If a, b, c are non-zero real numbers and if system of equations (a - 1 )x = y + z, (b

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3 and 4 .Determinants and Matrices

hard

If $a, b, c$ are non-zero real numbers and if the system of equations $(a - 1 )x = y + z,$ $(b - 1 )y = z + x ,$ $(c - 1 )z= x + y,$ has a non-trivial solution, then $ab + bc + ca$ equals

$a + b + c$

$abc$

$1$

$-1$

(JEE MAIN-2014)

Solution

Given system of equationa be written as

$\left( {a – 1} \right)x – y – z = 0$

$ – x + \left( {b – 1} \right)y – z = 0$

$ – x – y + \left( {c – 1} \right)z = 0$

For non-trivial solution, we have

$\begin{array}{*{20}{c}}
{a – 1}&{ – 1}&{ – 1}\\
{ – 1}&{b – 1}&{ – 1}\\
{ – 1}&{ – 1}&{c – 1}
\end{array} = 0$

${R_2} \to {R_2} – {R_3}$

$\begin{array}{*{20}{c}}
{a – 1}&{ – 1}&{ – 1}\\
0&b&{ – c}\\
{ – 1}&{ – 1}&{c – 1}
\end{array} = 0$

${C_2} \to {C_2} – {C_3}$

$\begin{array}{*{20}{c}}
{a – 1}&0&{ – 1}\\
0&{b + c}&{ – c}\\
{ – 1}&{ – c}&{c – 1}
\end{array} = 0$

Apply, ${R_3} \to {R_3} – {R_1}$

$\begin{array}{*{20}{c}}
{a – 1}&0&{ – 1}\\
0&{b + c}&{ – c}\\
{ – 1}&{ – c}&c
\end{array} = 0$

$ \Rightarrow \left( {a – 1} \right)\left[ {bc + {c^2} – {c^2}} \right] – 1\left[ {a\left( {b + c} \right)} \right] = 0$

$ \Rightarrow \left( {a – 1} \right)\left[ {bc} \right] – ab – ac = 0$

$ \Rightarrow abc – bc – ab – ac = 0$

$ \Rightarrow ab + bc + ca = abc$

Standard 12

Mathematics

Let $\theta \in\left(0, \frac{\pi}{2}\right)$. If the system of linear equations

$\left(1+\cos ^{2} \theta\right) x+\sin ^{2} \theta y+4 \sin 3 \theta z=0$

$\cos ^{2} \theta x+\left(1+\sin ^{2} \theta\right) y+4 \sin 3 \theta z=0$

$\cos ^{2} \theta x+\sin ^{2} \theta y+(1+4 \sin 3 \theta) z=0$

has a non-trivial solution, then the value of $\theta$ is :

hard

(JEE MAIN-2021)

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right| = $

medium

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)

Let $P $ and $Q $ be $3×3$ matrices $P \ne Q$. If ${P^3} = {Q^3},{P^2}Q = {Q^2}P$ then determinant of $\det \left( {{P^2} + {Q^2}} \right)$ is equal to :

hard

(AIEEE-2012)

Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$

medium

If $a, b, c$ are non-zero real numbers and if the system of equations $(a - 1 )x = y + z,$ $(b - 1 )y = z + x ,$ $(c - 1 )z= x + y,$ has a non-trivial solution, then $ab + bc + ca$ equals

Solution

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