If $left| begin{array}{*{20}{c}} { - 2a}&{a + b}&{a + c} {b + a}&{

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3 and 4 .Determinants and Matrices

hard

If $\left| \begin{array}{*{20}{c}}
{ - 2a}&{a + b}&{a + c}\\
{b + a}&{ - 2b}&{b + c}\\
{c + a}&{b + c}&{ - 2c}
\end{array}\right|$ $ = \alpha \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \ne 0$ then $\alpha $ is equal to

$a + b + c$

$abc$

$4$

$1$

(AIEEE-2012)

Solution

Let $\Delta = \left| {\begin{array}{*{20}{c}}
{ – 2a}&{a + b}&{a + c}\\
{b + a}&{ – 2b}&{b + c}\\
{c + a}&{b + c}&{ – 2c}
\end{array}} \right|$

Applying ${C_1} + {C_3}$ and ${C_2} + {C_3}$

$\Delta = \left| {\begin{array}{*{20}{c}}
{ – a + c}&{2a + b + c}&{a + c}\\
{2b + a + c}&{ – b + c}&{b + c}\\
{a – c}&{b – c}&{ – 2c}
\end{array}} \right|$

Now, applying ${R_1} + {R_3}$ and ${R_2} + {R_3}$

$\Delta = \left| {\begin{array}{*{20}{c}}
0&{2\left( {a + b} \right)}&{a – c}\\
{2\left( {a + b} \right)}&0&{b – c}\\
{a – c}&{b – c}&{ – 2c}
\end{array}} \right|$

On expanding, we get

$\Delta = – 2\left( {a + b} \right)\left\{ { – 2c\left[ {2\left( {a + b} \right)} \right] – \left( {a – c} \right)\left( {b – c} \right)} \right\}$ $ + \left( {a – c} \right)\left[ {2\left( {a + b} \right)\left( {b – c} \right)} \right]$

$\Delta = 8c\left( {a + b} \right)\left( {a + b} \right) + 4\left( {a + b} \right)\left( {a – c} \right)\left( {b – c} \right)$

$ = 4\left( {a + b} \right)\left[ {2ac + 2bc + ab – bc – ac + {c^2}} \right]$

$ = 4\left( {a + b} \right)\left[ {ac + bc + ab + {c^2}} \right]$

$ = 4\left( {a + b} \right)\left[ {c\left( {a + c} \right) + b\left( {a + c} \right)} \right]$

$ = 4\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$

$ = \alpha \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$

Hence, $\alpha = 4$

Standard 12

Mathematics

If the system of linear equations $x – 2y + kz = 1$ ; $2x + y + z = 2$ ; $3x – y – kz = 3$ Has a solution $(x, y, z) \ne 0$, then $(x, y)$ lies on the straight line whose equation is

hard

(JEE MAIN-2019)

In a square matrix $A$ of order $3, a_{i i}'s$ are the sum of the roots of the equation $x^2 – (a + b)x + ab= 0$; $a_{i , i + 1}'s$ are the product of the roots, $a_{i , i – 1}'s$ are all unity and the rest of the elements are all zero. The value of the det. $(A)$ is equal to

normal

Statement $-1$ : The system of linear equations

$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$

$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$

$x – \left( {\sin \,\alpha } \right)y – \left( {\cos \alpha } \right)z = 0$

has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$

Statement $-2$ : The equation in $\alpha $

$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ – \sin {\mkern 1mu} \alpha }&{ – \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$

has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$

hard

(JEE MAIN-2013)

If $p + q + r = 0 = a + b + c$, then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}{pa}&{qb}&{rc}\\{qc}&{ra}&{pb}\\{rb}&{pc}&{qa}\end{array}\,} \right|$ is

medium

If $\omega $ is cube root of unity, then root of the equation $\left| {\begin{array}{*{20}{c}}
{x + 2}&\omega &{{\omega ^2}} \\
\omega &{x + 1 + {\omega ^2}}&1 \\
{{\omega ^2}}&1&{x + 1 + \omega }
\end{array}} \right| = 0$ is

normal

If $\left| \begin{array}{*{20}{c}} { - 2a}&{a + b}&{a + c}\\ {b + a}&{ - 2b}&{b + c}\\ {c + a}&{b + c}&{ - 2c} \end{array}\right|$ $ = \alpha \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \ne 0$ then $\alpha $ is equal to

Solution

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