Matrices and Determinants

$\left| {\matrix{ {\sin x} & {\cos x} & {\cos x} \cr {\cos x} & {\sin x} & {\cos x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0, - {\pi \over 4} \le x \le {\pi \over 4}$

Apply : ${R_1} \to {R_1} - {R_2}$ & ${R_2} \to {R_2} - {R_3}$

$ \Rightarrow $ $\left| {\matrix{ {\sin x - \cos x} & {\cos x - \sin x} & 0 \cr 0 & {\sin x - \cos x} & {\cos x - \sin x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$

$ \Rightarrow $ ${(\sin x - \cos x)^2}\left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$

$ \Rightarrow $ ${(\sin x - \cos x)^2}(\sin x + 2\cos x) = 0$

$\therefore$ $x = {\pi \over 4}$

$P = \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right]$

${P^2} = \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]$

${P^3} = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr {{3 \over 2}} & 1 \cr } } \right]$

${P^4} = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right]$

$ \vdots $

$\therefore$ ${P^{50}} = \left[ {\matrix{ 1 & 0 \cr {25} & 1 \cr } } \right]$

$D = \left| {\matrix{ 2 & 3 & 6 \cr 1 & 2 & a \cr 3 & 5 & 9 \cr } } \right| = 3 - a$

$D = \left| {\matrix{ 2 & 3 & 8 \cr 1 & 2 & 5 \cr 3 & 5 & b \cr } } \right| = b - 13$

If a = 3, b $\ne$ 13, no solution.

$x + y + z = 6$ ..... (i)

$3x + 5y + 5z = 26$ .... (ii)

$x + 2y + \lambda z = \mu $ ..... (iii)

$5 \times (i) - (ii) \Rightarrow 2x = 4 \Rightarrow x = 2$

$\therefore$ from (i) and (iii)

$y + z = 4$ ..... (iv)

$2y + \lambda z = \mu - 2$ .....(v)

$(v) - 2 \times (iv)$

$ \Rightarrow (\lambda - 2)z = \mu - 10$

$ \Rightarrow z = {{\mu - 10} \over {\lambda - 2}}$ & $y = 4 - {{\mu - 10} \over {\lambda - 2}}$

$\therefore$ For no solution $\lambda$ = 2 and $\mu$ $\ne$ 10.

$A = \left[ {\matrix{ {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr } } \right]$

Let $x = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$

$AX = \left[ {\matrix{ {{a_{11}} + {a_{12}} + {a_{13}}} \cr {{a_{21}} + {a_{22}} + {a_{23}}} \cr {{a_{31}} + {a_{32}} + {a_{33}}} \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$

$\Rightarrow$ AX = X

Replace X by AX

A²X = AX = X

Replace X by AX

A³X = AX = X

Let ${A^3} = \left[ {\matrix{ {{x_1}} & {{x_2}} & {{x_3}} \cr {{y_1}} & {{y_2}} & {{y_3}} \cr {{z_1}} & {{z_2}} & {{z_3}} \cr } } \right]$

${A^3}\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = \left[ {\matrix{ {{x_1}} & {{x_2}} & {{x_3}} \cr {{y_1}} & {{y_2}} & {{y_3}} \cr {{z_1}} & {{z_2}} & {{z_3}} \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$

Sum of all the element = 3

$\left| {\matrix{ 3 & { - 1} & 4 \cr 1 & 2 & { - 3} \cr 6 & 5 & k \cr } } \right| = 0$

$\Rightarrow$ 3(2k + 15) + K + 18 $-$ 28 = 0

$\Rightarrow$ 7k + 35 = 0

$\Rightarrow$ k = $-$ 5

$A = \left[ {\matrix{ 2 & 3 \cr a & 0 \cr } } \right]$, ${A^T} = \left[ {\matrix{ 2 & a \cr 3 & 0 \cr } } \right]$

$A = {{A + {A^T}} \over 2} + {{A - {A^T}} \over 2}$

Let $P = {{A + {A^T}} \over 2}$ and $Q = {{A - {A^T}} \over 2}$

$Q = \left( {\matrix{ 0 & {{{3 - a} \over 2}} \cr {{{a - 3} \over 2}} & 0 \cr } } \right)$

Det (Q) = 9

$0 - \left( {{{3 - a} \over 2}} \right)\left( {{{a - 3} \over 2}} \right) = 9$

$ \Rightarrow {\left( {{{a - 3} \over 2}} \right)^2} = 9 \Rightarrow {(a - 3)^2} = 36$

$a - 3 = \pm \,6 \Rightarrow a = 9, - 3$

$P = \left[ {\matrix{ 2 & {{{a + 3} \over 2}} \cr {{{a + 3} \over 2}} & 0 \cr } } \right]$

$P = \left[ {\matrix{ 2 & 6 \cr 6 & 0 \cr } } \right]$ or $\left[ {\matrix{ 2 & 0 \cr 0 & 0 \cr } } \right]$

| P | = - 36 or 0

$\therefore$ | $-$36 + 0 | = 36

Given, system of linear equations

4x + $\lambda$y + 2z = 0

2x $-$ y + z = 0

$\mu$x + 2y + 3z = 0

For non-trivial solution, $\Delta$ = 0

$\left| {\matrix{ 4 & \lambda & 2 \cr 2 & { - 1} & 1 \cr \mu & 2 & 3 \cr } } \right| = 0$

$ \Rightarrow 4( - 3 - 2) - \lambda (6 - \mu ) + 2(4 + \mu ) = 0$

$ \Rightarrow - \lambda (6 - \mu ) - 2(6 - \mu ) = 0$

$ \Rightarrow (6 - \mu )(\lambda + 2) = 0$

$ \Rightarrow \lambda = - 2$ and $\mu \in R$ or $\mu$ = 6 and $\lambda \in R$.

