Question

Find inverse of a matrix using partition method
1 3 3
1 4 3
1 3 4

Answer 1

Step-by-step explanation:

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Answer 2

Final answer:

    $\left[\begin{array}{ccc}7&-3&-3\\-1&1&0\\-1&0&1\end{array}\right]$

Given that: We are given

$\left[\begin{array}{ccc}1&3&3\\1&4&3\\1&3&4\end{array}\right]$

To find: We have to find inverse of given matrix using partition method.

Explanation:

• The given matrix    

  $P = \left[\begin{array}{ccc}1&3&3\\1&4&3\\1&3&4\end{array}\right]$  

• To find inverse of a matrix using partition method first the given matrix covert in the form of,

$P=\left[\begin{array}{ccc}A&B\\C&d\end{array}\right]$    

• The inverse of the matrix,

$P = P^{-1} = \left[\begin{array}{ccc}X&Y\\Z&t\end{array}\right]$

Where, t = (d –CA⁻¹B)⁻¹

Y= - A⁻¹Bt

Z= - CA⁻¹t

X = A⁻¹ (I –BZ)

• Here,

$A=\left[\begin{array}{ccc}1&3\\1&4\end{array}\right] B = \left[\begin{array}{ccc}3\\3\end{array}\right] C = \left[\begin{array}{ccc}1&3\end{array}\right] d = \left[\begin{array}{ccc}4\end{array}\right]$

• First find the inverse of A by classical adjoint method.

           [ for a matrix $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ the inverse is $\left[\begin{array}{ccc}d&-b\\-c& a\end{array}\right]$]

$A^{-1} =\left[\begin{array}{ccc}4&-3\\-1&1\end{array}\right]$

• Now we can find the inverse of of the given matrix P

• t = (d –CA⁻¹B)⁻¹

$t = ( \left[\begin{array}{ccc}4\end{array}\right] - \left[\begin{array}{ccc}1&3\end{array}\right] \left[\begin{array}{ccc}4&-3\\-1&1\end{array}\right] \left[\begin{array}{ccc}3\\3\end{array}\right] )^{-1}$

$t = ( \left[\begin{array}{ccc}4\end{array}\right] - \left[\begin{array}{ccc}1&0\end{array}\right]\left[\begin{array}{ccc}3\\3\end{array}\right] )^{-1}$

t = (4-3)⁻¹ = 1

• Y= - A⁻¹Bt

$Y = -(\left[\begin{array}{ccc}4&-3\\-1&1\end{array}\right] \left[\begin{array}{ccc}3\\3\end{array}\right] )^{-1} 1$

$Y = - ( \left[\begin{array}{ccc}3\\0\end{array}\right]) = \left[\begin{array}{ccc}-3\\0\end{array}\right]$

• Z= -C A⁻¹t

$Z = -\left[\begin{array}{ccc}1&3\end{array}\right]\left[\begin{array}{ccc}4&-3\\-1&1\end{array}\right] 1$

$Z = -\left[\begin{array}{ccc}1&0\end{array}\right]=\left[\begin{array}{ccc}-1&0\end{array}\right]$

• X = A⁻¹ (I –BZ)

$X =\left[\begin{array}{ccc}4&-3\\-1&1\end{array}\right](\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]-\left[\begin{array}{ccc}3\\3\end{array}\right]\left[\begin{array}{ccc}-1&0\end{array}\right])$

$X =\left[\begin{array}{ccc}4&-3\\-1&1\end{array}\right](\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]-\left[\begin{array}{ccc}-3&0\\-3&0\end{array}\right])$

$X =\left[\begin{array}{ccc}4&-3\\-1&1\end{array}\right]\left[\begin{array}{ccc}4&0\\3&1\end{array}\right]$

$X =\left[\begin{array}{ccc}7&-3\\-1&1\end{array}\right]$

• Substitute the value of X,Y,Z,t in $\left[\begin{array}{ccc}X&Y\\Z&t\end{array}\right]$

• $P^{-1} = \left[\begin{array}{ccc}7&-3&-3\\-1&1&0\\-1&0&1\end{array}\right]$

To know more about the concept please go through the links

https://brainly.in/question/13357933

https://brainly.in/question/12357495

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