Let A = [ x 1 a m p ; y 1 a m p ; z 1 x 2 a m p ; y 2 a m p ; z 2 x 3 a m p ; y 3 a m p ; z 3 ] \text { Let } A=\left[\begin{array}{lll}x_1 & y_1 & z_1 \\x_2 & y_2 & z_2 \\x_3 & y_3 & z_3\end{array}\right] Let A = x 1 x 2 x 3 am p ; y 1 am p ; y 2 am p ; y 3 am p ; z 1 am p ; z 2 am p ; z 3
Given A [ 1 0 1 ] = [ 2 0 2 ] ..... (1) \text { Given } A\left[\begin{array}{l}1 \\0 \\1\end{array}\right]=\left[\begin{array}{l}2 \\0 \\2\end{array}\right] \quad \text{ ..... (1)} Given A 1 0 1 = 2 0 2 ..... (1)
∴ [ x 1 + z 1 x 2 + z 2 x 3 + z 3 ] = [ 2 0 2 ] \therefore\left[\begin{array}{l}\mathrm{x}_1+\mathrm{z}_1 \\\mathrm{x}_2+\mathrm{z}_2 \\\mathrm{x}_3+\mathrm{z}_3\end{array}\right]=\left[\begin{array}{l}2 \\0 \\2\end{array}\right] ∴ x 1 + z 1 x 2 + z 2 x 3 + z 3 = 2 0 2
∴ x 1 + z 1 a m p ; = 2 .... (2) x 2 + z 2 a m p ; = 0 .... (3) x 3 + z 3 a m p ; = 0 .... (4) \begin{aligned}\therefore \mathrm{x}_1+\mathrm{z}_1 & =2 \quad \text{.... (2)}\\\mathrm{x}_2+\mathrm{z}_2 & =0 \quad \text{.... (3)}\\\mathrm{x}_3+\mathrm{z}_3 & =0 \quad \text{.... (4)}\end{aligned} ∴ x 1 + z 1 x 2 + z 2 x 3 + z 3 am p ; = 2 .... (2) am p ; = 0 .... (3) am p ; = 0 .... (4)
a m p ; Given A [ − 1 0 1 ] = [ − 4 0 4 ] a m p ; ∴ [ − x 1 + z 1 − x 2 + z 2 − x 3 + z 3 ] = [ 4 0 4 ] \begin{aligned}& \text { Given } A\left[\begin{array}{l}-1 \\0 \\1\end{array}\right]=\left[\begin{array}{l}-4 \\0 \\4\end{array}\right] \\& \therefore\left[\begin{array}{l}-\mathrm{x}_1+\mathrm{z}_1 \\-\mathrm{x}_2+\mathrm{z}_2 \\-\mathrm{x}_3+\mathrm{z}_3\end{array}\right]=\left[\begin{array}{l}4 \\0 \\4\end{array}\right]\end{aligned} am p ; Given A − 1 0 1 = − 4 0 4 am p ; ∴ − x 1 + z 1 − x 2 + z 2 − x 3 + z 3 = 4 0 4
⇒ − x 1 + z 1 = − 4 ... (5) − x 2 + x 2 = 0 ... (6) − x 3 + z 3 = 4 \begin{array}{r}\Rightarrow-\mathrm{x}_1+\mathrm{z}_1=-4 \quad \text{... (5)}\\-\mathrm{x}_2+\mathrm{x}_2=0 \quad \text{... (6)}\\-\mathrm{x}_3+\mathrm{z}_3=4\end{array} ⇒ − x 1 + z 1 = − 4 ... (5) − x 2 + x 2 = 0 ... (6) − x 3 + z 3 = 4
a m p ; Given A [ 0 1 0 ] = [ 0 2 0 ] a m p ; ∴ [ y 1 y 2 y 3 ] = [ 0 2 0 ] \begin{aligned}& \text { Given } A\left[\begin{array}{l}0 \\1 \\0\end{array}\right]=\left[\begin{array}{l}0 \\2 \\0\end{array}\right] \\& \therefore\left[\begin{array}{l}\mathrm{y}_1 \\\mathrm{y}_2 \\\mathrm{y}_3\end{array}\right]=\left[\begin{array}{l}0 \\2 \\0\end{array}\right]\end{aligned} am p ; Given A 0 1 0 = 0 2 0 am p ; ∴ y 1 y 2 y 3 = 0 2 0
a m p ; ∴ y 1 = 0 , y 2 = 2 , y 3 = 0 a m p ; ∴ from ( 2 ) , ( 3 ) , ( 4 ) , ( 5 ) , ( 6 ) and ( 7 ) a m p ; x 1 = 3 x , x 2 = 0 , x 3 = − 1 a m p ; y 1 = 0 , y 2 = 2 , y 3 = 0 a m p ; z 1 = − 1 , z 2 = 0 , z 3 = 3 \begin{aligned}& \therefore \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\& \therefore \text { from }(2),(3),(4),(5),(6) \text { and }(7) \\& \mathrm{x}_1=3 \mathrm{x}, \mathrm{x}_2=0, \mathrm{x}_3=-1 \\& \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\& \mathrm{z}_1=-1, \mathrm{z}_2=0, \mathrm{z}_3=3\end{aligned} am p ; ∴ y 1 = 0 , y 2 = 2 , y 3 = 0 am p ; ∴ from ( 2 ) , ( 3 ) , ( 4 ) , ( 5 ) , ( 6 ) and ( 7 ) am p ; x 1 = 3 x , x 2 = 0 , x 3 = − 1 am p ; y 1 = 0 , y 2 = 2 , y 3 = 0 am p ; z 1 = − 1 , z 2 = 0 , z 3 = 3
∴ A = [ 3 a m p ; 0 a m p ; − 1 0 a m p ; 2 a m p ; 0 − 1 a m p ; 0 a m p ; 3 ] ∴ Now ( A − 31 ) [ x y z ] = [ − 1 2 3 ] ∴ [ 0 a m p ; 0 a m p ; − 1 0 a m p ; − 1 a m p ; 0 − 1 a m p ; 0 a m p ; 0 ] [ x y z ] = [ − 1 2 3 ] [ − z − y − x ] = [ 1 2 3 ] [ z = − 1 ] , [ y = − 2 ] , [ x = − 3 ] \begin{gathered}\therefore A=\left[\begin{array}{ccc}3 & 0 & -1 \\0 & 2 & 0 \\-1 & 0 & 3\end{array}\right] \\\therefore \text { Now }(A-31)\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{l}-1 \\2 \\3\end{array}\right] \\\therefore\left[\begin{array}{ccc}0 & 0 & -1 \\0 & -1 & 0 \\-1 & 0 & 0\end{array}\right]\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{r}-1 \\2 \\3\end{array}\right] \\{\left[\begin{array}{c}-z \\-y \\-x\end{array}\right]=\left[\begin{array}{l}1 \\2 \\3\end{array}\right]} \\{[z=-1],[y=-2],[x=-3]}\end{gathered} ∴ A = 3 0 − 1 am p ; 0 am p ; 2 am p ; 0 am p ; − 1 am p ; 0 am p ; 3 ∴ Now ( A − 31 ) x y z = − 1 2 3 ∴ 0 0 − 1 am p ; 0 am p ; − 1 am p ; 0 am p ; − 1 am p ; 0 am p ; 0 x y z = − 1 2 3 − z − y − x = 1 2 3 [ z = − 1 ] , [ y = − 2 ] , [ x = − 3 ]