Givem that the matrix A = ((3,1,5),(2,3,5),(5,1,6) ). If Adj. A = (((13),(-1),(-10)),((13),(-7),(-5)),((-13),2,7) ) (i) find A^(−1) (ii) Use the result in (i) to find the values of x, y and z that will satisfy the equations: 3x + y + 5z = 8 2x +3y + 5z = 0 5x + y + 6z = 13


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Question Number 205586 by necx122 last updated on 25/Mar/24

$G i v e m t h a t t h e m a t r i x A = (\begin{matrix} 3 & 1 & 5 \\ 2 & 3 & 5 \\ 5 & 1 & 6 \end{matrix}) .$ $I f A d j . A = (\begin{matrix} 13 & - 1 & - 10 \\ 13 & - 7 & - 5 \\ - 13 & 2 & 7 \end{matrix})$ $(i) f i n d A^{- 1}$ $(i i) U s e t h e r e s u l t i n (i) t o f i n d t h e$ $v a l u e s o f x, y a n d z t h a t w i l l s a t i s f y t h e$ $e q u a t i o n s :$ $3 x + y + 5 z = 8$ $2 x + 3 y + 5 z = 0$ $5 x + y + 6 z = 13$
	Answered by som(math1967) last updated on 25/Mar/24

	$A^{- 1} = \frac{1}{∣ A ∣} \times A d j . A$ $∣ A ∣= 3 (18 - 5) - 1 (12 - 25) + 5 (2 - 15)$ $= 39 + 13 - 65$ $= 13 (3 + 1 - 5) = - 13$ $∴ A^{- 1} = (\begin{matrix} - 1 & \frac{1}{13} & \frac{10}{13} \\ - 1 & \frac{7}{13} & \frac{5}{13} \\ 1 & \frac{- 2}{13} & \frac{- 7}{13} \end{matrix})$ $e q u a t i o n$ $A \times X = B$ $w h e r e = A = (\begin{matrix} 3 & 1 & 5 \\ 2 & 3 & 5 \\ 5 & 1 & 6 \end{matrix})$ $X = (\begin{matrix} x \\ y \\ z \end{matrix}) B = (\begin{matrix} 8 \\ 0 \\ 13 \end{matrix})$ $A^{- 1} \times A \times X = A^{- 1} \times B$ $I \times X = (\begin{matrix} - 8 + 0 + 10 \\ - 8 + 0 + 5 \\ 8 + 0 - 7 \end{matrix})$ $(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 2 \\ - 3 \\ 1 \end{matrix})$ $∴ x = 2, y = - 3, z = 1 a n s$
	Commented by necx122 last updated on 25/Mar/24

	$T h i s i s s o c l e a r a n d e x p r e s s i v e . T h a n k y o u$ $s o m u c h s i r .$
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