If A=12−332−22−11, then find A-1 and use it to solve following system of the equation:x+2y−3z=63x+2y−2z=32x−y+z=2ORUsing properties of determinants, prove that(b+c)2 a2 bc(c+a)2 b2 ca(a+b)2 c2 ab=(a−b)(b−c)(c−a)(a+b+c)a2+b2+c2

First, we will find,|A|=det A=12−332−22−11=1(2−2)−2(3+4)−3(−3−4)=0−14+21=7Now, We will find the cofactors of the matrix A:A11=2−2−11=0,A12=−3−221=−7,A13=322−1=−7,A21=−2−3−11=1,A22=1−321=7,A23=−122−1=5,A31=2−32−2=2,A32=−1−33−2=−7,A33=1232=−4Therefore, adjoint of matrix A will be given by:adjA=0−7−71752−7−4T=012−77−7−75−4∴A−1=1|A| adj A=17012−77−7−75−4Now, the system of equations can be reduced in the matrix form AX=B asWhere, X=xyz,B=632AX=B⇒X=A−1Bxyz=17012−77−7−75−4632=177−35−35=1−5−5x=1,y=−5 and z=−5Therefore, A−1=17012−77−7−75−4 and (x,y,z)=(1,−5,−5)ORGiven: L.H.S. =(b+c)2a2bc(c+a)2b2ca(a+b)2c2ab=b2+c2a2bcc2+a2b2caa2+b2c2ab ∵C1→C1−2C3=a2+b2+c2a2bca2+b2+c2b2caa2+b2+c2c2ab ∵c1→C1+C2=a2+b2+c2a2bc0b2−a2ca−bc0c2−a2ab−bc∵R1→R2−R1,R3→R3−R1=(b−a)(c−a)a2+b2+c2a2bc0b+a−c0c+a−b(∵get common (b-a) from R2and (c-a) from RR3)Expand along column 1 =a2+b2+c2(b−a)(c−a)−b2−ab+c2+ac=(a−b)(b−a)(c−a)(a+b+c)a2+b2+c2= R.H.S ∴ L.H.S. = R.H.S Therefore, which has proven.

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If $A = [\begin{array}{ccc} 1 & 2 & - 3 \\ 3 & 2 & - 2 \\ 2 & - 1 & 1 \end{array}]$ , then find A^-1 and use it to solve following system of the equation:
$\begin{aligned} x + 2 y - 3 z = 6 \\ 3 x + 2 y - 2 z = 3 \\ 2 x - y + z = 2 \end{aligned}$
OR
Using properties of determinants, prove that
$|\begin{array}{lll} (b + c)^{2} a^{2} bc \\ (c + a)^{2} b^{2} ca \\ (a + b)^{2} c^{2} ab \end{array}| = (a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2})$

