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Find the value of the determinant using the method of expansion by minors; expand on the third row

Find the value of the determinant using the method of expansion by minors; expand-example-1
User James Baxter
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1 Answer

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22 votes

For the matrix


\begin{bmatrix}{a_(11)} & {a_(12)} & {a_(13)} \\ {a_(21)} & {a_(22)} & {a_(23)} \\ {a_(31)} & {a_(32)} & {a_(33)}\end{bmatrix}

the determinant using the method of expansion by minors, expanding on the third row is:


\det \begin{bmatrix}{a_(11)} & {a_(12)} & {a_(13)} \\ {a_(21)} & {a_(22)} & {a_(23)} \\ {a_(31)} & {a_(32)} & {a_(33)}\end{bmatrix}=a_(31)\det \begin{bmatrix}{a_(12)} & {a_(13)} & {} \\ {a_(22)} & {a_(23)} & {} \\ {} & {} & {}\end{bmatrix}-a_(32)\det \begin{bmatrix}{a_(11)} & {a_(13)} & {} \\ {a_(21)} & {a_(23)} & {} \\ {} & {} & {}\end{bmatrix}+a_(33)\det \begin{bmatrix}{a_(12)} & {a_(12)} & {} \\ {a_(22)} & {a_(22)} & {} \\ {} & {} & {}\end{bmatrix}

Answer:

First, we compute the determinants of the minors:


\begin{gathered} \det \begin{bmatrix}{0} & {4} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=0+4=4, \\ \det \begin{bmatrix}{1} & {4} & {} \\ {1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=3-4=-1, \\ \det \begin{bmatrix}{1} & {0} & {} \\ {1} & {-1} & {} \\ {} & {} & {}\end{bmatrix}=-1-0=-1. \end{gathered}

Therefore:


\det \begin{bmatrix}{1} & {0} & {4} \\ {1} & {-1} & {3} \\ {0} & {5} & {-2}\end{bmatrix}=0*4-5*(-1)+(-2)*(-1)=5+2=7.

User Amit Parashar
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