Matrices and Determinants

$\Delta = 0 \Rightarrow \left| {\matrix{ 4 & k & 2 \cr k & 4 & 1 \cr 2 & 2 & 1 \cr } } \right| = 0$

$ \Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0$

$ \Rightarrow 8 - {k^2} + 2k + 4k - 16 = 0$

$ \Rightarrow {k^2} - 6k + 8 = 0$

$ \Rightarrow \left( {k - 4} \right)\left( {k - 2} \right) = 0,k = 4,2$

$\therefore$ $A' = A,B' = B$

Now $\,\,\,\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A'$

$ = \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right)$

Similarly $\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A$

So, $A\left( {BA} \right)\,\,\,\,$ and $A\left( {BA} \right)\,\,\,\,$ are symmetric matrices.

Again $\left( {AB} \right)' = B'A' = BA$

Now if $BA=AB$, then $AB$ is symmetric matrix.

$\left[ {\matrix{ 1 & {...} & {...} \cr {...} & 1 & {...} \cr {...} & {...} & 1 \cr } } \right]\,\,$ are $6$ non-singular matrices because $6$

blanks will be filled by $5$ zeros and $1$ one.

Similarly, $\left[ {\matrix{ {...} & {...} & 1 \cr {...} & 1 & {...} \cr 1 & {...} & {...} \cr } } \right]\,\,$ are $6$ non-singular matrices.

So, required cases are more than $7,$ non-singular $3 \times 3$ matrices.

Let $A = \left( {\matrix{ a & b \cr c & d \cr } } \right)$ where $a,b,c,d$ $ \ne 0$

${A^2} = \left( {\matrix{ a & b \cr c & d \cr } } \right)\left( {\matrix{ a & b \cr c & d \cr } } \right)$

$ \Rightarrow {A^2} = \left( {\matrix{ {{a^2} + bc} & {ab + bd} \cr {ac + cd} & {bc + {d^2}} \cr } } \right)$

$ \Rightarrow {a^2} + bc = 1,\,bc + {d^2} = 1$

$ab + bd = ac + cd = 0$

$c \ne 0\,\,\,\,\,b \ne 0$

$ \Rightarrow a + d = 0 \Rightarrow Tr\left( A \right) = 0$

$\left| A \right| = ad - bc = - {a^2} - bc = - 1$

$D = \left| {\matrix{ 1 & 2 & 1 \cr 2 & 3 & 1 \cr 3 & 5 & 2 \cr } } \right| = 0$

${D_1}\left| {\matrix{ 3 & 2 & 1 \cr 3 & 3 & 1 \cr 1 & 5 & 2 \cr } } \right| \ne 0$

$ \Rightarrow $ Given system, does not have any solution.

$ \Rightarrow $ No solution

We know that $\left| {adj\left( {adj\,\,A} \right)} \right| = {\left| {Adj\,\,A} \right|^{2 - 1}}$b

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left| A \right|^{2 - 1}} = \left| A \right|$

$\therefore$ $\,\,\,\,\,\,$ Both the statements are true and statement $-2$ is a correct explanation for statement - $1.$

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

$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {a - 1} & {{{\left( { - 1} \right)}^{n + 2}}a} \cr {b + 1} & {b - 1} & {{{\left( { - 1} \right)}^{n + 1}}b} \cr {c - 1} & {c + 1} & {{{\left( { - 1} \right)}^n}c} \cr } } \right| = 0$

(Taking transpose of second determinant)

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_1} \Leftrightarrow {C_3}$

$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| - \left| {\matrix{ {{{\left( { - 1} \right)}^{n + 2}}a} & {a - 1} & {a + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}\left( { - b} \right)} & {b - 1} & {b + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}c} & {c + 1} & {c - 1} \cr } } \right| = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} \Leftrightarrow {C_3}$

$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + {\left( 1 \right)^{n + 2}}\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} - {C_1},{C_3} - {C_1}$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & 1 & { - 1} \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$

${R_1} + {R_3}$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ {a + c} & 0 & 0 \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$

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

$ \Rightarrow 4b\left( {a + c} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right] = 0$

$ \Rightarrow 1 + {\left( { - 1} \right)^{n + 2}} = 0$ $\,\,\,\,\,$ as $\,\,\,\,\,b\left( {a + c} \right) \ne 0$

$ \Rightarrow n$ should be an odd integer.

The given equations are

$\matrix{ { - x + cy + bz = 0} \cr {cx - y + az = 0} \cr {bx + ay - z = 0} \cr } $

As $x,y,z$ are not all zero

$\therefore$ The above system should not have unique (zero) solution

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

$ \Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$

$ \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$

$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$

Let $A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$ $\,\,\,$ then ${A^2} = 1$

$ \Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0$

$ac + cd = 0\,\,\,\,bc + {d^2} = 1$

From these four relations,

${a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}$

and $\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = - d$

We can take $a = 1,b = 0,c = 0,d = - 1$

as one possible set of values, then

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

Clearly $A \ne I\,\,\,$ and $\,\,\,\,A \ne - I\,\,$ and $\,\,\,A = - 1$

$\therefore$ $\,\,\,\,\,$ Statement $1$ is true.

Also if $A \ne I\,\,\,\,\,tr\left( A \right) = 0$

$\therefore$ $\,\,\,\,\,$ Statement $2$ is false.

As all entries of square matrix $A$ are integers, therefore all co-factors should also be integers.

If det $A = \pm 1\,\,$ then ${A^{ - 1}}\,\,$ exists. Also all entries of ${A^{ - 1}}$ are integers.

$\left| {{A^2}} \right| = 25 \Rightarrow {\left| A \right|^2} = 25$

$ \Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}$

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 $AB=BA$

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

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

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

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

$ = \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.$