Answer on Question #85224 – Math – Linear Algebra
Question
Q2. If A = ( 1 2 6 4 11 7 9 13 3 ) A = \begin{pmatrix} 1 & 2 & 6 \\ 4 & 11 & 7 \\ 9 & 13 & 3 \end{pmatrix} A = ⎝ ⎛ 1 4 9 2 11 13 6 7 3 ⎠ ⎞
(a) Find the minors of 1, 2 and 6.
(b) Find the cofactors of 1, 2 and 6.
(c) Evaluate ∣ A ∣ |A| ∣ A ∣
(d) A − 1 A^{-1} A − 1
Solution
A = ( 1 2 6 4 11 7 9 13 3 ) A = \begin{pmatrix} 1 & 2 & 6 \\ 4 & 11 & 7 \\ 9 & 13 & 3 \end{pmatrix} A = ⎝ ⎛ 1 4 9 2 11 13 6 7 3 ⎠ ⎞
(a) Find the minors of 1, 2 and 6.
Minor of 1 is M 11 = ∣ 11 7 13 3 ∣ = 11 ⋅ 3 − 7 ⋅ 13 = − 58 M_{11} = \begin{vmatrix} 11 & 7 \\ 13 & 3 \end{vmatrix} = 11 \cdot 3 - 7 \cdot 13 = -58 M 11 = ∣ ∣ 11 13 7 3 ∣ ∣ = 11 ⋅ 3 − 7 ⋅ 13 = − 58
Minor of 2 is M 12 = ∣ 4 7 9 3 ∣ = 4 ⋅ 3 − 7 ⋅ 9 = − 51 M_{12} = \begin{vmatrix} 4 & 7 \\ 9 & 3 \end{vmatrix} = 4 \cdot 3 - 7 \cdot 9 = -51 M 12 = ∣ ∣ 4 9 7 3 ∣ ∣ = 4 ⋅ 3 − 7 ⋅ 9 = − 51
Minor of 6 is M 13 = ∣ 4 11 9 13 ∣ = 4 ⋅ 13 − 11 ⋅ 9 = − 47 M_{13} = \begin{vmatrix} 4 & 11 \\ 9 & 13 \end{vmatrix} = 4 \cdot 13 - 11 \cdot 9 = -47 M 13 = ∣ ∣ 4 9 11 13 ∣ ∣ = 4 ⋅ 13 − 11 ⋅ 9 = − 47
(b) Find the cofactors of 1, 2 and 6.
Cofactor of a i j = ( − 1 ) i + j M i j a_{ij} = (-1)^{i+j} M_{ij} a ij = ( − 1 ) i + j M ij
C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = − 58 C_{11} = (-1)^{1+1} M_{11} = M_{11} = -58 C 11 = ( − 1 ) 1 + 1 M 11 = M 11 = − 58 C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = 51 C_{12} = (-1)^{1+2} M_{12} = -M_{12} = 51 C 12 = ( − 1 ) 1 + 2 M 12 = − M 12 = 51 C 13 = ( − 1 ) 1 + 3 M 13 = M 13 = − 47 C_{13} = (-1)^{1+3} M_{13} = M_{13} = -47 C 13 = ( − 1 ) 1 + 3 M 13 = M 13 = − 47
(c) Evaluate ∣ A ∣ |A| ∣ A ∣
∣ A ∣ = a 11 C 11 + a 12 C 12 + a 13 C 13 |A| = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13} ∣ A ∣ = a 11 C 11 + a 12 C 12 + a 13 C 13 ∣ A ∣ = 1 ⋅ ( − 58 ) + 2 ⋅ 51 + 6 ⋅ ( − 47 ) = − 238 |A| = 1 \cdot (-58) + 2 \cdot 51 + 6 \cdot (-47) = -238 ∣ A ∣ = 1 ⋅ ( − 58 ) + 2 ⋅ 51 + 6 ⋅ ( − 47 ) = − 238
(d) A − 1 A^{-1} A − 1