By using C₁ $ \to $ C₁ $-$ C₂ and C₃ $ \to $ C₃ $-$ C₂ we get

$\left| {\matrix{ 1 & {{{\sin }^2}x} & 0 \cr { - 1} & {1 + {{\cos }^2}x} & { - 1} \cr 0 & {4\sin 2x} & 1 \cr } } \right| = 0$

Expanding by R₁ we get

$1(1 + {\cos ^2}x + 4\sin 2x) - {\sin ^2}x( - 1) = 0$

$ \Rightarrow 2 + 4\sin 2x = 0$

$ \Rightarrow \sin 2x = {{ - 1} \over 2}$

$ \Rightarrow 2x = n\pi + {( - 1)^n}\left( {{{ - \pi } \over 6}} \right),n \in Z$

$ \therefore $ $2x = {{7\pi } \over 6},{{11\pi } \over 6} $

$\Rightarrow x = {{7\pi } \over {12}},{{11\pi } \over 2}$

x³ + ax² + bx + c = 0

Roots are $\alpha$, $\beta$, $\gamma$.

For non-trivial solutions,

$\left| {\matrix{ \alpha & \beta & \gamma \cr \beta & \gamma & \alpha \cr \gamma & \alpha & \beta \cr } } \right| = 0$

$ \Rightarrow $ ${\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = 0$

$ \Rightarrow $ $(\alpha + \beta + \gamma )\left[ {{{\left( {\alpha + \beta + \alpha } \right)}^2} - 3\left( {\sum {\alpha \beta } } \right)} \right] = 0$

$ \Rightarrow $ $( - a)[{a^2} - 3b] = 0$

$ \Rightarrow $ ${a^2} = 3b$ ($ \because $ a $ \ne $ 0)

$ \Rightarrow {{{a^2}} \over b} = 3$

$A = {1 \over 5}((A + 2B) + 2(2A - B))$

$ = {1 \over 5}\left( {\left[ {\matrix{ 1 & 2 & 0 \cr 6 & { - 3} & 3 \cr { - 5} & 3 & 1 \cr } } \right] + \left[ {\matrix{ 4 & { - 2} & {10} \cr 4 & { - 2} & {12} \cr 0 & 2 & 4 \cr } } \right]} \right)$

$ = {1 \over 5}\left[ {\matrix{ 5 & 0 & {10} \cr {10} & { - 5} & {15} \cr { - 5} & 5 & 5 \cr } } \right] \Rightarrow tr(A) = 1$

Similarly,

$B = {1 \over 5}(2(A + 2B) - (2A - B))$

$ = {1 \over 5}\left( {\left[ {\matrix{ 2 & 4 & 0 \cr {12} & { - 6} & 6 \cr { - 10} & 6 & 2 \cr } } \right] - \left[ {\matrix{ 2 & { - 1} & 5 \cr 2 & { - 1} & 6 \cr 0 & 1 & 2 \cr } } \right]} \right)$

$ = {1 \over 5}\left[ {\matrix{ 0 & 6 & { - 5} \cr {10} & { - 5} & 0 \cr { - 10} & 5 & 0 \cr } } \right] \Rightarrow tr(B) = - 1$

$Tr(A) - Tr(B) = 1 - ( - 1) = 2$

$\left| {\matrix{ 3 & {4\sqrt 2 } & x \cr 4 & {5\sqrt 2 } & y \cr 5 & k & z \cr } } \right| = 0$

${R_1} \to {R_1} + {R_3} - 2{R_2}$

$ \Rightarrow $ $\left| {\matrix{ 0 & {4\sqrt 2 - k - 10\sqrt 2 } & 0 \cr 4 & {5\sqrt 2 } & y \cr 5 & k & z \cr } } \right| = 0$ { $ \because $ 2y = x + z}

$ \Rightarrow (k - 6\sqrt 2 )(4z - 5y) = 0$

$ \Rightarrow $ k = $6\sqrt 2 $ or 4z = 5y (Not possible $ \because $ x, y, z in A.P.)

So, k² = 72

$ \therefore $ Option (A)

$D = \left| {\matrix{ k & 1 & 1 \cr 1 & k & 1 \cr 1 & 1 & k \cr } } \right| = 0$

$ \Rightarrow k({k^2} - 1) - (k - 1) + (1 - k) = 0$

$ \Rightarrow (k - 1)({k^2} + k - 1 - 1) = 0$

$ \Rightarrow (k - 1)({k^2} + k - 2) = 0$

$ \Rightarrow (k - 1)(k - 1)(k + 2) = 0$

$ \Rightarrow k = 1,k = -2$

for k = 1 equation identical so k = $-$2 for no solution.

