-135

a₃₂ = -11
Alternative Method:
Given \(A =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right|\)
⇒ a₃₂ \(=\left| \begin{matrix} 5 & 8 \\ 2 & 1 \end{matrix} \right|\)
⇒ a₃₂ = 5-16 = -11

\(x=\frac { 13 }{ 15 }\)
Alternative Method:
Since, A is singular matrix
\(\left| \begin{matrix} 1+x & 7 \\ 3-x & 8 \end{matrix} \right| =0 \)
\(\Rightarrow 8(1+x)-7(3-x)=0 \)
\(\Rightarrow 8+8x-21+7x=0 \)
\(\Rightarrow 15x-13=0 \)
\(=\frac { 13 }{ 15 }\)

\(x=1 \)
Alternative Method:
\(A=\left| \begin{matrix} 3-2x & x+1 \\ 2 & 4 \end{matrix} \right|\)
Since A is a singular matrix. i.e., |A|=0
\(\left| \begin{matrix} 3-2x & x+1 \\ 2 & 4 \end{matrix} \right| =0 \)
\(\Rightarrow 4(3-2x)-2(x+1)=0 \)
\(\Rightarrow 12-8x-2x-2=0 \)
\(\Rightarrow -10x+10=0 \)
\(\Rightarrow x-1\)

\(2-20=2x^{ 2 }-24 \)
\(\Rightarrow x=\pm \sqrt { 3 }\)

k = 7
Alternative Method:
Since, |kA| = kⁿ|A|, where n is the order of matrix.
|3A| = 3³(A)
= 27|A|
⇒ k = 27

If points are collinear then area of triangle = 0.
\(\Rightarrow \frac{1}{2}\left|\begin{array}{ccc} 0 & 2 & 1 \\ 1 & x & 1 \\ 3 & 1 & 1 \end{array}\right|=0 \)
\(\Rightarrow \frac{1}{2}[0-2(1-3)+1(1-3 x)]=0 \)
\(\Rightarrow 4+1-3 x=0 \Rightarrow x=\frac{5}{3}\)

\(A^{-1}=\frac{1}{|A|} \text { adj } A\)
\(|A|=\left[\begin{array}{lr} 2 & 3 \\ 5 & -2 \end{array}\right]=-4-15=-19 \)
\(\Rightarrow A^{-1}=-\frac{1}{19}\left[\begin{array}{rr} -2 & -3 \\ -5 & 2 \end{array}\right]=\frac{1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]=\frac{1}{19} A \)

13

\(A=\left[ \begin{matrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{matrix} \right] \)
a₁₁ = (1 + 3) = 4
a₁₂ = -(-1 + 2) = -1
a₁₃ = (3 + 2) = 5
a₂₁ = -(-2 - 15) = 17
a₂₂ = (-1 - 10) = -11
a₂₃ = -(3 - 4) = 1
a₃₁ = (-2 + 5) = 3
a₃₂ = -(-1 - 5) = 6
a₃₃ = (-1 - 2) = -3
|A| = 1 x 4 + 2 x(-1) + 5 x (5)
= 4 - 2 + 25 = 27
\(\Rightarrow { A }^{ -1 }=\frac { 1 }{ 27 } \left[ \begin{matrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 4/27 & 17/27 & 1/9 \\ -1/27 & -11/27 & 2/9 \\ 5/27 & 1/27 & -1/9 \end{matrix} \right] \)
Given set of equations can be written as
x + 2y + 5z = 10
x - y - z = -2
and 2x + 3y - z = -11
\(\left[ \begin{matrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 10 \\ -2 \\ -11 \end{matrix} \right] \)
\(\Rightarrow AX=B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4/27 & 17/27 & 1/9 \\ -1/27 & -11/27 & 2/9 \\ 5/27 & 1/27 & -1/9 \end{matrix} \right] \left[ \begin{matrix} 10 \\ -2 \\ -11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} \frac { 40 }{ 27 } & -\frac { 34 }{ 27 } & -\frac { 33 }{ 27 } \\ -\frac { 10 }{ 27 } & \frac { 22 }{ 27 } & -\frac { 66 }{ 27 } \\ \frac { 50 }{ 27 } & -\frac { 2 }{ 27 } & \frac { 33 }{ 27 } \end{matrix} \right] =\left[ \begin{matrix} -\frac { 27 }{ 27 } \\ -\frac { 54 }{ 27 } \\ \frac { 81 }{ 27 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} -1 \\ -2 \\ 3 \end{matrix} \right] \)
\(\Rightarrow\) x = -1, y = -2 and z = 3