We can find inverse matrix by using formula
A − 1 = 1 ∣ A ∣ C T A^{-1} = \frac{1}{|A|} C^T A − 1 = ∣ A ∣ 1 C T
where C C C
C = ( C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 ) C = \left( \begin{array}{ccc} C _ {1 1} & C _ {1 2} & C _ {1 3} \\ C _ {2 1} & C _ {2 2} & C _ {2 3} \\ C _ {3 1} & C _ {3 2} & C _ {3 3} \end{array} \right) C = ⎝ ⎛ C 11 C 21 C 31 C 12 C 22 C 32 C 13 C 23 C 33 ⎠ ⎞
Find cofactors of all the elements
C 21 = ( − 1 ) 2 + 1 ∣ 2 6 13 3 ∣ = − ( 2 ⋅ 3 − 6 ⋅ 13 ) = 72 C _ {2 1} = (- 1) ^ {2 + 1} \left| \begin{array}{cc} 2 & 6 \\ 13 & 3 \end{array} \right| = -(2 \cdot 3 - 6 \cdot 13) = 72 C 21 = ( − 1 ) 2 + 1 ∣ ∣ 2 13 6 3 ∣ ∣ = − ( 2 ⋅ 3 − 6 ⋅ 13 ) = 72 C 22 = ( − 1 ) 2 + 2 ∣ 1 6 9 3 ∣ = ( 1 ⋅ 3 − 6 ⋅ 9 ) = − 51 C _ {2 2} = (- 1) ^ {2 + 2} \left| \begin{array}{cc} 1 & 6 \\ 9 & 3 \end{array} \right| = (1 \cdot 3 - 6 \cdot 9) = - 51 C 22 = ( − 1 ) 2 + 2 ∣ ∣ 1 9 6 3 ∣ ∣ = ( 1 ⋅ 3 − 6 ⋅ 9 ) = − 51 C 23 = ( − 1 ) 2 + 3 ∣ 1 2 9 13 ∣ = − ( 13 − 2 ⋅ 9 ) = 5 C _ {2 3} = (- 1) ^ {2 + 3} \left| \begin{array}{cc} 1 & 2 \\ 9 & 13 \end{array} \right| = -(13 - 2 \cdot 9) = 5 C 23 = ( − 1 ) 2 + 3 ∣ ∣ 1 9 2 13 ∣ ∣ = − ( 13 − 2 ⋅ 9 ) = 5 C 31 = ( − 1 ) 3 + 1 ∣ 2 6 11 7 ∣ = ( 2 ⋅ 7 − 6 ⋅ 11 ) = − 52 C _ {3 1} = (- 1) ^ {3 + 1} \left| \begin{array}{cc} 2 & 6 \\ 11 & 7 \end{array} \right| = (2 \cdot 7 - 6 \cdot 11) = - 52 C 31 = ( − 1 ) 3 + 1 ∣ ∣ 2 11 6 7 ∣ ∣ = ( 2 ⋅ 7 − 6 ⋅ 11 ) = − 52 C 32 = ( − 1 ) 3 + 2 ∣ 1 6 4 7 ∣ = − ( 7 − 6 ⋅ 4 ) = 17 C _ {3 2} = (- 1) ^ {3 + 2} \left| \begin{array}{cc} 1 & 6 \\ 4 & 7 \end{array} \right| = -(7 - 6 \cdot 4) = 17 C 32 = ( − 1 ) 3 + 2 ∣ ∣ 1 4 6 7 ∣ ∣ = − ( 7 − 6 ⋅ 4 ) = 17 C 33 = ( − 1 ) 3 + 3 ∣ 1 2 4 11 ∣ = ( 11 − 2 ⋅ 4 ) = 3 C _ {3 3} = (- 1) ^ {3 + 3} \left| \begin{array}{cc} 1 & 2 \\ 4 & 11 \end{array} \right| = (11 - 2 \cdot 4) = 3 C 33 = ( − 1 ) 3 + 3 ∣ ∣ 1 4 2 11 ∣ ∣ = ( 11 − 2 ⋅ 4 ) = 3
Construct Cofactor Matrix
