Matrices and Determinants

Given, $D = \left| {\matrix{ 1 & 1 & 1 \cr 1 & {1 + x} & 1 \cr 1 & 1 & {1 + y} \cr } } \right|$

Apply $\,\,\,{R^2} \to {R_2} - {R_1}$ $\,\,\,\,$

and $\,\,\,\,$ $R \to {R_3} - {R_1}$

$\therefore$ $\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$

Hence, $D$ is divisible by both $x$ and $y$

${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$

${A^2} - {B^2} = {A^2} + AB - BA - {B^2}$

$ \Rightarrow AB = BA$

$A = \left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right]\,\,\,\,B = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]$

$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$

$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$

Hence, $AB=BA$ only when $a=b$

$\therefore$ There can be infinitely many $B's$

for which $AB=BA$

$\begin{aligned} & A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array}\right] \\ & \text { Let, } \quad \mathrm{U}_{1}=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \\ & \mathrm{AU}_{1}=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow \quad\left[\begin{array}{c} x \\ 2 x+y \\ 3 x+2 y+z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow \quad x=1,2 x+y=0,3 x+2 y+z=0 \\ & y=-2,3-4+z=0 \\ & z=1 \text {. } \\ & \mathrm{U}_{1}=\left[\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right] \end{aligned}$

$\begin{aligned} \text { Similarly } U_{2} & =\left[\begin{array}{c} 2 \\ -1 \\ -4 \end{array}\right] U_{3}=\left[\begin{array}{c} 2 \\ -1 \\ -3 \end{array}\right] \\ \therefore \quad U & =\left[\begin{array}{ccc} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{array}\right] \\ |U| & =1(3-4)-2 \times 7+2 \times 5 \\ & =-15+18=3 \end{aligned}$

$ \mathrm{U}^{-1} \text { can be calculated by following shortcut } $

$ \begin{aligned} &\mathrm{U}^{-1}=\frac{1}{3}\left[\begin{array}{ccc} -1 & -2 & 0 \\ -7 & -5 & -3 \\ 9 & 6 & 3 \end{array}\right]\\ &\text { Sum of elements of } \mathrm{U}^{-1}=0 \end{aligned} $

$ \begin{aligned} \text { Solution } & =\left[\begin{array}{lll} 3 & 2 & 0 \end{array}\right] \cup\left[\begin{array}{l} 3 \\ 2 \\ 0 \end{array}\right] \\ & =\left[\begin{array}{lll} 3 & 2 & 0 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ 0 \end{array}\right] \\ & =\left[\begin{array}{lll} -1 & 4 & 4 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ 0 \end{array}\right]=-3+8=5 \end{aligned} $

$ax + y + z = \alpha - 1$

$x + \alpha \,y + z = \alpha - 1;$

$x + y + z\alpha = \alpha - 1$

$\Delta = \left| {\matrix{ \alpha & 1 & 1 \cr 1 & \alpha & 1 \cr 1 & 1 & \alpha \cr } } \right|$

$ = \alpha \left( {{\alpha ^2} - 1} \right) - 1\left( {\alpha - 1} \right) + 1\left( {1 - \alpha } \right)$

$ = \alpha \left( {\alpha - 1} \right)\left( {\alpha + 1} \right) - 1\left( {\alpha - 1} \right) - 1\left( {\alpha - 1} \right)$

For infinite solutions, $\Delta = 0$

$ \Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 1 - 1} \right] = 0$

$ \Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 2} \right] = 0$

$ \Rightarrow \alpha = - 2,1;$

But $\alpha \ne 1.\,\,\,$ $\therefore$ $\,\,\alpha = - 2$

As $\,\,\,\,{a_1},{a_2},{a_3},.........$ are in $G.P.$

$\therefore$ Using ${a_n} = a{r^{n - 1}},\,\,\,$ we get the given determinant,

as $\,\,\,\,\,\,\,\left| {\matrix{ {\log a{r^{n - 1}}} & {\log a{r^n}} & {\log a{r^{n + 1}}} \cr {\log a{r^{n + 2}}} & {\log a{r^{n + 3}}} & {\log a{r^{n + 4}}} \cr {\log a{r^{n + 5}}} & {\log a{r^{n + 6}}} & {\log a{r^{n + 7}}} \cr } } \right|$

Operating ${C_3} - {C_2}$ and ${C_2} - {C_1}$ and using

$\log m - \log n = \log {m \over n}\,\,\,\,$ we get

$ = \left| {\matrix{ {\log a{r^{n - 1}}} & {\log r} & {\log r} \cr {\log a{r^{n + 2}}} & {\log r} & {\log r} \cr {\log a{r^{n + 5}}} & {\log r} & {\log r} \cr } } \right| $

$=0$ (two columns being identical)

Given ${A^2} - A + I = 0$

${A^{ - 1}}{A^2} - {A^{ - 1}}A + {A^{ - 1}}.I = {A^{ - 1}}.0$

(Multiplying $\,\,\,{A^{ - 1}}$ on both sides)

$ \Rightarrow A - 1 + {A^{ - 1}} = 0$

or ${A^{ - 1}} = 1 - A$

Applying, ${C_1} \to {C_1} + {C_2} + {C_3}\,\,\,$ we get

$f\left( x \right) = \left| {\matrix{ {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \cr {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$

