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3x33 matrix with determinant = a11

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Calculate the determinant of the matrix.

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\\(|A|=\\) [[0]]

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Calculate the inverse matrix and input the entries correct to two decimal places.

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\\(A^{-1}=\\) [[1]]

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Hence solve the following system of equations.

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\\(\\begin{pmatrix} \\var{a11}&\\var{a12}&\\var{a13}\\\\ \\var{a21}&\\var{a22}&\\var{23}\\\\ \\var{a31}&\\var{a32}&\\var{a33}\\end{pmatrix}\\begin{pmatrix} x\\\\y\\\\z\\end{pmatrix}=\\begin{pmatrix} \\var{r}\\\\\\var{s}\\\\\\var{t}\\end{pmatrix}\\)

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\\(x=\\) [[2]]

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\\(y=\\) [[3]]

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\\(z=\\) [[4]]

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Given the matrix:

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\\(\\mathbf{A}=\\begin{pmatrix} \\var{a11}&\\var{a12}&\\var{a13}\\\\ \\var{a21}&\\var{a22}&\\var{a23}\\\\\\var{a31}&\\var{a32}&\\var{a33} \\end{pmatrix}\\)

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Given the matrix:

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\\(\\mathbf{A}=\\begin{pmatrix} {a11}&{a12}&{a13}\\\\ {a21}&{a22}&{a23}\\\\{a31}&{a32}&{a33} \\end{pmatrix}\\)

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The determinant of a 3x3 matrix is determined by the formula:

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\\(|\\mathbf{A}|={a11}*\\begin{vmatrix}{a22}&{a23}\\\\{a32}&{a33}\\end{vmatrix}-{a12}*\\begin{vmatrix}{a21}&{a23}\\\\{a31}&{a33}\\end{vmatrix}+{a13}*\\begin{vmatrix}{a21}&{a22}\\\\{a31}&{a32}\\end{vmatrix}\\)

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So in this example:

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\\(\\mathbf{A}=\\begin{pmatrix} \\var{a11}&\\var{a12}&\\var{a13}\\\\ \\var{a21}&\\var{a22}&\\var{a23}\\\\\\var{a31}&\\var{a32}&\\var{a33} \\end{pmatrix}\\)

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\\(|\\mathbf{A}|=\\var{a11}*\\begin{vmatrix}\\var{a22}&\\var{a23}\\\\\\var{a32}&\\var{a33}\\end{vmatrix}-\\var{a12}*\\begin{vmatrix}\\var{a21}&\\var{a23}\\\\\\var{a31}&\\var{a33}\\end{vmatrix}+\\var{a13}*\\begin{vmatrix}\\var{a21}&\\var{a22}\\\\\\var{a31}&\\var{a32}\\end{vmatrix}\\)

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\\(|\\mathbf{A}|=\\var{a11}*(\\var{a22}*\\var{a33}-\\var{a32}*\\var{a23})-\\var{a12}*(\\var{a21}*\\var{a33}-\\var{a31}*\\var{a23})+\\var{a13}*(\\var{a21}*\\var{a32}-\\var{a31}*\\var{a22})\\)

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\\(|\\mathbf{A}|=(\\simplify{{a11}*({a22}*{a33}-{a32}*{a23})})-(\\simplify{{a12}*({a21}*{a33}-{a31}*{a23})})+(\\simplify{{a13}*({a21}*{a32}-{a31}*{a22})})\\)

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\\(|\\mathbf{A}|=\\var{a11}\\)

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The adjoint of the matrix is given by:

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\\(\\mathbf{A}^*=\\begin{pmatrix}\\begin{vmatrix}\\var{a22}&\\var{a23}\\\\\\var{a32}&\\var{a33}\\end{vmatrix}&-\\begin{vmatrix}\\var{a21}&\\var{a23}\\\\\\var{a31}&\\var{a33}\\end{vmatrix}&\\begin{vmatrix}\\var{a21}&\\var{a22}\\\\\\var{a31}&\\var{a32}\\end{vmatrix}\\\\-\\begin{vmatrix}\\var{a12}&\\var{a13}\\\\\\var{a32}&\\var{a33}\\end{vmatrix}&\\begin{vmatrix}\\var{a11}&\\var{a13}\\\\\\var{a31}&\\var{a33}\\end{vmatrix}&-\\begin{vmatrix}\\var{a11}&\\var{a12}\\\\\\var{a31}&\\var{a32}\\end{vmatrix}\\\\\\begin{vmatrix}\\var{a12}&\\var{a13}\\\\\\var{a22}&\\var{a23}\\end{vmatrix}&-\\begin{vmatrix}\\var{a11}&\\var{a13}\\\\\\var{a21}&\\var{a23}\\end{vmatrix}&\\begin{vmatrix}\\var{a11}&\\var{a12}\\\\\\var{a21}&\\var{a11}\\end{vmatrix}\\end{pmatrix}^t\\)

