Find x,y and z\left( {\begin{array}{*{20}{c}}1&3&2\3&{ - 2}&5\2&{ - 3}&6\end{array}} ight)\,\left( {\begin{array}{*{20}{c}}x\y\z\end{array}} ight) = \left( {\begin{array}{*{20}{c}}6\5\7\end{array}} ight)

Question

Find x,y and z\left( {\begin{array}{*{20}{c}}1&3&2\3&{ - 2}&5\2&{ - 3}&6\end{array}} ight)\,\left( {\begin{array}{*{20}{c}}x\y\z\end{array}} ight) = \left( {\begin{array}{*{20}{c}}6\5\7\end{array}} ight)

Filo · Accepted Answer

Given, ⎝⎛​132​3−2−3​256​⎠⎞​⎝⎛​xyz​⎠⎞​=⎝⎛​657​⎠⎞​As we know for matrices A , X and B if AX=B then X=A−1BSo, we need to find the inverse of matrix A=⎝⎛​132​3−2−3​256​⎠⎞​For A−1 to exist ∣A∣=0∣A∣=1×∣∣​−2−3​56​∣∣​−3×∣∣​32​56​∣∣​+2×∣∣​32​−23​∣∣​⇒∣A∣=(−12+15)−3(18−10)+2(9+4)⇒∣A∣=3−3×8+2×13=3−24+26=5=0∴A−1 existsCofactors of A :-a11​=(−1)1+1×∣∣​−2−3​56​∣∣​=(−12+15)=3a12​=(−1)1+2×∣∣​32​56​∣∣​=−(18−10)=−8a13​=(−1)1+3×∣∣​32​−2−3​∣∣​=(−9+4)=−5a21​=(−1)2+1×∣∣​3−3​26​∣∣​=−(18+6)=−24a22​=(−1)2+2×∣∣​12​26​∣∣​=(6−4)=2a23​=(−1)2+3×∣∣​12​3−3​∣∣​=−(−3−6)=9a31​=(−1)3+1×∣∣​3−2​25​∣∣​=(15+4)=19a32​=(−1)3+2×∣∣​13​25​∣∣​=−(5−6)=1a33​=(−1)3+3×∣∣​13​3−2​∣∣​=(−2−9)=−11Thus , cofactor Matrix C=⎣⎡​3−2419​−821​−59−11​⎦⎤​So, AdjA=CT=⎣⎡​3−2419​−821​−59−11​⎦⎤​T=⎣⎡​3−8−5​−2429​191−11​⎦⎤​Now, A−1=∣A∣1​AdjA=51​×⎣⎡​3−8−5​−2429​191−11​⎦⎤​∴X=A−1B=51​⎣⎡​3−8−5​−2429​191−11​⎦⎤​⎣⎡​657​⎦⎤​X=51​⎣⎡​3×6−24×5+19×7−8×6+2×5+1×7−5×6+9×5−11×7​⎦⎤​=51​⎣⎡​31−31−62​⎦⎤​∴x=531​,y=5−31​,z=562​

Question Text	Find $x, y$ and $z$ \left( {\begin{array}{{20}{c}}1&3&2\\3&{ - 2}&5\\2&{ - 3}&6\end{array}} \right)\,\left( {\begin{array}{{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\5\\7\end{array}} \right)
Answer Type	Text solution:1
Upvotes	150

Find $x, y$ and $z$
\left( {\begin{array}{{20}{c}}1&3&2\\3&{ - 2}&5\\2&{ - 3}&6\end{array}} \right)\,\left( {\begin{array}{{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\5\\7\end{array}} \right)

Text solutionVerified

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Find x,y and z\left( {\begin{array}{*{20}{c}}1&3&2\\3&{ - 2}&5\\2&{ - 3}&6\end{array}} \right)\,\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\5\\7\end{array}} \right)

Text solutionVerified

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Find $x, y$ and $z$
\left( {\begin{array}{{20}{c}}1&3&2\\3&{ - 2}&5\\2&{ - 3}&6\end{array}} \right)\,\left( {\begin{array}{{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\5\\7\end{array}} \right)