\(={ (x+y+z) }^{ 3 }\left| \begin{matrix} { (y+z) }^{ 2 } & x-y-z & x-y-z \\ { y }^{ 2 } & x+z-y & 0 \\ { z }^{ 2 } & 0 & x+y-z \end{matrix} \right| \)
LHS = \(\left| \begin{matrix} { (y+z) }^{ 2 } & xy & zx \\ xy & { (x+z) }^{ 2 } & yz \\ xz & yz & { (x+y) }^{ 2 } \end{matrix} \right| \)
Apply \(({ R }_{ 1 }\rightarrow x{ R }_{ 1 },\quad { R }_{ 2 }y\rightarrow { R }_{ 2 },{ R }_{ 3 }z\rightarrow { R }_{ 3 })\)
\(=\frac { 1 }{ xyz } \left| \begin{matrix} { x(y+z) }^{ 2 } & { x }^{ 2 }y & z{ x }^{ 2 } \\ x{ y }^{ 2 } & { y(x+z) }^{ 2 } & { y }^{ 2 }z \\ x{ z }^{ 2 } & y{ z }^{ 2 } & { z(x+y) }^{ 2 } \end{matrix} \right| \)
\(=\left| \begin{matrix} { (y+z) }^{ 2 } & { x }^{ 2 } & { x }^{ 2 } \\ { y }^{ 2 } & { (x+z) }^{ 2 } & { y }^{ 2 } \\ x{ z }^{ 2 } & { z }^{ 2 } & { (x+y) }^{ 2 } \end{matrix} \right| \)
Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 },{ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\)
\(=\left| \begin{matrix} { (y+z) }^{ 2 } & { x }^{ 2 }-{ (y+z) }^{ 2 } & { x }^{ 2 }-{ (y+z) }^{ 2 } \\ { y }^{ 2 } & { (x+z) }^{ 2 }-{ y }^{ 2 } & 0 \\ { z }^{ 2 } & 0 & { (x+y) }^{ 2 }-{ z }^{ 2 } \end{matrix} \right| \)
Taking (x + y + z) common from C₂ and C₃ . 
\(={ (x+y+z) }^{ 3 }\left| \begin{matrix} { (y+z) }^{ 2 } & x-y-z & x-y-z \\ { y }^{ 2 } & x+z-y & 0 \\ { z }^{ 2 } & 0 & x+y-z \end{matrix} \right| \)
Apply \({ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }-{ R }_{ 3 }\)
\(={ (x+y+z) }^{ 2 }\left| \begin{matrix} { (y+z) }^{ 2 } & x-y-z & x-y-z \\ { y }^{ 2 } & x+z-y & 0 \\ { z }^{ 2 } & 0 & x+y-z \end{matrix} \right| \)
Apply, \({ C }_{ 2 }\rightarrow y{ C }_{ 2 },{ C }_{ 3 }\rightarrow z{ C }_{ 3 }\)
\(=\frac { { (x+y+z) }^{ 2 } }{ yz } \left| \begin{matrix} 2yz & 0 & 0 \\ { y }^{ 2 } & x+z-y & { y }^{ 2 } \\ { z }^{ 2 } & { z }^{ 2 } & zx+yz \end{matrix} \right| \)
\(=\frac { { (x+y+z) }^{ 2 } }{ yz } \left| (2yz).({ x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 }-{ y }^{ 2 }{ z }^{ 2 } \right| \)
\(=2xyz{ (x+y+z) }^{ 3 }=RHS\)

x=500, y=2000, z=3500

\(\text { Given }\left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right|=0, a \neq 0\)
\(\Rightarrow\left|\begin{array}{rrc} a & 0 & x \\ -a & a & x \\ 0 & -a & x+a \end{array}\right|=0\)
\(\left[\text { by performing } C_{1} \rightarrow C_{1}-C_{2} \text { and } C_{2} \rightarrow C_{2}-C_{3}\right]\)
\(\Rightarrow\left|\begin{array}{ccc} a & 0 & x \\ 0 & a & 2 x \\ 0 & -a & x+a \end{array}\right|=0\left[\text { by performing } R_{2} \rightarrow R_{2}+R_{1}\right]\)
\(\Rightarrow a\left(a x+a^{2}+2 a x\right)=0 \text { [on expanding along } \left.C_{1}\right] \)
\(\Rightarrow 3 a x+a^{2}=0 \Rightarrow x=-\frac{a}{3}, a \neq 0\)