${A^2} = \left[ {\matrix{ 0 & {\sin \alpha } \cr {\sin \alpha } & 0 \cr } } \right]\left[ {\matrix{ 0 & {\sin \alpha } \cr {\sin \alpha } & 0 \cr } } \right] = \left[ {\matrix{ {{{\sin }^2}\alpha } & 0 \cr 0 & {{{\sin }^2}\alpha } \cr } } \right]$

${A^2} - {1 \over 2}I = \left[ {\matrix{ {{{\sin }^2}\alpha } & 0 \cr 0 & {{{\sin }^2}\alpha } \cr } } \right] - \left[ {\matrix{ {{1 \over 2}} & 0 \cr 0 & {{1 \over 2}} \cr } } \right] = \left[ {\matrix{ {{{\sin }^2}\alpha - {1 \over 2}} & 0 \cr 0 & {{{\sin }^2}\alpha - {1 \over 2}} \cr } } \right]$

Given, $\left| {{A^2} - {1 \over 2}I} \right| = 0$

$ \Rightarrow \left| {\matrix{ {{{\sin }^2}\alpha - {1 \over 2}} & 0 \cr 0 & {{{\sin }^2}\alpha - {1 \over 2}} \cr } } \right| = 0$

$ \Rightarrow {\left( {{{\sin }^2}\alpha - {1 \over 2}} \right)^2} = 0 $

$\Rightarrow {\sin ^2}\alpha = {1 \over 2} \Rightarrow \sin \alpha = {1 \over {\sqrt 2 }}, - {1 \over {\sqrt 2 }}$

$ \therefore $ $\alpha = {\pi \over 4}$

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

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

${A^4} = 4\left[ {\matrix{ { - 1} & 1 \cr 1 & { - 1} \cr } } \right]\left[ {\matrix{ { - 1} & 1 \cr 1 & { - 1} \cr } } \right] = 4\left[ {\matrix{ 2 & { - 2} \cr { - 2} & 2 \cr } } \right] = 8\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]$

${A^8} = 64\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right] = 64\left[ {\matrix{ 2 & { - 2} \cr { - 2} & 2 \cr } } \right] = 128\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]$

$128\left[ {\matrix{ 1 & { - 1} \cr { - 1} & 1 \cr } } \right]\left[ {\matrix{ x \cr y \cr } } \right] = \left[ {\matrix{ 8 \cr {64} \cr } } \right]$

$128\left[ {\matrix{ {x - y} \cr { - x + y} \cr } } \right] = \left[ {\matrix{ 8 \cr {64} \cr } } \right] $

$\Rightarrow 128(x - y) = 8$

$ \Rightarrow x - y = {1 \over {16}}$ .... (1)

and $128( - x + y) = 64 \Rightarrow x - y = {{ - 1} \over 2}$ .... (2)

$ \Rightarrow $ No solution (from eq. (1) & (2))

$D = \left| {\matrix{ 1 & 2 & { - 3} \cr 2 & 6 & { - 11} \cr 1 & { - 2} & 7 \cr } } \right|$

= 20 $-$ 2(25) $-$3($-$10)

= 20 $-$ 50 + 30 = 0

${D_1} = \left| {\matrix{ a & 2 & { - 3} \cr b & 6 & { - 11} \cr c & { - 2} & 7 \cr } } \right|$

= 20a $-$ 2(7b + 11c) $-$3($-$2b $-$ 6c)

= 20a $-$ 14b $-$ 22c + 6b +18c

= 20a $-$ 8b $-$ 4c

= 4(5a $-$ 2b $-$ c)

${D_2} = \left| {\matrix{ 1 & a & { - 3} \cr 2 & b & { - 11} \cr 1 & c & 7 \cr } } \right|$

= 7b + 11c $-$ a(25) $-$3(2c $-$ b)

= 7b + 11c $-$ 25a $-$ 6c + 3b

= $-$25a + 10b + 5c

= $-$5(5a $-$ 2b $-$ c)

${D_3} = \left| {\matrix{ 1 & 2 & a \cr 2 & 6 & b \cr 1 & { - 2} & c \cr } } \right|$

= 6c + 2b $-$ 2(2c $-$ b) $-$ 10a

= $-$10a + 4b + 2c

= $-$2(5a $-$ 2b $-$ c)

for infinite solution

$D = {D_1} = {D_2} = {D_3} = 0$

$ \Rightarrow $ 5a = 2b + c

Let $A = \left[ {\matrix{ a & b \cr b & c \cr } } \right]$

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

$ = {a^2} + 2{b^2} + {c^2} = 1$

$a = 1,b = 0,c = 0$

$a = 0,b = 0,c = 1$

$a = - 1,b = 0,c = 0$

$c = - 1,b = 0,a = 0$

Given, $\Delta $ = $\left| {\matrix{ {(a + 1)(a + 2)} & {a + 2} & 1 \cr {(a + 2)(a + 3)} & {a + 3} & 1 \cr {(a + 3)(a + 4)} & {a + 4} & 1 \cr } } \right|$

R₂ $ \to $ R₂ $-$ R₁ and R₃ $ \to $ R₃ $-$ R₁

$\Delta$ = $\left| {\matrix{ {(a + 1)(a + 2)} & {a + 2} & 1 \cr {(a + 2)(a + 3 - a - 1)} & 1 & 0 \cr {{a^2} + 7a + 12 - {a^2} - 3a - 2} & 2 & 0 \cr } } \right|$

