Question

inverse of the matrix​

Answer 1

$\sf\large\red{\underline{\underline{Question}}}:$

$\textsf{Find the inverse of matrix A} = \left[\begin{array}{ccc}1&3&3\\1&4&3\\1&3&4\end{array}\right]$

$\sf\large\red{\underline{\underline{Solution}}}:$

$\red\bigstar\;\boxed{\sf A^{-1}=\dfrac{adj. A}{|A|}}$

We know that,

adj. A is transpose of Co-factor matrix.

Co-factor :

$\longmapsto \sf\pink{A_{11}} =\left|\begin{array}{cc}4&3\\3&4\end{array}\right| = 16-9=7$

$\longmapsto \sf\pink{A_{12}} =-\left|\begin{array}{cc}1&3\\1&4\end{array}\right| = -(4-3)= -1$

$\longmapsto \sf\pink{A_{13}} =\left|\begin{array}{cc}1&4\\1&3\end{array}\right| = 3-4=-1$

Similarly,

• A₂₁ = -3 ; A₂₂ = 1 ; A₂₃ = 0
• A₃₁ = -3 ; A₃₂ = 0 ; A₃₃ = 1

Therefore,

$\sf \dag \;\green{Co-factor \;matrix} =\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]$

So, $\sf \pink{adj.A} = \left[\begin{array}{ccc}7&-3&-3\\-1&1&0\\-1&0&1\end{array}\right]$

Now,

Determinant :

|A| = $\dashrightarrow \left|\begin{array}{ccc}1&3&3\\1&4&3\\1&3&4\end{array}\right| = 1(16-9)-3(4-3)+3(3-4)$

$\dashrightarrow\left|\begin{array}{ccc}1&3&3\\1&4&3\\1&3&4\end{array}\right| = 7-3-3 = 1$

Inverse of matrix A :

$\red\bigstar\;\boxed{\sf A^{-1}=\dfrac{adj. A}{|A|}}$

$\longmapsto \sf A^{-1} = \dfrac{ \left[\begin{array}{ccc}7&-3&-3\\-1&1&0\\-1&0&1\end{array}\right]}{1}$

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

Therefore,

Inverse of matrix A $\sf = \left[\begin{array}{ccc}7&-3&-3\\-1&1&0\\-1&0&1\end{array}\right]$