\(\begin{bmatrix} 0& sin \alpha &cos \alpha \\ sin \alpha & 0 & sin \beta \\ cos \alpha & -sin\beta & 0 \end{bmatrix}\)= 0M₁₁ - sin\(\alpha\) M₁₂ + cos\(\alpha\) M₁₃
= 0 − sin\(\alpha\) (0 − cos\(\alpha\) sin\(\beta\) ) + cos\(\alpha\) (−sin\(\alpha\) sin\(\beta\) − 0) = 0.

Given A^T = \(\left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{matrix} \right] \)
(ii) Verify (A + B)^T= A^T + B^T = B^T = B^T + A^T
(A)^T =\({ \left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 4 & -1 & 2 \\ 5 & 0 & 3 \end{matrix} \right] \)
A = \(\left[ \begin{matrix} 4 & -2 & 2 \\ 5 & 0 & 3 \end{matrix} \right] \)
Now, A + B = \(\left[ \begin{matrix} 4 & -1 & 2 \\ 5 & 0 & 3 \end{matrix} \right] +\left[ \begin{matrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{matrix} \right] =\left[ \begin{matrix} 6 & -2 & 3 \\ 12 & 5 & 1 \end{matrix} \right] \)
\(\therefore\) (A + B)^T= \(\left[ \begin{matrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{matrix} \right] \) ---(1)
B^T = \(\left[ \begin{matrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{matrix} \right] \)
\(\therefore\) A^T + B^T = \(\left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] +\left[ \begin{matrix} 2 & 7 \\ -1 & 5 \\ 1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{matrix} \right] \)----(2)
B^T + A^T = \(\left[ \begin{matrix} 2 & 7 \\ -1 & 5 \\ 1 & -2 \end{matrix} \right] +\left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] =\left[ \begin{matrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{matrix} \right] \) ----------------------(3)
From (1), (2) and (3), (A + B)^T = A^T + B^T = B^T+ A^T

3B +  4C − D =\(\begin{bmatrix} 6 & 9 & 0\\3 & -3 & 15\end{bmatrix}\)+\(\begin{bmatrix} -4 & -8 & 12\\-4 & 0 & 8\end{bmatrix}\)+\(\begin{bmatrix} 0 & -4 & 1\\-5 & -6 & 5\end{bmatrix}\)=\(\begin{bmatrix} 2 & -3 & 13\\-6 & -9 & 28\end{bmatrix}\).

\(\text { (i) Civen } A=\left[\begin{array}{ccc} 0 & 1 & -2 \\ -1 & 0 & x^3 \\ 2 & -3 & 0 \end{array}\right] \text { is skew-symmetric. }\)
\(A^T=-A\)
\(aA^T=\left(\begin{array}{ccc} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & x^3 & 0 \end{array}\right)\)
\(A^T=-A\)
\(\left[\begin{array}{ccc} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & x^3 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 1 & 0 & -x^3 \\ -2 & 3 & 0 \end{array}\right]\)
\(x^3=3\)
\(\therefore x=3^{1 / 3}\)

Given A = \(\begin{bmatrix} 1 &0 &0 \\0 & 1 & 0 \\a &b &-1 \end{bmatrix}\)
A² = A \(\times\) A =\(​​​​\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{matrix} \right] =\left[ \begin{matrix} 1+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ a+0-1 & 0+b-b & 0+0+1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\therefore\) A² is a unit matrix.

AB =\(\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}\)\(\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 16 & 2 & -1 \\ -2 & 9 & 9 \\ 7 & 1 & 14 \end{bmatrix}\)
(AB)^T =\(\begin{bmatrix} 16 & -2 & 7 \\ 2 & 9 & 1 \\ 22 & 9 & 14 \end{bmatrix}\)...(1)
B^T = \(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\), A^T=\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)
B^TA^T=\(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\)\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)=\(\begin{bmatrix} 16 & -2 & 7 \\ 2 & 9 & 1 \\ 22 & 9 & 14 \end{bmatrix}\)...(2)
From (1) and (2), (AB)^T = B^TA^T.

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

\(A^{T}=\left[\begin{array}{ll} a & y \\ x & a \end{array}\right] \quad \therefore A A^{T}=\left[\begin{array}{ll} a & x \\ y & a \end{array}\right]\left[\begin{array}{ll} a & y \\ x & a \end{array}\right] \)
\(=\left[\begin{array}{ll} a^{2}+x^{2} & a y+a x \\ a y+a x & y^{2}+a^{2} \end{array}\right] \)
\(\operatorname{det}\left(A A^{T}\right)=\left|\begin{array}{ll} a^{2}+x^{2} & a y+a x \\ a y+a x & y^{2}+a^{2} \end{array}\right| \)
\(=\left(a^{2}+x^{2}\right)\left(a^{2}+y^{2}\right)-(a y+a x)(a y+a x) \)
\(=a^{4}+a^{2} y^{2}+a^{2} x^{2}+x^{2} y^{2}-a^{2} y^{2}-a^{2} x y -a^{2} x y-a^{2} x^{2}\)
\(=a^{4}+1-2 a^{2} x y=a^{4}-2 a^{2}+1(\because x y=1) \)
\(=\left(a^{2}-1\right)^{2} \)

\(2 X=\left[\begin{array}{ll} 3 & 8 \\ 7 & 2 \end{array}\right]-\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 2 & 6 \\ 4 & -2 \end{array}\right]\)
\(X=\left[\begin{array}{cc} 1 & 3 \\ 2 & -1 \end{array}\right]\)

\(A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \)
\(a_{11}=\frac{1}{2}(3-2)=\frac{1}{2} ; a_{12}=\frac{1}{2}(3-4)=\frac{-1}{2} \)
\(a_{21}=\frac{1}{2}(3(2)-2)=\frac{4}{2}=2 ; a_{22}=\frac{1}{2}(6-4)=\frac{2}{2}=1 \)
\(\therefore A=\left[\begin{array}{ll} \frac{1}{2} & -\frac{1}{2} \\ 2 & 1 \end{array}\right] \)