LHS: \(\begin{vmatrix}a+b+2c&a&b\\c&b+c+2a&b\\c&c&c+a+2b \end{vmatrix}\)
= \((a+b+c)^2\begin{vmatrix} 1&-1&0\\0&1&-1\\c&a&c+a+2b\end{vmatrix}\)
= \((a+b+c)^2\begin{vmatrix}1&0&0\\0&1&-1\\ c&a+c&c+a+2b\end{vmatrix}\)
= \((a+b+c)^2\begin{vmatrix} 1&-1\\a+c&c+a+2b\end{vmatrix}\)
= (a+b+c)²[(c+a+2b)+(a+c)]
= (a+b+c)²(2a+2b+2c)
= 2(a+b+c)²(a+b+c)
= 2(a+b+c)³ = RHS

|A| = \(\left| \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right| \)
= \((1)\left| \begin{matrix} 1 & -3 \\ 4 & -9 \end{matrix} \right| -1\begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix}-2\begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}\)
= 1(-9 + 12) - (-18 + 15) -2 (8 - 5)
= 3 + 3 - 6 = 0
= 6 - 6 = 0
= 0

\(M_{11}=\begin{vmatrix} 0&4\\5&-7\end{vmatrix}=-0-20=-20\)
\(A_{11}=(-1)^{1+1}M_{11}=(-1)^2(-20)=-20\)
\(M_{12}=\begin{vmatrix}6&4\\1&-7 \end{vmatrix}=-42-4=-46\)
\(A_{12}=(-1)^{1+2}M_{12}=(-1)^3(-46)=(-1)(-46)=46\)
\(M_{13}=\begin{vmatrix}6&0\\1&5 \end{vmatrix}=30-0=30\)
\(A_{13}=(-1)^{1+3}M_{13}=(-1)^4(30)=30\)
\(M_{21}=\begin{vmatrix} -3&5\\5&-7\end{vmatrix}=21-25=-4\)
\(A_{21}=(-1)^{ 2+1}M_{21}=(-1)^3(-4=(-1 )(-4)=4)\)
\(M_{22}=\begin{vmatrix} 2&5\\1&-7\end{vmatrix}=-14-15=-19\)
\(A_{22}=(-1)^{2+2}M_{22}=(1)^4(-19)=-19\)
\(M_{23}=\begin{vmatrix} 2&-3\\1&5\end{vmatrix}=10+3=13\)
\(A_{23}=(-1)^{2+3}M_{23}=(-1)^513=-13\)
\(M_{31}=\begin{vmatrix}-3&5\\0&4 \end{vmatrix}=-12-0=-12\)
\(A_{31}=(-1)^{3+1}M_{31}=(-1)^4(-12)=-12\)
\(M_{32}=\begin{vmatrix} 2&5\\6&4\end{vmatrix}=8-30=-22\)
\(A_{32}=(-1)^{3+2}M_{32}=(-1)^5(-22)=(-1)(-22)=22\)
\(M_{33}=\begin{vmatrix} 2&-3\\6&0\end{vmatrix}=0+18=18\)
\(A_{33}=(-1)^{3+3}M_{33}=(-1)^6(18)=18\)
(ii)\(a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{32}\)
\(=(2)(-12)+(-3)(22)+(5)(18)=-24-66+90=0\)

Expanding R₁, we get:
\(\Delta =0\left| \begin{matrix} 0 & sin\beta \\ -sin\beta & 0 \end{matrix} \right| -sin\alpha \begin{vmatrix} -sin\alpha & sin\beta \\ cos\alpha & 0 \end{vmatrix}-cos\alpha \begin{vmatrix} -sin\alpha & 0 \\ cos\alpha & -sin\beta \end{vmatrix}\)
= 0 - sin \(\alpha\) (-0-sin \(\beta\) cos \(\alpha\))-cos \(\alpha\) (sin \(\alpha\) sin \(\beta\) - 0)
= sin \(\alpha\) sin \(\beta\) cos \(\alpha\) - cos \(\alpha\) sin \(\alpha\) sin \(\beta\) = 0

Note that in the third column, two entries are zero. So expanding along third column (C₃), we get
\(\Delta =\left| \begin{matrix} -1 & 3 \\ 4 & 1 \end{matrix} \right| -0\left| \begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix} \right| +0\left| \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right| \)
= 4 (–1 – 12) – 0 + 0 = – 52

We have 
\(\begin{vmatrix} 2 &4 \\-1 & 2 \end{vmatrix}\) = 2(2) – 4(–1)
= 4 + 4
= 8.