C = ( C 11 C 12 C 13 C 21 C 22 C 23 C 31 C 32 C 33 ) = ( − 58 51 − 47 72 − 51 5 − 52 17 3 ) C = \left( \begin{array}{ccc} C _ {1 1} & C _ {1 2} & C _ {1 3} \\ C _ {2 1} & C _ {2 2} & C _ {2 3} \\ C _ {3 1} & C _ {3 2} & C _ {3 3} \end{array} \right) = \left( \begin{array}{ccc} - 5 8 & 5 1 & - 4 7 \\ 7 2 & - 5 1 & 5 \\ - 5 2 & 1 7 & 3 \end{array} \right) C = ⎝ ⎛ C 11 C 21 C 31 C 12 C 22 C 32 C 13 C 23 C 33 ⎠ ⎞ = ⎝ ⎛ − 58 72 − 52 51 − 51 17 − 47 5 3 ⎠ ⎞
Transpose of the cofactor matrix (adjugate matrix)
C T = ( − 58 72 − 52 51 − 51 17 − 47 5 3 ) C ^ {T} = \left( \begin{array}{ccc} - 5 8 & 7 2 & - 5 2 \\ 5 1 & - 5 1 & 1 7 \\ - 4 7 & 5 & 3 \end{array} \right) C T = ⎝ ⎛ − 58 51 − 47 72 − 51 5 − 52 17 3 ⎠ ⎞ Thus A − 1 = − 1 238 ( − 58 72 − 52 51 − 51 17 − 47 5 3 ) = ( 58 / 238 − 72 / 238 52 / 238 − 51 / 238 51 / 238 − 17 / 238 47 / 238 − 5 / 238 − 3 / 238 ) \text{Thus } A ^ {- 1} = \frac {- 1}{238} \left( \begin{array}{ccc} - 5 8 & 7 2 & - 5 2 \\ 5 1 & - 5 1 & 1 7 \\ - 4 7 & 5 & 3 \end{array} \right) = \left( \begin{array}{ccc} 5 8 / 238 & - 7 2 / 238 & 5 2 / 238 \\ - 5 1 / 238 & 5 1 / 238 & - 1 7 / 238 \\ 4 7 / 238 & - 5 / 238 & - 3 / 238 \end{array} \right) Thus A − 1 = 238 − 1 ⎝ ⎛ − 58 51 − 47 72 − 51 5 − 52 17 3 ⎠ ⎞ = ⎝ ⎛ 58/238 − 51/238 47/238 − 72/238 51/238 − 5/238 52/238 − 17/238 − 3/238 ⎠ ⎞
**Answer:**
(a) The minors of 1,2 and 6 are M 11 = − 58 M_{11} = -58 M 11 = − 58 M 12 = − 51 M_{12} = -51 M 12 = − 51 M 13 = − 47 M_{13} = -47 M 13 = − 47
(b) The cofactors of 1,2 and 6 are C 11 = − 58 C_{11} = -58 C 11 = − 58 C 12 = 51 C_{12} = 51 C 12 = 51 C 13 = − 47 C_{13} = -47 C 13 = − 47
(c) ∣ A ∣ = − 238 |A| = -238 ∣ A ∣ = − 238
(d) A − 1 = ( 58 / 238 − 72 / 238 52 / 238 − 51 / 238 51 / 238 − 17 / 238 47 / 238 − 5 / 238 − 3 / 238 ) . A^{-1} = \left( \begin{array}{ccc}58 / 238 & -72 / 238 & 52 / 238 \\ -51 / 238 & 51 / 238 & -17 / 238 \\ 47 / 238 & -5 / 238 & -3 / 238 \end{array} \right). A − 1 = ⎝ ⎛ 58/238 − 51/238 47/238 − 72/238 51/238 − 5/238 52/238 − 17/238 − 3/238 ⎠ ⎞ .
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#340153 on Dec 2023