$ = \left| {\matrix{ {{a^2} + 3a + 2} & {a + 2} & 1 \cr {2(a + 2)} & 1 & 0 \cr {4a + 10} & 2 & 0 \cr } } \right|$

$ = 4(a + 2) - 4a - 10$

$ = 4a + 8 - 4a - 10 = - 2$

$A = \left[ {\matrix{ {{R_{11}}} & {{R_{12}}} & {{R_{13}}} \cr {{R_{21}}} & {{R_{22}}} & {{R_{23}}} \cr {{R_{31}}} & {{R_{32}}} & {{R_{33}}} \cr } } \right]$

$2A = \left[ {\matrix{ {2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \cr {2{R_{21}}} & {2{R_{22}}} & {2{R_{23}}} \cr {2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \cr } } \right]$

${R_2} \to 2{R_2} + 5{R_3}$

$B = \left[ {\matrix{ {2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \cr {4{R_{21}} + 10{R_{31}}} & {4{R_{22}} + 10{R_{32}}} & {4{R_{23}} + 10{R_{33}}} \cr {2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \cr } } \right]$

${R_2} \to {R_2} - 5{R_3}$

$B = \left[ {\matrix{ {2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \cr {4{R_{21}}} & {4{R_{22}}} & {4{R_{23}}} \cr {2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \cr } } \right]$

$\left| B \right| = \left[ {\matrix{ {2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \cr {4{R_{21}}} & {4{R_{22}}} & {4{R_{23}}} \cr {2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \cr } } \right]$

$\left| B \right| = 2 \times 2 \times 4\left| {\matrix{ {{R_{11}}} & {{R_{12}}} & {{R_{13}}} \cr {{R_{21}}} & {{R_{22}}} & {{R_{23}}} \cr {{R_{31}}} & {{R_{32}}} & {{R_{33}}} \cr } } \right|$

$ = 16 \times 4$

$ = 64$

$\left[ {\matrix{ 1 & { - \alpha } \cr \alpha & \beta \cr } } \right]\left[ {\matrix{ 1 & \alpha \cr { - \alpha } & \beta \cr } } \right] = \left[ {\matrix{ {1 + {\alpha ^2}} & {\alpha - \alpha \beta } \cr {\alpha - \alpha \beta } & {{\alpha ^2} + {\beta ^2}} \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$

1 + $\alpha$² = 1

$\alpha$² = 0

$\alpha$² + $\beta$² = 1

$\beta$² = 1

$\alpha$⁴ = 0

$\beta$⁴ = 1

$\alpha$⁴ + $\beta$⁴ = 1

$\Delta = \left| {\matrix{ 2 & 3 & 2 \cr 3 & 2 & 2 \cr 1 & { - 1} & 4 \cr } } \right| = - 20 \ne 0$ $ \therefore $ unique solution

${\Delta _x} = \left| {\matrix{ 9 & 3 & 2 \cr 9 & 2 & 2 \cr 8 & { - 1} & 4 \cr } } \right| = 0$

${\Delta _y} = \left| {\matrix{ 2 & 9 & 2 \cr 3 & 9 & 2 \cr 1 & 8 & 4 \cr } } \right| = - 20$

${\Delta _z} = \left| {\matrix{ 2 & 3 & 9 \cr 3 & 2 & 9 \cr 1 & { - 1} & 8 \cr } } \right| = - 40$

$ \therefore $ $x = {{{\Delta _x}} \over \Delta } = 0$

$y = {{{\Delta _y}} \over \Delta } = 1$

$z = {{{\Delta _z}} \over \Delta } = 2$

Unique solution : (0, 1, 2)

A^T = A, B^T = $-$B

Let A²B² $-$ B²A² = P

P^T = (A²B² $-$ B²A²)^T = (A²B²)^T $-$ (B²A²)^T

= (B²)^T (A²)^T $-$ (A²)^T (B²)^T

= B²A² $-$ A²B²

$ \Rightarrow $ P is skew-symmetric matrix

$\left[ {\matrix{ 0 & a & b \cr { - a} & 0 & c \cr { - b} & { - c} & 0 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

$ \therefore $ ay + bz = 0 ..... (1)

$-$ax + cz = 0 .... (2)

$-$bx $-$cy = 0 ..... (3)

From equation 1, 2, 3

$\Delta$ = 0 & $\Delta$₁ = $\Delta$₂ = $\Delta$₃ = 0

$ \therefore $ equation have infinite number of solution

$x - 2y + 0.z = 1$

$x - y + kz = - 2$

$0.x + ky + 4z = 6$

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

For unique solution $4 - {k^2} \ne 0$

$ \Rightarrow $ k $ \ne $ $ \pm $ 2

For k = 2 :

$x - 2y + 0.z = 1$

$x - y + 2z = - 2$

$0.x + 2y + 4z = 6$

$\Delta x = \left| {\matrix{ 1 & { - 2} & 0 \cr 2 & { - 1} & 2 \cr 6 & 2 & 4 \cr } } \right| = ( - 8) + 2[ - 20]$

$\Delta x = - 48 \ne 0$

For k = 2, $\Delta x \ne 0$

$ \therefore $ For K = 2; The system has no solution.

