D_1$** $$D_1=\begin{vmatrix} a & a+1 & a-1\\ -b & b+1 & b-1\\ c & c-1 & c+1 \end{vmatrix}.$$ Apply column operations: $$C_2\to C_2-C_1,\qquad C_3\to C_3-C_1.$$ Then $$D_1=\begin{vmatrix} a & 1 & -1\\ -b & 2b+1 & 2b-1\\ c & -1 & 1 \end{vmatrix}.$$ Now expand along the first row: $$D_1=a\begin{vmatrix}2b+1 & 2b-1\\ -1 & 1\end{vmatrix}-1\begin{vmatrix}-b & 2b-1\\ c & 1\end{vmatrix}+(-1)\begin{vmatrix}-b & 2b+1\\ c & -1\end{vmatrix}.$$ Compute minors: $$\begin{vmatrix}2b+1 & 2b-1\\ -1 & 1\end{vmatrix}=(2b+1)(1)-(2b-1)(-1)=4b,$$ $$\begin{vmatrix}-b & 2b-1\\ c & 1\end{vmatrix}=-b-c(2b-1),$$ $$\begin{vmatrix}-b & 2b+1\\ c & -1\end{vmatrix}=b-c(2b+1).$$ So $$D_1=4ab-\big(-b-c(2b-1)\big)-\big(b-c(2b+1)\big).$$ Simplifying, $$D_1=4ab+b+c(2b-1)-b+c(2b+1)=4ab+4bc=4b(a+c).$$ Hence, $$D_1=4b(a+c).$$ --- 3. **Evaluate $D_2$ for parity of $n$** $$D_2=\begin{vmatrix} a+1 & b+1 & c-1\\ a-1 & b-1 & c+1\\ (-1)^{n+2}a & (-1)^{n+1}b & (-1)^n c \end{vmatrix}.$$ Since $$(-1)^{n+2}=(-1)^n, \qquad (-1)^{n+1}=-(-1)^n,$$ we consider two cases. --- ### Case 1: $n$ even Then $(-1)^n=1$, so the third row becomes $$(a,-b,c).$$ Thus $$D_2=\begin{vmatrix} a+1 & b+1 & c-1\\ a-1 & b-1 & c+1\\ a & -b & c \end{vmatrix}.$$ Now do row operations: $$R_1\to R_1-R_3,\qquad R_2\to R_2-R_3.$$ Then $$D_2=\begin{vmatrix} 1 & 2b+1 & -1\\ -1 & 2b-1 & 1\\ a & -b & c \end{vmatrix}.$$ Next, add first two rows: $$R_1\to R_1+R_2,$$ so $$D_2=\begin{vmatrix} 0 & 4b & 0\\ -1 & 2b-1 & 1\\ a & -b & c \end{vmatrix}.$$ Expand along the first row: $$D_2=4b(-1)^{1+2}\begin{vmatrix}-1 & 1\\ a & c\end{vmatrix} =-4b\big((-1)c-a\big)=4b(a+c).$$ So for even $n$, $$D_2=4b(a+c).$$ Therefore, $$D_1+D_2=4b(a+c)+4b(a+c)=8b(a+c).$$ Given $b(a+c)\ne 0$, this is **not zero**. Hence even $n$ is not possible. --- ### Case 2: $n$ odd Then $(-1)^n=-1$, so the third row becomes $$(-a,b,-c).$$ Thus $$D_2=\begin{vmatrix} a+1 & b+1 & c-1\\ a-1 & b-1 & c+1\\ -a & b & -c \end{vmatrix}.$$ Now apply row operations: $$R_1\to R_1+R_3,\qquad R_2\to R_2+R_3.$$ Then $$D_2=\begin{vmatrix} 1 & 2b+1 & -1\\ -1 & 2b-1 & 1\\ -a & b & -c \end{vmatrix}.$$ Again, $$R_1\to R_1+R_2,$$ so $$D_2=\begin{vmatrix} 0 & 4b & 0\\ -1 & 2b-1 & 1\\ -a & b & -c \end{vmatrix}.$$ Expand along the first row: $$D_2=4b(-1)^{1+2}\begin{vmatrix}-1 & 1\\ -a & -c\end{vmatrix}.$$ Now $$\begin{vmatrix}-1 & 1\\ -a & -c\end{vmatrix}=(-1)(-c)-1(-a)=a+c.$$ Thus $$D_2=-4b(a+c).$$ Therefore, $$D_1+D_2=4b(a+c)-4b(a+c)=0,$$ which satisfies the condition. --- 4. **Conclusion** The equation holds exactly when $n$ is **odd**. So the correct option is $$\boxed{\text{B: any odd integer}}.$$

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