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\\(\\mathbf{A}^*=\\begin{pmatrix}\\simplify{({a22}*{a33}-{a32}*{a23})}&\\simplify{({a23}*{a31}-{a21}*{a33})}&\\simplify{({a21}*{a32}-{a22}*{a31})}\\\\\\simplify{(-{a12}*{a33}+{a32}*{a13})}&\\simplify{({a11}*{a33}-{a13}*{a31})}&\\simplify{({a12}*{a31}-{a32}*{a11})}\\\\\\simplify{({a12}*{a23}-{a22}*{a13})}&\\simplify{({a21}*{a13}-{a11}*{a23})}&\\simplify{({a22}*{a11}-{a12}*{a21})}\\end{pmatrix}^t\\)

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\\(\\mathbf{A}^*=\\begin{pmatrix}\\simplify{({a22}*{a33}-{a32}*{a23})}&\\simplify{(-{a12}*{a33}+{a32}*{a13})}&\\simplify{({a12}*{a23}-{a22}*{a13})}\\\\\\simplify{({a23}*{a31}-{a21}*{a33})}&\\simplify{({a11}*{a33}-{a13}*{a31})}&\\simplify{({a21}*{a13}-{a11}*{a23})}\\\\\\simplify{({a21}*{a32}-{a22}*{a31})}&\\simplify{({a12}*{a31}-{a32}*{a11})}&\\simplify{({a22}*{a11}-{a12}*{a21})}\\end{pmatrix}\\)

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The inverse matrix is defined by \\(A^{-1}=\\frac{1}{|A|}A^*\\)

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\\(A^{-1}=\\frac{1}{\\var{a11}}\\begin{pmatrix}\\simplify{({a22}*{a33}-{a32}*{a23})}&\\simplify{(-{a12}*{a33}+{a32}*{a13})}&\\simplify{({a12}*{a23}-{a22}*{a13})}\\\\\\simplify{({a23}*{a31}-{a21}*{a33})}&\\simplify{({a11}*{a33}-{a13}*{a31})}&\\simplify{({a21}*{a13}-{a11}*{a23})}\\\\\\simplify{({a21}*{a32}-{a22}*{a31})}&\\simplify{({a12}*{a31}-{a32}*{a11})}&\\simplify{({a22}*{a11}-{a12}*{a21})}\\end{pmatrix}\\)

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Using the inverse matrix method gives:

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\\(\\begin{pmatrix} x\\\\y\\\\z\\end{pmatrix}=\\frac{1}{\\var{a11}}\\begin{pmatrix}\\simplify{({a22}*{a33}-{a32}*{a23})}&\\simplify{(-{a12}*{a33}+{a32}*{a13})}&\\simplify{({a12}*{a23}-{a22}*{a13})}\\\\\\simplify{({a23}*{a31}-{a21}*{a33})}&\\simplify{({a11}*{a33}-{a13}*{a31})}&\\simplify{({a21}*{a13}-{a11}*{a23})}\\\\\\simplify{({a21}*{a32}-{a22}*{a31})}&\\simplify{({a12}*{a31}-{a32}*{a11})}&\\simplify{({a22}*{a11}-{a12}*{a21})}\\end{pmatrix}\\begin{pmatrix} \\var{r}\\\\\\var{s}\\\\\\var{t}\\end{pmatrix}\\)

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\\(\\begin{pmatrix} x\\\\y\\\\z\\end{pmatrix}=\\begin{pmatrix}\\var{b11}&\\var{b12}&\\var{b13}\\\\\\var{b21}&\\var{b22}&\\var{b23}\\\\\\var{b31}&\\var{b32}&\\var{b33}\\end{pmatrix}\\begin{pmatrix} \\var{r}\\\\\\var{s}\\\\\\var{t}\\end{pmatrix}\\)

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\\(\\begin{pmatrix} x\\\\y\\\\z\\end{pmatrix}=\\begin{pmatrix}(\\var{b11})*\\var{r}+(\\var{b12})*\\var{s}+(\\var{b13})*\\var{t}\\\\(\\var{b21})*\\var{r}+(\\var{b22})*\\var{s}+(\\var{b23})*\\var{t}\\\\(\\var{b31})*\\var{r}+(\\var{b32})*\\var{s}+(\\var{b33})*\\var{t}\\end{pmatrix}\\)

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\\(\\begin{pmatrix} x\\\\y\\\\z\\end{pmatrix}=\\begin{pmatrix} \\var{r1}\\\\\\var{s1}\\\\\\var{t1}\\end{pmatrix}\\)

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This question tests learner's knowledge of the inverse matrix method for a 3x3 matrix.

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