$\Delta = \left| {\matrix{ 3 & { - 2} & { - k} \cr 1 & { - 4} & { - 2} \cr 1 & 2 & { - 1} \cr } } \right| = 0$

$3(4 + 4) + 2( - 2 + 2) - k(4 + 4) = 0$

$ \Rightarrow k = 3$

${\Delta _x} = \left| {\matrix{ {10} & { - 2} & { - 3} \cr 6 & { - 4} & { - 2} \cr {5m} & 2 & { - 1} \cr } } \right| \ne 0$

$10(4 + 4) + 2( - 6 + 10m) - 3(12 + 20m) \ne 0$

$80 - 12 + 20m - 36 - 60m \ne 0$

$40m \ne 32 \Rightarrow m \ne {4 \over 5}$

${\Delta _y} = \left| {\matrix{ 3 & {10} & { - 3} \cr 2 & 6 & { - 2} \cr 1 & {5m} & { - 1} \cr } } \right| \ne 0$

$3( - 6 + 10m) - 10( - 2 + 2) - 3(10m - 6) \ne 0$

$ - 18 + 30m - 30m + 18 \ne 0 \Rightarrow 0$

${\Delta _z} = \left| {\matrix{ 3 & { - 2} & {10} \cr 2 & { - 4} & 6 \cr 1 & 2 & {5m} \cr } } \right| \ne 0$

$3( - 20m - 12) + 2(10m - 6) + 10(4 + 4) - 40m + 32 \ne 0 \Rightarrow m \ne {4 \over 5}$

For option (a)

$PEP = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & 2 & 3 \cr 2 & 3 & 4 \cr 8 & {13} & {18} \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$

$ = \left[ {\matrix{ 1 & 2 & 3 \cr 8 & {13} & {18} \cr 2 & 3 & 4 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 3 & 2 \cr 8 & {18} & {13} \cr 2 & 4 & 3 \cr } } \right] = F$

and ${P^2} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$

Hence, option (a) is correct.

For option (b)

$\left| {EQ + PF{Q^{ - 1}}} \right| = \left| {EQ} \right| + \left| {PF{Q^{ - 1}}} \right|$ .... (i)

$\because$ | E | = 0 and | F | = 0 and | Q | $\ne$ 0

$\therefore$ $\left| {EQ} \right| = \left| E \right|\left| Q \right| = 0$

and $\left| {PF{Q^{ - 1}}} \right| = {{\left| P \right|\left| F \right|} \over {\left| Q \right|}} = 0$

Let $R = EQ + PF{Q^{ - 1}}$ .... (ii)

$ \Rightarrow RQ = E{Q^2} + PF = E{Q^2} + {P^2}EP = E{Q^2} + EP$ [$\because$ P² = I]

$ = E({Q^2} + P)$

$ \Rightarrow \left| {RQ} \right| = \left| {E({Q^2} + P)} \right|$

$ \Rightarrow \left| R \right|\left| Q \right| = \left| E \right|\left| {{Q^2} + P} \right| = 0$ [$\because$ | E | = O]

$ \Rightarrow \left| R \right| = 0$ (as $\left| Q \right| \ne 0$) ..... (iii)

From Eqs. (ii) and (iii), we get Eq. (i) is true.

Hence, option (b) is correct.

For option (c)

$\left| {{{(EF)}^3}} \right| > {\left| {EF} \right|^2}$

i.e. 0 > 0 which is false.

For option (d)

$\because$ ${P^2} = I \Rightarrow {P^{ - 1}} = P$

$\therefore$ ${P^{ - 1}}FP = PFP = PPEPP = E$

So, $E + {P^{ - 1}}FP = E + E = 2E$

$ \Rightarrow Tr(E + {P^{ - 1}}FP) = Tr(2E) = 2Tr(E)$ ..... (iv)

and ${P^{ - 1}}EP + F$

$ \Rightarrow PEP + F = 2PEP$ [$\because$ F = PEP]

$\therefore$ $Tr(2PEP) = 2Tr(PEP) = 2Tr(F) = 2Tr(E)$ ..... (v)

From Eqs. (iv) and (v) option (d) is also correct.

$\because$ I $-$ EF = G^$-$1 $\Rightarrow$ G $-$ GEF = I ..... (i)

and G $-$ EFG = I ..... (ii)

Clearly, GEF = EFG $\to$ option (c) is correct.

Also, (I $-$ FE) (I + FGE)

= I $-$ FE + FGE $-$ FEFGE

= I $-$ FE + FGE $-$ F(G $-$ I) E

= I $-$ FE + FGE $-$ FGE + FE

= I $\to$ option (b) is correct but option (d) is incorrect.

$\because$ (I $-$ FE) (I $-$ FGE) = I $-$ FE $-$ FGE + F(G $-$ I) E

= I $-$ 2FE

Now, (I $-$ FE) ($-$ FGE) = $-$ FE

$\Rightarrow$ | I $-$ FE | | FGE | = | FE |

$\to$ option (a) is correct.

