જો left| begin{array}{*{20}{c}}
{ - 2a}& | vedclass

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Question

Let $\Delta  = \left| {\begin{array}{*{20}{c}}
{ - 2a}&{a + b}&{a + c}\
{b + a}&{ - 2b}&{b + c}\
{c + a}&{b + c}&am

Vedclass · Accepted Answer

$4$

Vedclass · Answer

Let $\Delta  = \left| {\begin{array}{*{20}{c}}
{ - 2a}&{a + b}&{a + c}\
{b + a}&{ - 2b}&{b + c}\
{c + a}&{b + c}&{ - 2c}
\end{array}} ight|$

Applying ${C_1} + {C_3}$ and ${C_2} + {C_3}$

$\Delta  = \left| {\begin{array}{*{20}{c}}
{ - a + c}&{2a + b + c}&{a + c}\
{2b + a + c}&{ - b + c}&{b + c}\
{a - c}&{b - c}&{ - 2c}
\end{array}} ight|$

Now, applying ${R_1} + {R_3}$ and ${R_2} + {R_3}$

$\Delta  = \left| {\begin{array}{*{20}{c}}
0&{2\left( {a + b} ight)}&{a - c}\
{2\left( {a + b} ight)}&0&{b - c}\
{a - c}&{b - c}&{ - 2c}
\end{array}} ight|$

On expanding, we get

$\Delta  =  - 2\left( {a + b} ight)\left\{ { - 2c\left[ {2\left( {a + b} ight)} ight] - \left( {a - c} ight)\left( {b - c} ight)} ight\}$ $ + \left( {a - c} ight)\left[ {2\left( {a + b} ight)\left( {b - c} ight)} ight]$

$\Delta  = 8c\left( {a + b} ight)\left( {a + b} ight) + 4\left( {a + b} ight)\left( {a - c} ight)\left( {b - c} ight)$

$ = 4\left( {a + b} ight)\left[ {2ac + 2bc + ab - bc - ac + {c^2}} ight]$

$ = 4\left( {a + b} ight)\left[ {ac + bc + ab + {c^2}} ight]$

$ = 4\left( {a + b} ight)\left[ {c\left( {a + c} ight) + b\left( {a + c} ight)} ight]$

$ = 4\left( {a + b} ight)\left( {b + c} ight)\left( {c + a} ight)$

$ = \alpha \left( {a + b} ight)\left( {b + c} ight)\left( {c + a} ight)$

Hence, $\alpha  = 4$

જો $\left| \begin{array}{*{20}{c}}
{ - 2a}&{a + b}&{a + c}\\
{b + a}&{ - 2b}&{b + c}\\
{c + a}&{b + c}&{ - 2c}
\end{array}\right|$ $ = \alpha \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \ne 0$ તો $\alpha $ મેળવો.

Similar Questions

જો $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$ હોય, તો સાબિત કરો કે $|3 A|=27|A|$.

સમીકરણની સંહતિ ${x_1} - {x_2} + {x_3} = 2,$ $\,3{x_1} - {x_2} + 2{x_3} = - 6$ અને $3{x_1} + {x_2} + {x_3} = - 18$ નો ઉકેલ . . . .

રેખીય સમીકરણની સિસ્ટમ $x + y + z = 2, 2x + 3y + 2z = 5$, $2x + 3y + (a^2 -1)\,z = a + 1$ તો

જો $\left| \begin{array}{*{20}{c}} { - 2a}&{a + b}&{a + c}\\ {b + a}&{ - 2b}&{b + c}\\ {c + a}&{b + c}&{ - 2c} \end{array}\right|$ $ = \alpha \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \ne 0$ તો $\alpha $ મેળવો.

Similar Questions

જો $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$ હોય, તો સાબિત કરો કે $|3 A|=27|A|$.

સમીકરણની સંહતિ ${x_1} - {x_2} + {x_3} = 2,$ $\,3{x_1} - {x_2} + 2{x_3} = - 6$ અને $3{x_1} + {x_2} + {x_3} = - 18$ નો ઉકેલ . . . .

રેખીય સમીકરણની સિસ્ટમ $x + y + z = 2, 2x + 3y + 2z = 5$, $2x + 3y + (a^2 -1)\,z = a + 1$ તો

જો $\left| \begin{array}{*{20}{c}}
{ - 2a}&{a + b}&{a + c}\\
{b + a}&{ - 2b}&{b + c}\\
{c + a}&{b + c}&{ - 2c}
\end{array}\right|$ $ = \alpha \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \ne 0$ તો $\alpha $ મેળવો.