If A=⎣⎡231−3215−4−2⎦⎤, find A−1. Using A−1 solve the system of equations2x−3y+5z=113x+2y−4z=5x+y− 2z=3

Question

If A=⎣⎡​231​−321​5−4−2​⎦⎤​, find A−1. Using A−1 solve the system of equations2x−3y+5z=113x+2y−4z=5x+y− 2z=3

Filo · Accepted Answer

Given system of equations2x−3y+5z=113x+2y−4z=5x+y−2z=3This can be written as AX=Bwhere A=⎣⎡​231​−321​5−4−2​⎦⎤​,X=⎣⎡​xyz​⎦⎤​,B=⎣⎡​1153​⎦⎤​Here, ∣A∣=2(−4+4)+3(−6+4)+5(3−2)⇒∣A∣=−6+5=−1Since, ∣A∣= 0Hence, the system of equations is consistent and has a unique solution given by X==A−1BA−1=∣A∣adjA​ and adjA=CTC11​=(−1)1+1 ∣∣​21​−4−2​∣∣​⇒C11​=−4+4=0C12​=(−1)1+2 ∣∣​31​−4−2​∣∣​⇒C12​=−(−6+4)=2C13​=(−1)1+3 ∣∣​31​21​∣∣​⇒C13​=3−2=1C21​=(−1)2+1 ∣∣​−31​5−2​∣∣​⇒C21​=−(6−5)=−1C22​=(−1)2+2 ∣∣​21​5−2​∣∣​⇒C22​=−4−5=−9C23​=(−1)2+3 ∣∣​21​−31​∣∣​⇒C23​=−(2+3)=−5C31​=(−1)3+1 ∣∣​−32​5−4​∣∣​⇒C31​=12−10=2C32​=(−1)3+2 ∣∣​23​5−4​∣∣​⇒C32​=−(−8−15)=23C33​=(−1)3+3 ∣∣​23​−32​∣∣​⇒C33​=4+9=13Hence, the co-factor matrix is C=⎣⎡​0−12​2−923​1−513​⎦⎤​⇒adjA=CT=⎣⎡​021​−1−9−5​22313​⎦⎤​⇒A−1=∣A∣adjA​=−11​⎣⎡​021​−1−9−5​22313​⎦⎤​⇒A−1=⎣⎡​0−2−1​195​−2−23−13​⎦⎤​Solution is given by ⎣⎡​xyz​⎦⎤​=⎣⎡​0−2−1​195​−2−23−13​⎦⎤​⎣⎡​1153​⎦⎤​⎣⎡​xyz​⎦⎤​=⎣⎡​5−6−22+45−69−11+25−39​⎦⎤​⎣⎡​xyz​⎦⎤​=41​⎣⎡​8412​⎦⎤​⎣⎡​xyz​⎦⎤​=⎣⎡​−1−46−25​⎦⎤​Hence, x=−1,y=−46,z=−25

Question Text	If $A = ⎣ ⎡ 231 - 3 21 5 - 4 - 2 ⎦ ⎤$ , find $A^{- 1}$ . Using $A^{- 1}$ solve the system of equations $2 x - 3 y + 5 z = 11$ $3 x + 2 y - 4 z = 5$ $x + y - 2 z = 3$
Answer Type	Text solution:1
Upvotes	150

If $A = ⎣ ⎡ 231 - 3 21 5 - 4 - 2 ⎦ ⎤$ , find $A^{- 1}$ . Using $A^{- 1}$ solve the system of equations
$2 x - 3 y + 5 z = 11$
$3 x + 2 y - 4 z = 5$
$x + y - 2 z = 3$

Text solutionVerified

Practice questions from similar books

If A=⎣⎡​231​−321​5−4−2​⎦⎤​, find A−1. Using A−1 solve the system of equations2x−3y+5z=113x+2y−4z=5x+y− 2z=3

Text solutionVerified

Practice questions from similar books

If $A = ⎣ ⎡ 231 - 3 21 5 - 4 - 2 ⎦ ⎤$ , find $A^{- 1}$ . Using $A^{- 1}$ solve the system of equations
$2 x - 3 y + 5 z = 11$
$3 x + 2 y - 4 z = 5$
$x + y - 2 z = 3$