Answer on Question #82906 - Math - Linear Algebra
To solve an equation of the form A b ˙ = u ⃗ \mathbf{A}\dot{\boldsymbol{b}} = \vec{\boldsymbol{u}} A b ˙ = u A \mathbf{A} A [ 2 1 − 1 1 − 2 2 3 − 1 3 ] \begin{bmatrix} 2 & 1 & -1 \\ 1 & -2 & 2 \\ 3 & -1 & 3 \end{bmatrix} ⎣ ⎡ 2 1 3 1 − 2 − 1 − 1 2 3 ⎦ ⎤
b ˙ \dot{\pmb{b}} b b ˙ [ x y z ] \left[ \begin{array}{l}x\\ y\\ z \end{array} \right] ⎣ ⎡ x y z ⎦ ⎤ u ⃗ \vec{u} u [ 11 − 2 5 ] \left[ \begin{array}{l}11\\ -2\\ 5 \end{array} \right] ⎣ ⎡ 11 − 2 5 ⎦ ⎤
we multiply both sides of the equation (from the right side) by A − 1 \mathbf{A}^{-1} A − 1 A A − 1 = A − 1 A = I \mathbf{A A}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I} AA − 1 = A − 1 A = I A − 1 = 1 ∣ A ∣ A d j ( A ) \mathbf{A}^{-1} = \frac{1}{|A|} Adj(A) A − 1 = ∣ A ∣ 1 A d j ( A )
where ∣ A ∣ |A| ∣ A ∣ A A A
∣ A ∣ = 2 ( ∣ − 2 2 ∣ ∣ − 1 3 ∣ ) − ( ∣ 1 2 ∣ ∣ 3 3 ∣ ) − ( ∣ 1 − 2 ∣ ∣ 3 − 1 ∣ ) = − 10 | A | = 2 \left( \begin{array}{cc} | - 2 & 2 | \\ | - 1 & 3 | \end{array} \right) - \left( \begin{array}{cc} | 1 & 2 | \\ | 3 & 3 | \end{array} \right) - \left( \begin{array}{cc} | 1 & - 2 | \\ | 3 & - 1 | \end{array} \right) = - 1 0 ∣ A ∣ = 2 ( ∣ − 2 ∣ − 1 2∣ 3∣ ) − ( ∣1 ∣3 2∣ 3∣ ) − ( ∣1 ∣3 − 2∣ − 1∣ ) = − 10
Adj(A) is adjugate of A and it's obtained as follows :
Step 1. Matrix of minors of A
[ ∣ − 2 2 ∣ ∣ − 1 3 ∣ ] [ 1 2 ∣ 3 3 ∣ ] [ 1 − 2 ∣ 3 − 1 ∣ ] = [ − 4 − 3 5 2 9 − 5 0 5 − 5 ] \left[ \begin{array}{cc} | - 2 & 2 | \\ | - 1 & 3 | \end{array} \right] \left[ \begin{array}{cc} 1 & 2 | \\ 3 & 3 | \end{array} \right] \left[ \begin{array}{cc} 1 & - 2 | \\ 3 & - 1 | \end{array} \right] = \left[ \begin{array}{cc} - 4 & - 3 & 5 \\ 2 & 9 & - 5 \\ 0 & 5 & - 5 \end{array} \right] [ ∣ − 2 ∣ − 1 2∣ 3∣ ] [ 1 3 2∣ 3∣ ] [ 1 3 − 2∣ − 1∣ ] = ⎣ ⎡ − 4 2 0 − 3 9 5 5 − 5 − 5 ⎦ ⎤
Step 2. Change signs (multiply position-wise!)
