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Consider the matrix A =(1 1 1 3 4 3 3 3 4) Find the determinant |A| and the inverse matrix A^-1.

User Hampus
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1 Answer

3 votes

Answer:


A)\,\,det(A)=1


B)\,\,A^(-1)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]

Explanation:


det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|

Expanding with first row


det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1

To find inverse we first find cofactor matrix


C_(1,1)=(-1)^(1+1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|=7\\\\C_(1,2)=(-1)^(1+2)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|=-3\\\\C_(1,3)=(-1)^(1+3)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|=-3\\\\C_(2,1)=(-1)^(2+1)\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=-1\\\\C_(2,2)=(-1)^(2+2)\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\C_(2,3)=(-1)^(2+3)\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\


C_(3,1)=(-1)^(3+1)\left\Big|\begin{array}{cc}1&1\\4&3\end{array}\right\Big|=-1\\\\C_(3,2)=(-1)^(3+2)\left\Big|\begin{array}{cc}1&1\\3&3\end{array}\right\Big|=0\\\\\\C_(3,3)=(-1)^(3+3)\left\Big|\begin{array}{cc}1&1\\3&4\end{array}\right\Big|=1\\\\

Cofactor matrix is


C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^(T)\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^(-1)=(adj(A))/(det(A))\\\\A^(-1)=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^(-1)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]

User Henry Vonfire
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