$\operatorname{det} A=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=k$ (given)

$\begin{aligned} \text { Then, } \operatorname{det} B & =\left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2+2 a_1 & b_2+2 b_1 & c_2+2 c_1 \\ a_3 & b_3 & c_3 \end{array}\right| \\ & =\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|+\left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ 2 a_1 & 2 b_1 & 2 c_1 \\ a_3 & b_3 & c_3 \end{array}\right| \\ & =k+0 \\ \Rightarrow \operatorname{det} B & =k \end{aligned}$

$\begin{aligned} A & =\left[\begin{array}{llll}\sqrt{2020} & \sqrt{2022} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{4040} & \sqrt{4042} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{6060} & \sqrt{6063} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{8080} & \sqrt{8084} & \sqrt{8088} & \sqrt{8092}\end{array}\right] \\ & =\left[\begin{array}{cccc}\sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{2} \cdot \sqrt{2020} & \sqrt{2} \sqrt{2021} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{3} \sqrt{2020} & \sqrt{3} \sqrt{2021} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{4} \sqrt{2020} & \sqrt{4} \sqrt{2021} & \sqrt{8088} & \sqrt{8092}\end{array}\right] \\ R_2 & \rightarrow R_2-\sqrt{2} R_1, R_3 \rightarrow R_3-\sqrt{3} R_1, R_4 \rightarrow R_4-\sqrt{4} R_1 \\ A & \sim\left[\begin{array}{cccc}\sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]\end{aligned}$

$\because$ A has one non-zero row

$\therefore$ Rank (A) = 1

Option (a) is correct.

The given equation is:

$ \left| \begin{array}{lll} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{array} \right| = 0 $

We can rewrite the third column as (1) + (the cube term), so this is the same as:

$ \left| \begin{array}{lll} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array} \right| + \left| \begin{array}{lll} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{array} \right| = 0 $

Let’s call the first determinant as $A$ and the second as $B$.

Step 1: Find the value of $|A|$

Let: $ A = \left| \begin{array}{lll} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{array} \right| $

Subtract first row from second and third, $ R_2 \rightarrow R_2 - R_1,\quad R_3 \rightarrow R_3 - R_1 $ so, $ \left| \begin{array}{ccc} x & x^2 & 1 \\ y-x & y^2 - x^2 & 0 \\ z-x & z^2 - x^2 & 0 \end{array} \right| $

Now, since last column for $R_2$ and $R_3$ is 0, expand along third column:

$|A| = (y-x)(z^2 - x^2) - (z-x)(y^2 - x^2)$

$= (y-x)(z-x)(z+x) - (z-x)(y-x)(y+x)$

$= (y-x)(z-x)(z+x - y-x)$

$= (y-x)(z-x)(z-y)$

Step 2: Find the value of $|B|$

$ B = \left| \begin{array}{ccc} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{array} \right| $

Factor $x, y, z$ from each row: $ = x y z \left| \begin{array}{ccc} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{array} \right| $

Now, do the same row operation as before: $ R_2 \rightarrow R_2 - R_1,\quad R_3 \rightarrow R_3 - R_1 $

So determinant becomes: $ x y z \left| \begin{array}{ccc} 1 & x & x^2 \\ 0 & y-x & y^2 - x^2 \\ 0 & z-x & z^2 - x^2 \end{array} \right| $

Expand this along the first column:

$|B| = x y z \left\{ (y-x)(z^2 - x^2) - (z-x)(y^2 - x^2) \right\}$

$= x y z (y-x)(z-x)(z-y)$

Step 3: Use the original equation

We have: $ |A| + |B| = 0 $ So, $ (y-x)(z-x)(z-y) + x y z (y-x)(z-x)(z-y) = 0 $

Factor out $(y-x)(z-x)(z-y)$: $ (y-x)(z-x)(z-y)(1 + x y z) = 0 $

Because $x, y, z$ are all different, $(y-x)(z-x)(z-y) \neq 0$. So, $ 1 + x y z = 0 $

Therefore, $ x y z = -1 $

Let $n=2$ (say) and $A=\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right]$, Then $\operatorname{adj} A=\left[\begin{array}{cc}c & -b \\ 0 & a\end{array}\right]$ which is a upper triangular matrix. Then, for any $n \times n$ matrix $A$, $\operatorname{adj} A$ is also upper triangular matrix.

$f(x)=\left|\begin{array}{ccc}x & x^2 & x^3 \\ 1 & 2 x & 3 x^2 \\ 0 & 2 & 6 x\end{array}\right|$

Expand along column 1

$\begin{aligned} f(x) & =x\left(12 x^2-6 x^2\right)-\left(6 x^3-2 x^3\right) \\ & =x\left(6 x^2\right)-4 x^3=6 x^3-4 x^3=2 x^3 \\ \therefore \quad f(x) & =2 x^3 \Rightarrow f^{\prime}(x)=6 x^2 \\ f^{\prime \prime}(x) & =12 x \end{aligned}$

Then, $f^{\prime \prime}(x): f^{\prime}(x)=12 x: 6 x^2=2: x$

Trace of A = Sum of diagonal elements of A

Given, $A=\left[\begin{array}{ccc}1 & -5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9\end{array}\right]$

$=1+7+9=17$

$\begin{aligned} & \text { }-x+y \cos C+z \cos B=0 \\ & x \cos C-y+z \cos A=0 \\ & x \cos B+y \cos A-Z=0 \\ & \Delta=\left|\begin{array}{ccc}-1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1\end{array}\right| \\ & =-1\left(1-\cos ^2 A\right)+\cos C(\cos A \cos B+\cos C) \\ & +\cos B(\cos A \cos C+\cos B) \\ & \end{aligned}$

