Answer to Question #85241 – Math – Linear Algebra
Question
Find the inverse, if possible, for the following matrices:
(a) ( 8 − 5 11 23 ) \left( \begin{array}{cc} 8 & -5 \\ 11 & 23 \end{array} \right) ( 8 11 − 5 23 )
(b) ( 5 − 7 6 − 11 6 2 2 4 − 7 ) \left( \begin{array}{ccc} 5 & -7 & 6 \\ -11 & 6 & 2 \\ 2 & 4 & -7 \end{array} \right) ⎝ ⎛ 5 − 11 2 − 7 6 4 6 2 − 7 ⎠ ⎞
Solution
(a) Let A = ( 8 − 5 11 23 ) A = \left( \begin{array}{cc} 8 & -5 \\ 11 & 23 \end{array} \right) A = ( 8 11 − 5 23 )
Then we get,
Det A = ∣ 8 − 5 11 23 ∣ = ( 8 ) ( 23 ) − ( − 5 ) ( 11 ) = 184 + 55 = 239 \operatorname{Det} A = \left| \begin{array}{cc} 8 & -5 \\ 11 & 23 \end{array} \right| = (8)(23) - (-5)(11) = 184 + 55 = 239 Det A = ∣ ∣ 8 11 − 5 23 ∣ ∣ = ( 8 ) ( 23 ) − ( − 5 ) ( 11 ) = 184 + 55 = 239
Therefore,
Inverse of A = A − 1 = Adjoint of A Det A \text{Inverse of } A = A^{-1} = \frac{\text{Adjoint of } A}{\text{Det } A} Inverse of A = A − 1 = Det A Adjoint of A A − 1 = 1 239 ( 23 − 11 5 8 ) T A^{-1} = \frac{1}{239} \left( \begin{array}{cc} 23 & -11 \\ 5 & 8 \end{array} \right)^T A − 1 = 239 1 ( 23 5 − 11 8 ) T A − 1 = 1 239 ( 23 5 − 11 8 ) A^{-1} = \frac{1}{239} \left( \begin{array}{cc} 23 & 5 \\ -11 & 8 \end{array} \right) A − 1 = 239 1 ( 23 − 11 5 8 ) A − 1 = ( 23 239 5 239 − 11 239 8 239 ) A^{-1} = \left( \begin{array}{cc} \frac{23}{239} & \frac{5}{239} \\ \frac{-11}{239} & \frac{8}{239} \end{array} \right) A − 1 = ( 239 23 239 − 11 239 5 239 8 )
(b) Let B = ( 5 − 7 6 − 11 6 2 2 4 − 7 ) B = \left( \begin{array}{ccc} 5 & -7 & 6 \\ -11 & 6 & 2 \\ 2 & 4 & -7 \end{array} \right) B = ⎝ ⎛ 5 − 11 2 − 7 6 4 6 2 − 7 ⎠ ⎞
Then we get,
Det B = ∣ 5 − 7 6 − 11 6 2 2 4 − 7 ∣ \operatorname{Det} B = \left| \begin{array}{ccc} 5 & -7 & 6 \\ -11 & 6 & 2 \\ 2 & 4 & -7 \end{array} \right| Det B = ∣ ∣ 5 − 11 2 − 7 6 4 6 2 − 7 ∣ ∣
Det B B B
= 5 [ 6 ( − 7 ) − 2 ( 4 ) ] + 7 [ ( − 11 ) ( − 7 ) − 2 ( 2 ) ] + 6 [ ( − 11 ) ( 4 ) − 2 ( 6 ) ] = 5 ( − 42 − 8 ) + 7 ( 77 − 4 ) + 6 ( − 44 − 12 ) = 5 ( − 50 ) + 7 ( 73 ) + 6 ( − 56 ) \begin{array}{l}
= 5[6(-7) - 2(4)] + 7[(-11)(-7) - 2(2)] + 6[(-11)(4) - 2(6)] \\
= 5(-42 - 8) + 7(77 - 4) + 6(-44 - 12) = 5(-50) + 7(73) + \\
6(-56)
\end{array} = 5 [ 6 ( − 7 ) − 2 ( 4 )] + 7 [( − 11 ) ( − 7 ) − 2 ( 2 )] + 6 [( − 11 ) ( 4 ) − 2 ( 6 )] = 5 ( − 42 − 8 ) + 7 ( 77 − 4 ) + 6 ( − 44 − 12 ) = 5 ( − 50 ) + 7 ( 73 ) + 6 ( − 56 ) = − 250 + 511 − 336 = − 75 \begin{array}{l}
= -250 + 511 - 336 \\
= -75 \\
\end{array} = − 250 + 511 − 336 = − 75
Adjoint B
= ( ( − 1 ) 1 + 1 ∣ 6 2 4 − 7 ∣ ( − 1 ) 1 + 2 ∣ − 11 2 2 − 7 ∣ ( − 1 ) 1 + 3 ∣ − 11 6 2 4 ∣ ( − 1 ) 2 + 1 ∣ − 7 6 4 − 7 ∣ ( − 1 ) 2 + 2 ∣ 5 6 2 − 7 ∣ ( − 1 ) 2 + 3 ∣ 5 − 7 2 4 ∣ ( − 1 ) 3 + 1 ∣ − 7 6 6 2 ∣ ( − 1 ) 3 + 2 ∣ 5 6 − 11 2 ∣ ( − 1 ) 3 + 3 ∣ 5 − 7 − 11 6 ∣ ) = ( − 50 − 73 − 56 − 25 − 47 − 34 − 50 − 76 − 47 ) T = ( − 50 − 25 − 50 − 73 − 47 − 76 − 56 − 34 − 47 ) \begin{array}{l}
= \left( \begin{array}{cccccc}
