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Question

Using the properties of determinants, show that:
$$\begin{vmatrix} { a }^{ 2 }+1 & ab & ac \\ ab & { b }^{ 2 }+1 & bc \\ ca & cb & { c }^{ 2 }+1 \end{vmatrix}=1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$$


Solution

Consider, $$\begin{vmatrix} { a }^{ 2 }+1 & ab & ac \\ ab & { b }^{ 2 }+1 & bc \\ ca & cb & { c }^{ 2 }+1 \end{vmatrix}$$

Taking $$a, b,c $$ common from $$R_1,R_2, R_3$$ respectively
$$=abc\begin{vmatrix} a+\dfrac { 1 }{ a }  & b & c \\ a & b+\dfrac { 1 }{ b }  & c \\ a & b & c+\dfrac { 1 }{ c }  \end{vmatrix}$$

Multiplying $$C_1, C_2, C_3$$ by $$a, b ,c$$ respectively
$$=\displaystyle \frac { abc }{ abc } \begin{vmatrix} { a }^{ 2 }+1 & { b }^{ 2 } & { c }^{ 2 } \\ { a }^{ 2 } & { b }^{ 2 }+1 & { c }^{ 2 } \\ { a }^{ 2 } & { b }^{ 2 } & { c }^{ 2 }+1 \end{vmatrix}$$

$$C_1 \rightarrow C_1+C_2+C_3$$
$$=\begin{vmatrix} 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } & { b }^{ 2 } & { c }^{ 2 } \\ 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } & { b }^{ 2 }+1 & { c }^{ 2 } \\ 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } & { b }^{ 2 } & { c }^{ 2 }+1 \end{vmatrix}$$

Taking $$(1+a^2+b^2+c^2)$$ common from $$C_1$$
$$=(1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\begin{vmatrix} 1 & { b }^{ 2 } & { c }^{ 2 } \\ 1 & { b }^{ 2 }+1 & { c }^{ 2 } \\ 1 & { b }^{ 2 } & { c }^{ 2 }+1 \end{vmatrix}$$

$$R_1 \rightarrow R_1-R_2, R_2\rightarrow R_2-R_3$$
$$=(1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\begin{vmatrix} 0 & -1 & 0 \\ 0 & { b }^{ 2 } & -1 \\ 1 & { b }^{ 2 } & { c }^{ 2 }+1 \end{vmatrix}$$

Expanding along first column,
$$=(1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) \begin{vmatrix} -1 & 0 \\ { b }^{ 2 } & -1 \end{vmatrix}$$
$$=1+a^2+b^2+c^2$$

Mathematics
NCERT
Standard XII

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