Question

#QualityQuestion
@Matrix

Determine the inverse of each of the matrixes, if it exists.
${\begin{bmatrix} 2 & -3 & 3\\ 2 & 2 & 3 \\ 3& -2 & -2 \end{bmatrix} }$
​

Answer 1

Steps by Sarru's rule:

$\tt\left(\begin{matrix} 2 & -3 & 3 \\2 & 2 & 3 \\3 & -2 & -2 \end{matrix}\right)$

Find the determinant of the matrix using the method of diagonals.

$⇉det(\left(\begin{matrix}2&-3&3\\2&2&3\\3&-2&-2\end{matrix}\right))$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$⇉\left(\begin{matrix}2&-3&3&2&-3\\2&2&3&2&2\\3&-2&-2&3&-2\end{matrix}\right)$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$⇉ \tt \:2\times 2\left(-2\right)-3\times 3\times 3+3\times 2\left(-2\right)=-47$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$⇉\sf \: 3\times 2\times 3-2\times 3\times 2-2\times 2\left(-3\right)=18$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

⇉-47-18

⇉-65

_____________________

$\bold\blue{Celestial}\bold\green{Centrix}$

Answer 2

Question: Determine the inverse of each of the matrixes, if it exists.

a11 = 2, a12 = -3, a33 = 3

a21 = 2, a22 = 2, a23 = 3

a31 = 3, a32 = -2, a33 = -2

Answer:

As determinant ≠ 0, inverse exists.

$\frac{ 1}{65}{\begin{bmatrix} - 2& 12& 15 \\ - 13& 13 & 0\\ 10& 5 & - 10\end{bmatrix} }$

Step-by-step explanation:

$\sf Using, \boxed{A {}^{ - 1} =\frac{1}{ |A| }(adj A)}$

$\sf adj A _{ } = {\begin{bmatrix} A _{11 } & A _{ 12} & A _{ 13} \\A _{21} & A _{22 } & A _{ 23}\\A _{ 31} &A _{ 32} & A _{ 33} \end{bmatrix} } {}^{ {}^{ {}^{ {}^{ {}^{ {}^{ ^{t} } } } } } }$

Solving for adj A :

$A _{11} = {\begin{vmatrix} a _{22 } & a _{ 23} \\ a_{ 32} & a _{ 33} \end{vmatrix} } ={\begin{vmatrix}2& 3 \\- 2&- 2 \end{vmatrix} }= 2 \\ A _{12} = - {\begin{vmatrix} a _{21 } & a _{ 23} \\ a_{ 31} & a _{ 33} \end{vmatrix} } = - {\begin{vmatrix}2& 3 \\3&- 2 \end{vmatrix} }= 13 \\ A _{13} = {\begin{vmatrix} a _{21} & a _{ 22} \\ a_{ 31} & a _{ 32} \end{vmatrix} }={\begin{vmatrix}2& 2 \\3&- 2 \end{vmatrix} } = - 10 \\ A _{21} = - {\begin{vmatrix} a _{12} & a _{13}\\ a_{ 32} & a _{ 33} \end{vmatrix} } = - {\begin{vmatrix} - 3& 3 \\- 2&- 2 \end{vmatrix} } = - 12 \\ A _{22} = {\begin{vmatrix} a _{11 } & a _{13} \\ a_{ 31} & a _{ 33} \end{vmatrix} } ={\begin{vmatrix}2& 3 \\3&- 2 \end{vmatrix} } = - 13 \\ A _{23} = - {\begin{vmatrix} a _{11} & a _{ 12} \\ a_{ 31} & a _{ 32} \end{vmatrix} }= - {\begin{vmatrix}2& - 3 \\3&- 2 \end{vmatrix} } = - 5 \\ A _{31} = {\begin{vmatrix} a _{12} & a _{13} \\ a_{ 22} & a _{ 23} \end{vmatrix} } ={\begin{vmatrix} - 3& 3 \\2&3 \end{vmatrix} } = - 15 \\ A _{32} = - {\begin{vmatrix} a _{11} & a _{13} \\ a_{ 21} & a _{ 23} \end{vmatrix} } = - {\begin{vmatrix} 2& 3 \\2&3 \end{vmatrix} } = 0 \\ A _{33} = {\begin{vmatrix} a _{11} & a _{12} \\ a_{ 21} & a _{ 22} \end{vmatrix} } ={\begin{vmatrix} 2& - 3 \\2&2\end{vmatrix} } = 10$

$\sf Therefore,$

$adj A _{ } = {\begin{bmatrix} A _{11 } & A _{ 12} & A _{ 13} \\A _{21} & A _{22 } & A _{ 23}\\A _{ 31} &A _{ 32} & A _{ 33} \end{bmatrix} } {}^{ {}^{ {}^{ {}^{ {}^{ {}^{ ^{t} } } } } } } \\ \\adj A _{ } = {\begin{bmatrix} 2& 13& - 10 \\ - 12& - 13 & - 5\\ - 15 &0 & 10\end{bmatrix} } {}^{ {}^{ {}^{ {}^{ {}^{ {}^{ ^{t} } } } } } } \\ \\adj A _{ } = {\begin{bmatrix} 2& -12& - 15 \\ 13& - 13 & 0\\ - 10& - 5 & 10\end{bmatrix} }$

$\sf Moreover,$

$|A|=2(-4+6)- (-3)(-4-9)+3(-4 -6) =-65$ ≠ 0

$\sf Hence,$

$A {}^{ - 1} = \frac{1}{ |A| } (adj A) \\ \\ A {}^{ - 1} = - \frac{ 1}{65}{\begin{bmatrix} 2& -12& - 15 \\ 13& - 13 & 0\\ - 10& - 5 & 10\end{bmatrix} } \\ \\ A {}^{ - 1} = \frac{ 1}{65}{\begin{bmatrix} - 2& 12& 15 \\ - 13& 13 & 0\\ 10& 5 & - 10\end{bmatrix} }$