$ = \left| {\matrix{ 1 & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr 1 & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \cr 1 & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$

$\left[ \, \right.$ As given that ${a^2} + {b^2} + {c^2} = - 2$ $\left. {} \right]$

$\therefore$ ${a^2} + {b^2} + {c^2} + 2 = 0$

Applying ${R_1} \to {R_1} - {R_2},\,\,\,{R_2} \to {R_2} - {R_3}$

$\therefore$ $f\left( x \right) = \left| {\matrix{ 0 & {x - 1} & 0 \cr 0 & {1 - x} & {x - 1} \cr 1 & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$

$f\left( x \right) = {\left( {x - 1} \right)^2}$

Hence degree $=2.$

Given that $10B$ $\,\,\, = \left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$

$ \Rightarrow B = {1 \over {10}}\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$

Also since, $B = {A^{ - 1}} \Rightarrow AB = I$

$ \Rightarrow {1 \over {10}}\left[ {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$

$ \Rightarrow {1 \over {10}}\left[ {\matrix{ {10} & 0 & {5 - 2} \cr 0 & {10} & { - 5 + \alpha } \cr 0 & 0 & {5 + \alpha } \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$

$ \Rightarrow {{5 - \alpha } \over {10}} = 0$

$ \Rightarrow \alpha = 5$

$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$

$ = \left| {\matrix{ {\log {a_1}r{}^{n - 1}} & {\log {a_1}r{}^n} & {\log {a_1}r{}^{n + 1}} \cr {\log {a_1}r{}^{n + 2}} & {\log {a_1}r{}^{n + 3}} & {\log {a_1}r{}^{n + 4}} \cr {\log {a_1}r{}^{n + 5}} & {\log {a_1}r{}^{n + 6}} & {\log {a_1}r{}^{n + 7}} \cr } } \right|$

$ = \left| {\matrix{ {\log {a_1} + \left( {n - 1} \right)\log r} & {\log {a_1} + n\log r} & {\log {a_1} + \left( {n + 1} \right)\log r} \cr {\log {a_1} + \left( {n + 2} \right)\log r} & {\log {a_1} + \left( {n + 3} \right)\log r} & {\log {a_1} + \left( {n + 4} \right)\log r} \cr {\log {a_1} + \left( {n + 5} \right)\log r} & {\log {a_1} + \left( {n + 6} \right)\log r} & {\log {a_1} + \left( {n + 7} \right)\log r} \cr } } \right|$

$ = 0\left[ \, \right.$ Apply $\,\,\,\,{c_2} \to {c_2} - {1 \over 2}{c_1} - {1 \over 2}{c_3}\,\left. \, \right]$

$A = \left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]$

clearly $\,\,\,A \ne 0.\,$ Also $\,\,\left| A \right| = - 1 \ne 0$

$\therefore$ ${A^{ - 1}}\,\,$ exists, further

$\left( { - 1} \right)I = \left[ {\matrix{ { - 1} & 0 & 0 \cr 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr } } \right] \ne A$

Also ${A^2} = \left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]\left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] = I$

$\Delta = \left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr } } \right|$

$ = 1\left( {{\omega ^{3n}} - 1} \right) - {\omega ^n}\left( {{\omega ^{2n}} - {\omega ^{2n}}} \right) + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^{2n}}\left( {{\omega ^n} - {\omega ^{4n}}} \right)$

$ = {\omega ^{3n}} - 1 - 0 + {\omega ^{3n}} - {\omega ^{6n}}$

$ = 1 - 1 + 1 - 1 = 0$ $\left[ {} \right.$ as $\,\,\,\,\,$ ${\omega ^{3n}} = 1$ $\left. {} \right]$

For homogeneous system of equations to have non zero solution, $\Delta = 0$

$\left| {\matrix{ 1 & {2a} & a \cr 1 & {3b} & b \cr 1 & {4c} & c \cr } } \right| = 0\,{C_2} \to {C_2} - 2{C_3}$

$\left| {\matrix{ 1 & 0 & a \cr 1 & b & b \cr 1 & {2c} & c \cr } } \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}$

$\left| {\matrix{ 1 & 0 & a \cr 0 & b & {b - a} \cr 0 & {2c - b} & {c - b} \cr } } \right| = 0$

$b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0$

On simplification, ${2 \over b} = {1 \over a} + {1 \over c}$

$\therefore$ $a,b,c$ are in Harmonic Progression.

${A^2} = \left[ {\matrix{ \alpha & \beta \cr \beta & \alpha \cr } } \right] = \left[ {\matrix{ a & b \cr b & a \cr } } \right]\left[ {\matrix{ a & b \cr b & a \cr } } \right]$

$ = \left[ {\matrix{ {{a^2} + {b^2}} & {2ab} \cr {2ab} & {{a^2} + {b^2}} \cr } } \right]$

$\alpha = {a^2} + {b^2};\,\,\beta = 2ab$

We have $\left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr {ax + b} & {bx + c} & 0 \cr } } \right|$

By $\,\,\,{R_3} \to {R_3} - \left( {x{R_1} + {R_2}} \right);$

$ = \left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr 0 & 0 & { - \left( {a{x^2} + 2bx + C} \right)} \cr } } \right|$

$ = \left( {a{x^2} + 2bx + c} \right)\left( {{b^2} - ac} \right)$

$ = \left( + \right)\left( - \right) = - ve.$