Applying C₁\(\rightarrow\) C₁ - C₂ , we get \(\begin{vmatrix} 4&(a^x-a^{-x})^2 &1 \\ 4 & (b^x-b^{-x})^2 & 1 \\ 4 & (c^x-c^{-x})^2 & 1 \end{vmatrix}\) = 0, Since C₁ and C₃ are proportional.

\(A=\left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]\)
\(\therefore A^T=\left[\begin{array}{ccc} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right]\)
\(A+A^T=\left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]+\left[\begin{array}{ccc} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{array}\right]\)
\(A-A^T=\left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]-\left[\begin{array}{ccc} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{array}\right]\)
\((1) \Rightarrow A=\frac{1}{2}\left(A+A^T\right)+\frac{1}{2}\left(A-A^T\right)\)
\(A=\frac{1}{2}\left[\begin{array}{ccc} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{array}\right]+\frac{1}{2}\left[\begin{array}{ccc} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{array}\right]\)

\(\text { (i) } \mathrm{A}=\left[\begin{array}{ll} 4 & -2 \\ 3 & -5 \end{array}\right]\)
\(We \ can \ write \ A=\frac{1}{2}\left(A+A^T\right)+\frac{1}{2}\left(A-A^T\right) (1)\)
\(A^T=\left[\begin{array}{cc} 4 & 3 \\ -2 & -5 \end{array}\right]\)
\(A+A^{\top}=\left[\begin{array}{cc} 4 & -2 \\ 3 & -5 \end{array}\right]+\left[\begin{array}{cc} 4 & 3 \\ -2 & -5 \end{array}\right]=\left[\begin{array}{cc} 8 & 1 \\ 1 & -10 \end{array}\right]\)
\(A-A^T=\left[\begin{array}{cc} 4 & -2 \\ 3 & -5 \end{array}\right]-\left[\begin{array}{cc} 4 & 3 \\ -2 & -5 \end{array}\right]=\left[\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right]\)
\(\therefore A=\frac{1}{2}\left[\begin{array}{cc} 8 & 1 \\ 1 & -10 \end{array}\right]+\frac{1}{2}\left[\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right]\)

Given a_ij = i-j
Let A = [a_ij] _{3 \(\times\) 3}
Ingeneralwe can wrlte \(\Lambda=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{21} & a_{13} \\ a_{31} & a_{32} & a_{33} \end{array}\right]\)
\(a_{i j}=1-j\)
\(\therefore\) a₁₁ = 1 - 1 = 0 a₂₁ = 2 - 1 = 1  a₃₁ = 3 - 1 = 2
a₁₂ = 1 - 2 = 0  a₂₂ = 2 - 2 = 0 a₃₂ = 3 - 2 = 1
a₁₃ = 1 - 3 = -2 a₂₃ = 2 - 3 = -1 a₃₃ = 3 - 3 = 0
\(\Rightarrow A=\left[ \begin{matrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{matrix} \right] \)
\(\Rightarrow A^{T}=\left[ \begin{matrix} 0 & -1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{matrix} \right] =-A\)
Since A^T = -A  \(\Rightarrow\) A is a skew-symmetric matrix.

\(\begin{vmatrix} log_364 &log_43 \\ log_38 & log_49 \end{vmatrix}\times \begin{vmatrix} log_23 & log_83 \\ log_34 & log_34 \end{vmatrix}\) = \(\left| \begin{matrix} { log }_{ 3 }64.{ { log }_{ 2 }3+log }_{ 4 }3.{ log }_{ 3 }4 & { log }_{ 3 }64.{ log }_{ 8 }3+{ log }_{ 4 }3.{ log }_{ 3 }4 \\ { log }_{ 3 }8.{ log }_{ 2 }3+{ log }_{ 4 }9.{ log }_{ 3 }4 & { log }_{ 3 }8.{ log }_{ 8 }3+{ log }_{ 4 }9.{ log }_{ 3 }4 \end{matrix} \right| \)
= \(\left| \begin{matrix} { log }_{ 2 }64+1 & { log }_{ 8 }64+1 \\ { log }_{ 2 }8+{ log }_{ 3 }9 & 1+{ log }_{ 3 }9 \end{matrix} \right| [\therefore { log }_{ y }x.{ log }_{ x }y=1]\)
= \(\left| \begin{matrix} { log }_{ 2 }{ 2 }^{ 6 }+1 & { log }_{ 8 }{ 8 }^{ 2 }+1 \\ { log }_{ 2 }{ 2 }^{ 3 }+{ log }_{ 3 }{ 3 }^{ 2 } & 1+{ log }_{ 3 }{ 3 }^{ 2 } \end{matrix} \right| [\therefore { log }_{ x }x=1]\)
= \(\left| \begin{matrix} 6+1 & 2+1 \\ 3+2 & 1+2 \end{matrix} \right| =\left| \begin{matrix} 7 & 3 \\ 5 & 3 \end{matrix} \right| \) = 21 - 15 = 6