$\begin{aligned} & =-1+\cos ^2 A+2 \cos A \cos B \cos C \\ & \qquad+\cos ^2 C+\cos ^2 B \\ & =-\sin ^2 A+\cos ^2 B+\cos ^2 C+2 \cos A \cos B \cos C \\ & \text { Clearly, } \Delta>0 . \\ & \Rightarrow \text { System of equations has a non-zero solution. }\end{aligned}$

$\begin{aligned} \text { Let } A & =\left[\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right] \\ A^{-1} & =\frac{\operatorname{adj}(A)}{|A|} \\ & =\left(\frac{1}{\sec ^2 \theta}\right)\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right] \end{aligned}$

So, $\frac{1}{\sec ^2 \theta}\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$

$\begin{aligned} &=\frac{1}{\sec ^2 \theta}\left[\begin{array}{cc} 1-\tan ^2 \theta & -2 \tan \theta \\ 2 \tan \theta & -\tan ^2 \theta+1 \end{array}\right] \\ & {\left[\begin{array}{ll} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] } \\ & \Rightarrow \quad a=\cos 2 \theta, b=\sin 2 \theta \end{aligned}$

$\begin{aligned} & {\left[\begin{array}{ccc}a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b\end{array}\right] \quad\left[R_1 \rightarrow R_1+R_3\right]} \\ & =(a+b+c)\left[\begin{array}{ccc}1 & 1 & 1 \\ a-b & b-c & c-a \\ b+c & c+a & a+b\end{array}\right] \\ & =(a+b+c)\left[\begin{array}{ccc}1 & 1 & 1 \\ a+c & b+a & b+c \\ b+c & c+a & a+b\end{array}\right] \quad {\left[R_2 \rightarrow R_2+R_3\right]} \\ & \end{aligned}$

$\begin{aligned} C_2 & \rightarrow C_2-C_1 \text { and } C_3 \rightarrow C_3-C_1 \\ & =(a+b+c)\left[\begin{array}{ccc}1 & 0 & 0 \\ a+c & b-c & b-a \\ b+c & a-b & a-c\end{array}\right] \\ & =(a+b+c)[(b-c)(a-c)-(b-a)(a-b)] \\ & =(a+b+c)\left(a b-b c-c a+c^2-b a+b^2+a^2-a b\right) \\ & =(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) \\ & =a^3+b^3+c^3-3 a b c\end{aligned}$

$\begin{aligned} & \text { }\left|\begin{array}{ccc} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{array}\right| \\ & =(b+c)[(c+a)(a+b)-b c]-a\left(a b+b^2-b c\right) \\ & =(b+c)\left(a c+b c+a^2+a b-b c\right) \quad+a\left(b c-c^2-a c\right) \\ & -a^2 b-a b^2+a b c+a b c-a c^2-a^2 c \\ & =a b c+a^2 b+a b^2+a c^2+a^2 c+a b c -a^2 b-a b^2+a b c+a b c-a c^2-a^2 c\\ & =4 a b c \end{aligned}$

Option (c) is correct.

${D^T} = ABC$

and $S = ABCD$

${S^2} = (ABCD)(ABCD) = ({D^T}\,.\,D)({D^T}D)$

$ = {({D^T}D)^2}$ ..... (i)

$\because$ $S = ABCD = {D^T}D$

${S^T} = {({D^T}D)^T} = {D^T}D$

$\therefore$ From Eq. (i), we get

${S^2} = {({S^T})^2}$

$\therefore$ ${S^2} = {({S^T})^2} = {({S^2})^T}$

$\begin{aligned} A & =\left[\begin{array}{lll} a^2 & 15 & 31 \\ 12 & b^2 & 41 \\ 35 & 61 & c^2 \end{array}\right], \\ B & =\left[\begin{array}{ccc} 2 a & 3 & 5 \\ 2 & 2 b & 8 \\ 1 & 4 & 2 c-3 \end{array}\right] \end{aligned}$

Given, trace of $A=\operatorname{trace}$ of $B$

$\begin{aligned} & \Rightarrow a^2+b^2+c^2=2 a+2 b+2 c-3 \\ & \Rightarrow\left(a^2-2 a+1\right)+\left(b^2-2 b+1\right)+\left(c^2-2 c+1\right)=0 \\ & \Rightarrow(a-1)^2+(b-1)^2+(c-1)^2=0 \\ & \Rightarrow a=1, b=1, c=1 \end{aligned}$

Then product of diagonal elements of $B$

$=(2 a)(2 b)(2 c-3)=(2)(2)(2-3)=-4$

$\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$

$ +\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n-1} b & (-1)^n c \end{array}\right|=0$

$\begin{aligned} & \because\left.|A|=\mid A^T\right] \\ & \Rightarrow\left|\begin{array}{ccc} a & -b & c \\ a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \end{array}\right| \\ &+(-1)^n\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ a & -b & c \end{array}\right|=0 \end{aligned}$

$\begin{aligned} & R_2 \leftrightarrow R_3 \text { and } R_3 \leftrightarrow R_1 \text { in 2nd determinant } \\ & \Rightarrow\left|\begin{array}{ccc} a & -b & c \\ a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \end{array}\right| \\ & +(-1)^n\left|\begin{array}{ccc} a & -b & c \\ a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \end{array}\right|=0 \\ \end{aligned}$

Both determinants are equal and their sum is zero when $n$ is an odd integer.

