Question

Let a, b, c be such that b (a + c) $\ne$0. If $\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| + \left| \begin{gathered} a + 1{\text{ }}b + 1{\text{ }}c - 1 \\\ a - 1{\text{ }}b - 1{\text{ }}c + 1 \\\ {\left( { - 1} \right)^{n + 2}}a{\text{ }}{\left( { - 1} \right)^{n + 1}}b{\text{ }}{\left( { - 1} \right)^n}c \\\ \end{gathered} \right| = 0$, then the value of n is
(A). Zero
(B). Any even integer
(C). Any odd integer
(D). Any integer

Answer 1

Before attempting this question, one should have prior knowledge about the concept of determinants and also remember to use row and column transformation method, use this information to approach the solution of the question.

Complete step-by-step answer :

a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| + \left| \begin{gathered} a + 1{\text{ }}b + 1{\text{ }}c - 1 \\\ a - 1{\text{ }}b - 1{\text{ }}c + 1 \\\ {\left( { - 1} \right)^{n + 2}}a{\text{ }}{\left( { - 1} \right)^{n + 1}}b{\text{ }}{\left( { - 1} \right)^n}c \\\ \end{gathered} \right| = 0$$ Let $$D = \left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right|$$ and $$A = \left| \begin{gathered} a + 1{\text{ }}b + 1{\text{ }}c - 1 \\\ a - 1{\text{ }}b - 1{\text{ }}c + 1 \\\ {\left( { - 1} \right)^{n + 2}}a{\text{ }}{\left( { - 1} \right)^{n + 1}}b{\text{ }}{\left( { - 1} \right)^n}c \\\ \end{gathered} \right|$$ Taking ${\left( { - 1} \right)^n}$as common for A $$\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| + {\left( { - 1} \right)^n}\left| \begin{gathered} a + 1{\text{ }}b + 1{\text{ }}c - 1 \\\ a - 1{\text{ }}b - 1{\text{ }}c + 1 \\\ {\text{ }}a{\text{ }} - b{\text{ }}c \\\ \end{gathered} \right| = 0$$ Now using the property of transpose i.e.${\left| A \right|^T} = \left| A \right|$ $${\left| \begin{gathered} a + 1{\text{ }}b + 1{\text{ }}c - 1 \\\ a - 1{\text{ }}b - 1{\text{ }}c + 1 \\\ {\text{ }}a{\text{ }} - b{\text{ }}c \\\ \end{gathered} \right|^T} = \left| \begin{gathered} a + 1{\text{ }}a - 1{\text{ }}a \\\ b + 1{\text{ }}b - 1{\text{ }} - b \\\ c - 1{\text{ }}c + 1{\text{ }}c \\\ \end{gathered} \right|$$ Therefore, $$\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| + {\left( { - 1} \right)^n}\left| \begin{gathered} a + 1{\text{ }}a - 1{\text{ }}a \\\ b + 1{\text{ }}b - 1{\text{ }} - b \\\ c - 1{\text{ }}c + 1{\text{ }}c \\\ \end{gathered} \right| = 0$$ Now, using the rows and column transformation method in the determinant A As we use ${C_2} \leftrightarrow {C_3}$in the determinant A the sign of determinant A will change $${\left( { - 1} \right)^n}\left( - \right)\left| \begin{gathered} a + 1{\text{ }}a{\text{ }}a - 1 \\\ b + 1{\text{ }} - b{\text{ }}b - 1 \\\ c - 1{\text{ }}c{\text{ }}c + 1 \\\ \end{gathered} \right| = 0$$ As we use ${C_1} \leftrightarrow {C_2}$in the determinant A the sign of determinant A will change $${\left( { - 1} \right)^n}\left( - \right)\left( - \right)\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| = 0$$ $ \Rightarrow $$${\left( { - 1} \right)^n}\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| = 0$$ So, $$\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| + {\left( { - 1} \right)^n}\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right|$$ Since, $$D = \left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right|$$ Therefore, $${\left( { - 1} \right)^n}\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| = {\left( { - 1} \right)^n}D$$ So, $$\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| + {\left( { - 1} \right)^n}\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| = D + {\left( { - 1} \right)^n}D$$ $ \Rightarrow $$$D + {\left( { - 