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,$$=(1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) \begin{vmatrix} -1 & 0 \\ { b }^{ 2 } & -1 \end{vmatrix}$$$$=1+a^2+b^2+c^2$$MathematicsNCERTStandard XII

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