\(LHS=\begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix}^2=\begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix}\times \begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix}\)
\(=\begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix} \times(-1)(-1)\begin{vmatrix} 1 &x &x \\ -x & -1 &-x \\- x &-x &-1 \end{vmatrix} \)
= \(\begin{vmatrix} 1 &x &x \\ x & 1 &x \\ x &x &1 \end{vmatrix} \times \begin{vmatrix} 1 &x &x \\ -x & -1 &-x \\- x &-x &-1 \end{vmatrix} \)
= \(\begin{vmatrix} 1-x^2-x^2 &x-x-x^2 &x-x^2-x \\ x-x-x^2 & x^2-1-x^2 &x^2-x-x \\ x-x^2-x &x^2-x-x &x^2-x^2-1 \end{vmatrix}\)
= \(\begin{vmatrix} 1-2x^2 &-x^2 &-x^2 \\- x^2 & -1 &x^2-2x \\ -x^2 &x^2-2x &-1 \end{vmatrix}\)
 = RHS.

LHS = \(\begin{vmatrix} 0 & c &b \\ c & 0 &a \\ b & a & 0 \end{vmatrix}^2=\begin{vmatrix} 0 & c &b \\ c & 0 &a \\ b & a & 0 \end{vmatrix}\times \begin{vmatrix} 0 & c &b \\ c & 0 &a \\ b & a & 0 \end{vmatrix}\)
\(=\begin{vmatrix} 0+c^2+b^2 & 0+0+ab &0+ac+0 \\ 0+0+ab & c^2+0+a^2 &bc+0+0 \\ 0+ac+0 & bc+0+0 & b^2+a^2+0 \end{vmatrix}\)
\(=\begin{vmatrix} c^2+b^2 & ab & ac \\ ab & c^2+a^2 & bc \\ ab & bc & b^2+a^2 \end{vmatrix}\) = \(\left|\begin{array}{ccc} b^{2}+c^{2} & a b & a c \\ a b & c^{2}+a^{2} & b c \\ a c & b c & a^{2}+b^{2} \end{array}\right|\) = RHS.

Given \(A=\left[\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right]\left[\begin{array}{lll} + & - & + \\ - & + & - \\ + & - & + \end{array}\right]\)
Cofactors of elements of second row
\(|A|=-2\left|\begin{array}{ll} 3 & 8 \\ 2 & 3 \end{array}\right|+0-1\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|\)
= 2(9-16)-1(10-3)
= -2(-7) - 1(7) = 14 - 7= 7
\(|A|=7\)

Given \(\left|\begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2 x & 5 x^2 \end{array}\right|=0\)
\(\Rightarrow\left|\begin{array}{ccc} 1 & 4 & 20 \\ 0 & 6 & 15 \\ 0 & -2-2 x & 15-5 x^2 \end{array}\right|=0 \quad \begin{aligned} &R_1 \rightarrow R_1 \\ &R_2 \rightarrow R_1-R_2 \\ &R_3 \rightarrow R_2-R_3 \end{aligned}\)
expand by C₁
\(1\left[6\left(5-5 x^2\right)-15(-2-2 x)\right]-0+0=0\)
\(-30 x^2+30 x+60=0\)
\((\div 30)-x^2+x+2=0\)
\(x^2-x-2=0\)
\((x-2)(x+1)=0\)
\(\therefore x=2 \text { (or) } x=-1\)

Gift pack matrix
\(A=\left[\begin{array}{ccc} \text { Pack I } & \text { PII } & \text { PIII } \\ 100 & 200 & 250 \\ 100 & 100 & 250 \\ 50 & 100 & 150 \end{array}\right] \begin{aligned} &\text { Cashew } \\ &\text { raisins } \\ &\text { almonds } \end{aligned}\)
Given Cost of 50 gm of cashew is Rs. 50
\(\therefore\)1 gm of cashew is Rs. 1
Cost of 50 gm of raisin is Rs. 10
1 gm of raisin is Rs. 1/5
Cost of 50gm of almonds is Rs. 60
1 gm of raisin is Rs. 6/5
Cost matrix \(B=\left(1, \frac{1}{5}, \frac{6}{5}\right)\)
Cost of package is AB.
\(A B=\left(\begin{array}{lll} 1 & \frac{1}{5} & \frac{6}{5} \end{array}\right)\left[\begin{array}{ccc} 100 & 200 & 250 \\ 100 & 100 & 250 \\ 50 & 100 & 150 \end{array}\right]\)
\(=\left[\begin{array}{c} 100+20+60 \\ 200+20+120 \\ 250+50+180 \end{array}\right]=\left[\begin{array}{c} 180 \\ 340 \\ 480 \end{array}\right]\)
Pack I Cost Rs. 180
Pack II Cost Rs. 340
Pack III Cost Rs. 480