(-1)^{1+1} \left| \begin{array}{cc} 6 & 2 \\ 4 & -7 \end{array} \right| & (-1)^{1+2} \left| \begin{array}{cc} -11 & 2 \\ 2 & -7 \end{array} \right| & (-1)^{1+3} \left| \begin{array}{cc} -11 & 6 \\ 2 & 4 \end{array} \right| \\
(-1)^{2+1} \left| \begin{array}{cc} -7 & 6 \\ 4 & -7 \end{array} \right| & (-1)^{2+2} \left| \begin{array}{cc} 5 & 6 \\ 2 & -7 \end{array} \right| & (-1)^{2+3} \left| \begin{array}{cc} 5 & -7 \\ 2 & 4 \end{array} \right| \\
(-1)^{3+1} \left| \begin{array}{cc} -7 & 6 \\ 6 & 2 \end{array} \right| & (-1)^{3+2} \left| \begin{array}{cc} 5 & 6 \\ -11 & 2 \end{array} \right| & (-1)^{3+3} \left| \begin{array}{cc} 5 & -7 \\ -11 & 6 \end{array} \right| \end{array} \right) \\
= \left( \begin{array}{ccc} -50 & -73 & -56 \\ -25 & -47 & -34 \\ -50 & -76 & -47 \end{array} \right)^T \\
= \left( \begin{array}{ccc} -50 & -25 & -50 \\ -73 & -47 & -76 \\ -56 & -34 & -47 \end{array} \right) \\
\end{array} = ⎝ ⎛ ( − 1 ) 1 + 1 ∣ ∣ 6 4 2 − 7 ∣ ∣ ( − 1 ) 2 + 1 ∣ ∣ − 7 4 6 − 7 ∣ ∣ ( − 1 ) 3 + 1 ∣ ∣ − 7 6 6 2 ∣ ∣ ( − 1 ) 1 + 2 ∣ ∣ − 11 2 2 − 7 ∣ ∣ ( − 1 ) 2 + 2 ∣ ∣ 5 2 6 − 7 ∣ ∣ ( − 1 ) 3 + 2 ∣ ∣ 5 − 11 6 2 ∣ ∣ ( − 1 ) 1 + 3 ∣ ∣ − 11 2 6 4 ∣ ∣ ( − 1 ) 2 + 3 ∣ ∣ 5 2 − 7 4 ∣ ∣ ( − 1 ) 3 + 3 ∣ ∣ 5 − 11 − 7 6 ∣ ∣ ⎠ ⎞ = ⎝ ⎛ − 50 − 25 − 50 − 73 − 47 − 76 − 56 − 34 − 47 ⎠ ⎞ T = ⎝ ⎛ − 50 − 73 − 56 − 25 − 47 − 34 − 50 − 76 − 47 ⎠ ⎞
We know that,
Inverse of B = B − 1 = Adjoint of B Det B \text{Inverse of } B = B^{-1} = \frac{\text{Adjoint of } B}{\text{Det } B} Inverse of B = B − 1 = Det B Adjoint of B
Therefore,
B − 1 = 1 ( − 75 ) ( − 50 − 25 − 50 − 73 − 47 − 76 − 56 − 34 − 47 ) B^{-1} = \frac{1}{(-75)} \left( \begin{array}{ccc} -50 & -25 & -50 \\ -73 & -47 & -76 \\ -56 & -34 & -47 \end{array} \right) B − 1 = ( − 75 ) 1 ⎝ ⎛ − 50 − 73 − 56 − 25 − 47 − 34 − 50 − 76 − 47 ⎠ ⎞ B − 1 = ( 2 3 1 3 2 3 73 75 47 75 76 75 56 75 34 75 47 75 ) . B^{-1} = \left( \begin{array}{ccc} \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{73}{75} & \frac{47}{75} & \frac{76}{75} \\ \frac{56}{75} & \frac{34}{75} & \frac{47}{75} \end{array} \right). B − 1 = ⎝ ⎛ 3 2 75 73 75 56 3 1 75 47 75 34 3 2 75 76 75 47 ⎠ ⎞ .
Answer:
a) ( 8 − 5 11 23 ) − 1 = ( 23 239 5 239 − 11 239 8 239 ) \left( \begin{array}{cc} 8 & -5 \\ 11 & 23 \end{array} \right)^{-1} = \left( \begin{array}{cc} \frac{23}{239} & \frac{5}{239} \\ \frac{-11}{239} & \frac{8}{239} \end{array} \right) ( 8 11 − 5 23 ) − 1 = ( 239 23 239 − 11 239 5 239 8 )
b) ( 5 − 7 6 − 11 6 2 2 4 − 7 ) − 1 = ( 2 3 1 3 2 3 73 75 47 75 76 75 56 75 34 75 47 75 ) \left( \begin{array}{ccc} 5 & -7 & 6 \\ -11 & 6 & 2 \\ 2 & 4 & -7 \end{array} \right)^{-1} = \left( \begin{array}{ccc} \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{73}{75} & \frac{47}{75} & \frac{76}{75} \\ \frac{56}{75} & \frac{34}{75} & \frac{47}{75} \end{array} \right) ⎝ ⎛ 5 − 11 2 − 7 6 4 6 2 − 7 ⎠ ⎞ − 1 = ⎝ ⎛ 3 2 75 73 75 56 3 1 75 47 75 34 3 2 75 76 75 47 ⎠ ⎞
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#340153 on Dec 2023