1} \right)^n}D = 0$$ $$ \Rightarrow D\left( {1 + {{\left( { - 1} \right)}^n}} \right) = 0$$ Now, applying the row and column transformation in$D = \left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| = 0$ ${C_2} \to {C_2} - {C_3}$ $D = \left| \begin{gathered} a{\text{ }}a + 1 - \left( {a - 1} \right){\text{ }}a - 1 \\\ \- b{\text{ }}b + 1 - \left( {b - 1} \right){\text{ }}b - 1 \\\ c{\text{ }}c - 1 - \left( {c + 1} \right){\text{ }}c + 1 \\\ \end{gathered} \right| = 0$ $ \Rightarrow $ $D = \left| \begin{gathered} a{\text{ }}2{\text{ }}a - 1 \\\ \- b{\text{ }}2{\text{ }}b - 1 \\\ c{\text{ }} - 2{\text{ }}c + 1 \\\ \end{gathered} \right| = 0$ ${R_1} \to {R_1} + {R_3}$, ${R_2} \to {R_2} + {R_3}$ $D = \left| \begin{gathered} a + \left( { - b} \right){\text{ }}2 + \left( { - 2} \right){\text{ }}a - 1 + \left( {c + 1} \right) \\\ \- b + c{\text{ }}2 + \left( { - 2} \right){\text{ }}b - 1 + \left( {c + 1} \right) \\\ c{\text{ }} - 2{\text{ }}c + 1 \\\ \end{gathered} \right| = 0$ $ \Rightarrow $ $D = \left| \begin{gathered} a + \left( { - b} \right){\text{ }}0{\text{ }}a + c \\\ \- b + c{\text{ }}0{\text{ }}b + c \\\ c{\text{ }} - 2{\text{ }}c + 1 \\\ \end{gathered} \right| = 0$ Let’s use the formula of determinant to find the value of D i.e. $\left| D \right| = \left| \begin{gathered} a{\text{ }}b{\text{ }}c \\\ e{\text{ }}f{\text{ }}g \\\ h{\text{ }}i{\text{ }}j \\\ \end{gathered} \right| = a\left( {fj - gi} \right) - b\left( {ej - gh} \right) + c\left( {ei - fh} \right)$and expanding along 2nd column Therefore, $\left| D \right| = \left| \begin{gathered} a + c{\text{ }}0{\text{ }}a - 1 \\\ \- b + c{\text{ }}0{\text{ }}b - 1 \\\ {\text{ }}c{\text{ }} - 2{\text{ }}c + 1 \\\ \end{gathered} \right| = a + c\left( {0\left( {c + 1} \right) - \left( {b + c} \right)\left( { - 2} \right)} \right) - 0\left( {\left( { - b + c} \right)\left( {c + 1} \right) - \left( {b + c} \right)\left( c \right)} \right) + \left( {a + c} \right)\left( {\left( { - b + c} \right)\left( { - 2} \right) - 0\left( c \right)} \right)$ $ \Rightarrow $ $\left| D \right| = 2\left\\{ {\left( {a + c} \right)\left( {b + c} \right) - \left( {a + c} \right)\left( {c - b} \right)} \right\\}$ $ \Rightarrow $ $\left| D \right| = 2\left( {a + c} \right)2b$ $ \Rightarrow $ $\left| D \right| = 4b\left( {a + c} \right)$ $ \Rightarrow $ $\left| D \right| = 4b\left( {a + c} \right) \ne 0$ as it is given that b (a + c)$ \ne $0 Therefore, $$D\left( {1 + {{\left( { - 1} \right)}^n}} \right) = 0$$ $ \Rightarrow $$$1 + {\left( { - 1} \right)^n} = 0$$ So, for the given determinant i.e. $$\left| \begin{gathered} a{\text{ }}a + 1{\text{ }}a - 1 \\\ \- b{\text{ }}b + 1{\text{ }}b - 1 \\\ c{\text{ }}c - 1{\text{ }}c + 1 \\\ \end{gathered} \right| + \left| \begin{gathered} a + 1{\text{ }}b + 1{\text{ }}c - 1 \\\ a - 1{\text{ }}b - 1{\text{ }}c + 1 \\\ {\left( { - 1} \right)^{n + 2}}a{\text{ }}{\left( { - 1} \right)^{n + 1}}b{\text{ }}{\left( { - 1} \right)^n}c \\\ \end{gathered} \right|$$ to be equal to 0 value of n should be any odd integer Therefore, n = any odd integer Hence, option C is the correct option. **Note** : In the above solution we came across the term “determinant” which can be explained as the value which is obtained from the elements that exist in a square matrix. Determinant of any square matrix is represented as $\left| D \right|$there are some properties shown by the matrix such as the determinant of any square matrix remains the same when rows of the determinant are interchanged with rows to column and column to rows, when each element of row or column is zero then the determinant is zero, etc.