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Detailed Solution

First, we will find,
$| A | = \det A = |\begin{array}{ccc} 1 & 2 & - 3 \\ 3 & 2 & - 2 \\ 2 & - 1 & 1 \end{array}| = 1 (2 - 2) - 2 (3 + 4) - 3 (- 3 - 4) = 0 - 14 + 21 = 7$
Now, We will find the cofactors of the matrix A:
$\begin{array}{l} A_{11} = |\begin{array}{lc} 2 & - 2 \\ - 1 & 1 \end{array}| = 0, \\ A_{12} = - |\begin{array}{cc} 3 & - 2 \\ 2 & 1 \end{array}| = - 7, \\ A_{13} = |\begin{array}{cc} 3 & 2 \\ 2 & - 1 \end{array}| = - 7, \\ A_{21} = - |\begin{array}{cc} 2 & - 3 \\ - 1 & 1 \end{array}| = 1, \\ A_{22} = |\begin{array}{cc} 1 & - 3 \\ 2 & 1 \end{array}| = 7, \\ A_{23} = - |\begin{array}{cc} 1 & 2 \\ 2 & - 1 \end{array}| = 5, \\ A_{31} = |\begin{array}{ll} 2 & - 3 \\ 2 & - 2 \end{array}| = 2, \\ A_{32} = - |\begin{array}{ll} 1 & - 3 \\ 3 & - 2 \end{array}| = - 7, \\ A_{33} = |\begin{array}{ll} 1 & 2 \\ 3 & 2 \end{array}| = - 4 \end{array}$
Therefore, adjoint of matrix A will be given by:
$\begin{aligned} adjA = {[\begin{array}{lcc} 0 & - 7 & - 7 \\ 1 & 7 & 5 \\ 2 & - 7 & - 4 \end{array}]}^{T} = [\begin{array}{ccc} 0 & 1 & 2 \\ - 7 & 7 & - 7 \\ - 7 & 5 & - 4 \end{array}] \\ ∴ A^{- 1} = \frac{1}{| A |} adj A = \frac{1}{7} [\begin{array}{ccc} 0 & 1 & 2 \\ - 7 & 7 & - 7 \\ - 7 & 5 & - 4 \end{array}] \end{aligned}$
Now, the system of equations can be reduced in the matrix form AX=B as
Where, $X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{l} 6 \\ 3 \\ 2 \end{array}]$
$\begin{array}{l} AX = B \Rightarrow X = A^{- 1} B \\ [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{7} [\begin{array}{ccl} 0 & 1 & 2 \\ - 7 & 7 & - 7 \\ - 7 & 5 & - 4 \end{array}] [\begin{matrix} 6 \\ 3 \\ 2 \end{matrix}] = \frac{1}{7} [\begin{array}{c} 7 \\ - 35 \\ - 35 \end{array}] = [\begin{array}{c} 1 \\ - 5 \\ - 5 \end{array}] \\ x = 1, y = - 5 and z = - 5 \end{array}$
Therefore, $A^{- 1} = \frac{1}{7} [\begin{array}{ccc} 0 & 1 & 2 \\ - 7 & 7 & - 7 \\ - 7 & 5 & - 4 \end{array}] and (x, y, z) = (1, - 5, - 5)$

Given:
$\begin{array}{l} L.H.S. = |\begin{array}{lll} (b + c)^{2} & a^{2} & bc \\ (c + a)^{2} & b^{2} & ca \\ (a + b)^{2} & c^{2} & ab \end{array}| \\ = |\begin{array}{ccc} b^{2} + c^{2} & a^{2} & bc \\ c^{2} + a^{2} & b^{2} & ca \\ a^{2} + b^{2} & c^{2} & ab \end{array}| (∵ C_{1} \to C_{1} - 2 C_{3}) \\ = |\begin{array}{ccc} a^{2} + b^{2} + c^{2} & a^{2} & bc \\ a^{2} + b^{2} + c^{2} & b^{2} & ca \\ a^{2} + b^{2} + c^{2} & c^{2} & ab \end{array}| (∵ c_{1} \to C_{1} + C_{2}) \\ = |\begin{array}{ccc} a^{2} + b^{2} + c^{2} & a^{2} & bc \\ 0 & b^{2} - a^{2} & ca - bc \\ 0 & c^{2} - a^{2} & ab - bc \end{array}| (∵ R_{1} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}) \end{array}$
$= (b - a) (c - a) |\begin{array}{ccc} a^{2} + b^{2} + c^{2} & a^{2} & bc \\ 0 & b + a & - c \\ 0 & c + a & - b \end{array}| (∵$ get common (b-a) from $R_{2}$ and (c-a) from R $R_{3}$ )
Expand along column 1
$\begin{array}{l} = (a^{2} + b^{2} + c^{2}) (b - a) (c - a) (- b^{2} - ab + c^{2} + ac) \\ = (a - b) (b - a) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2}) \\ = R.H.S \\ ∴ L.H.S. = R.H.S \end{array}$
Therefore, which has proven.