$\begin{array}{lr}\text { }\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3 x & x^2\end{array}\right|=0 \\ \Rightarrow 1\left(6 x^2-12 x\right)+3\left(x^2-4\right)+1(3 x-6)=0 \\ \Rightarrow 6 x^2-12 x+3 x^2-12+3 x-6=0 \\ \Rightarrow 9 x^2-9 x-18=0 \\ \Rightarrow x^2-x-2=0\end{array}$

$\begin{aligned} & \text { } A=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] \\ & \text { Then, } A^T=\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right] \\ & \because \quad A A^T=20 I \\ & \Rightarrow \quad\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right]=20\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{array}\right]=\left[\begin{array}{cc} 20 & 0 \\ 0 & 20 \end{array}\right] \\ & \therefore \quad a^2+b^2=20 \\ & \Rightarrow \quad(a+b)^2-2 a b=20 \\ & \Rightarrow \quad(a+b)^2=20+2 \times \frac{5}{2} \quad\left[\because a b=\frac{5}{2}\right] \\ & \text { or } \quad(a+b)^2=25 \\ & or\quad a+b= \pm 5 \\ \end{aligned}$

Equation whose roots are $a$ and $b$ is given by $x^2+(a+b) x+a b=0$

or $\quad x^2 \pm 5 x+\frac{5}{2}=0$ or $2 x^2 \pm 10 x+5=0$

$\begin{aligned} & \text { (c) } A=\left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{array}\right] \\ & \therefore|A|=1(1+3)+1(2+3)+1(2-1)=10 \quad ... (\mathrm{i})\\ & \text { and } 10 B=\left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{array}\right] \\ & \therefore|10 B|=4(2 \alpha)-2(-15-\alpha)+2(10) \\ & \Rightarrow 10^3|B|=10 \alpha+50 \quad\left[\because|k A|=k^n|A|\right] \ldots \mathrm{(ii)}\\ & \because B=A^{-1}\\ & \Rightarrow|B|=\left|A^{-1}\right|=|A|^{-1} \\ & \Rightarrow|B|=\frac{1}{|A|}=\frac{1}{10} \quad [\text { [rom Eq. (i)} \end{aligned}$

[from Eq. (i)] From Eq. (ii), we get

$1000~.~\frac{1}{10}=10\alpha+50$

or $\alpha=\frac{50}{10}=5$

Let $A=\left[\begin{array}{ccc}4 & 2 & 1-x \\ 5 & k & 1 \\ 6 & 3 & 1+x\end{array}\right]$

Given that, rank of matrix $A$ is 1 .

$\Rightarrow$ All the minors of order 2 are zero.

$\therefore {M_{33}} = 0$

$ \Rightarrow \left| {\matrix{ 4 & 2 \cr 5 & k \cr } } \right| = 0$

or $4 k-10=0$ or $k=\frac{5}{2}$

and $M_{11}=\left|\begin{array}{cc}k & 1 \\ 3 & 1+x\end{array}\right|=0$

or $\left|\begin{array}{cc}\frac{5}{2} & 1 \\ 3 & 1+x\end{array}\right|=0$

or $\frac{5}{2}+\frac{5}{2} x-3=0$

or $\frac{5}{2} x =\frac{1}{2}$

or $x =\frac{1}{5}$

$a_1, a_2, \ldots, a_9$, are in GP.

Then, $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\ldots \frac{a_9}{a_8}=r$ [common ratio]

Now, $\left|\begin{array}{lll}\log a_1 & \log a_2 & \log a_3 \\ \log a_4 & \log a_5 & \log a_6 \\ \log a_7 & \log a_8 & \log a_9\end{array}\right|$

$C_3 \rightarrow C_3-C_2, C_2 \rightarrow C_2-C_1 $

$\left| {\matrix{ {\log {a_1}} & {\log {{{a_2}} \over {{a_1}}}} & {\log {{{a_3}} \over {{a_2}}}} \cr {\log {a_4}} & {\log {{{a_5}} \over {{a_4}}}} & {\log {{{a_6}} \over {{a_5}}}} \cr {\log {a_7}} & {\log {{{a_8}} \over {{a_7}}}} & {\log {{{a_9}} \over {{a_8}}}} \cr } } \right|$

or $\left| {\matrix{ {\log {a_1}} & {\log r} & {\log r} \cr {\log {a_4}} & {\log r} & {\log r} \cr {\log {a_7}} & {\log r} & {\log r} \cr } } \right|=0$

[$\because$ C$_2$ and C$_3$ are identical]

$\begin{aligned} & \text { } \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{c}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathbf{a} \cdot \mathbf{a}=4+1+9=14 \\ & \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}=2+3-3=2\end{aligned}$

$\begin{aligned} & \mathbf{a} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=6-1-6=-1 \\ & \text { b. } \mathbf{b}=1+9+1=11 \\ & \mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{b}=3-3+2=2 \\ & \mathbf{c} \cdot \mathbf{c}=9+1+4=14 \\ & \text { Now, }\left|\begin{array}{lll}\mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c}\end{array}\right|=\left|\begin{array}{rrr}14 & 2 & -1 \\ 2 & 11 & 2 \\ -1 & 2 & 14\end{array}\right| \\ & =14(154-4)-2(28+2)-1(4+11) \\ & =2100-60-15 \\ & =2025 \\ & \end{aligned}$