[ − 4 − 3 5 2 9 − 5 0 5 − 5 ] [ + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 ] = [ − 4 3 5 − 2 9 5 0 − 5 − 5 ] \left[ \begin{array}{ccc} - 4 & - 3 & 5 \\ 2 & 9 & - 5 \\ 0 & 5 & - 5 \end{array} \right] \left[ \begin{array}{ccc} + 1 & - 1 & + 1 \\ - 1 & + 1 & - 1 \\ + 1 & - 1 & + 1 \end{array} \right] = \left[ \begin{array}{ccc} - 4 & 3 & 5 \\ - 2 & 9 & 5 \\ 0 & - 5 & - 5 \end{array} \right] ⎣ ⎡ − 4 2 0 − 3 9 5 5 − 5 − 5 ⎦ ⎤ ⎣ ⎡ + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 ⎦ ⎤ = ⎣ ⎡ − 4 − 2 0 3 9 − 5 5 5 − 5 ⎦ ⎤
Step 3. Transpose
[ − 4 3 5 − 2 9 5 0 − 5 − 5 ] T = [ − 4 − 2 0 3 9 − 5 5 5 − 5 ] \left[ \begin{array}{ccc} - 4 & 3 & 5 \\ - 2 & 9 & 5 \\ 0 & - 5 & - 5 \end{array} \right] ^ {T} = \left[ \begin{array}{ccc} - 4 & - 2 & 0 \\ 3 & 9 & - 5 \\ 5 & 5 & - 5 \end{array} \right] ⎣ ⎡ − 4 − 2 0 3 9 − 5 5 5 − 5 ⎦ ⎤ T = ⎣ ⎡ − 4 3 5 − 2 9 5 0 − 5 − 5 ⎦ ⎤ A − 1 = 1 − 10 [ − 4 − 2 0 3 9 − 5 5 5 − 5 ] = [ 0.4 0.2 0 − 0.3 − 0.9 . 5 − 0.5 − 0.5 0.5 ] \mathbf {A} ^ {- 1} = \frac {1}{- 1 0} \left[ \begin{array}{c c c} - 4 & - 2 & 0 \\ 3 & 9 & - 5 \\ 5 & 5 & - 5 \end{array} \right] = \left[ \begin{array}{c c c} 0. 4 & 0. 2 & 0 \\ - 0. 3 & - 0. 9 & . 5 \\ - 0. 5 & - 0. 5 & 0. 5 \end{array} \right] A − 1 = − 10 1 ⎣ ⎡ − 4 3 5 − 2 9 5 0 − 5 − 5 ⎦ ⎤ = ⎣ ⎡ 0.4 − 0.3 − 0.5 0.2 − 0.9 − 0.5 0 .5 0.5 ⎦ ⎤ A − 1 A b ⃗ = A − 1 u ⃗ \mathbf {A} ^ {- 1} \mathbf {A} \vec {\boldsymbol {b}} = \mathbf {A} ^ {- 1} \vec {\boldsymbol {u}} A − 1 A b = A − 1 u [ 0.4 0.2 0 − 0.3 − 0.9 . 5 − 0.5 − 0.5 0.5 ] [ 2 1 − 1 1 − 2 2 3 − 1 3 ] [ x y z ] = [ 0.4 0.2 0 − 0.3 − 0.9 . 5 − 0.5 − 0.5 0.5 ] [ 11 − 2 5 ] \left[ \begin{array}{c c c} 0. 4 & 0. 2 & 0 \\ - 0. 3 & - 0. 9 & . 5 \\ - 0. 5 & - 0. 5 & 0. 5 \end{array} \right] \left[ \begin{array}{c c c} 2 & 1 & - 1 \\ 1 & - 2 & 2 \\ 3 & - 1 & 3 \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{c c c} 0. 4 & 0. 2 & 0 \\ - 0. 3 & - 0. 9 & . 5 \\ - 0. 5 & - 0. 5 & 0. 5 \end{array} \right] \left[ \begin{array}{c} 1 1 \\ - 2 \\ 5 \end{array} \right] ⎣ ⎡ 0.4 − 0.3 − 0.5 0.2 − 0.9 − 0.5 0 .5 0.5 ⎦ ⎤ ⎣ ⎡ 2 1 3 1 − 2 − 1 − 1 2 3 ⎦ ⎤ ⎣ ⎡ x y z ⎦ ⎤ = ⎣ ⎡ 0.4 − 0.3 − 0.5 0.2 − 0.9 − 0.5 0 .5 0.5 ⎦ ⎤ ⎣ ⎡ 11 − 2 5 ⎦ ⎤
by applying inverse matrix we obtain solution : [ 4 1 − 2 ] \begin{bmatrix} 4 \\ 1 \\ -2 \end{bmatrix} ⎣ ⎡ 4 1 − 2 ⎦ ⎤
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#339914 on Dec 2023