Consider the following hypothesis test: H_0: μ ≤ 3432 H_a: μ > 3432 A sample of 96 provided a sample mean x = 3648 and sample standard deviation…

The cofactor $c_{ij}$cij​ of an element $a_{ij}$aij​ in a matrix is given by:

$$c_{ij} = (-1)^{i+j} \cdot M_{ij}$$cij​=(−1)i+j⋅Mij​

where $M_{ij}$Mij​ is the minor — the determinant of the smaller matrix obtained after deleting row $i$i and column $j$j.

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Given matrix:

$$A = \begin{bmatrix} 1 & 2 & -1 \\ 0 & -3 & 2 \\ 4 & 2 & 3 \end{bmatrix}$$A=​104​2−32​−123​​

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To find $c_{21}$c21​, delete Row 2 and Column 1:

$$M_{21} = \begin{vmatrix} 2 & -1 \\ 2 & 3 \end{vmatrix}$$M21​=​22​−13​​

$$M_{21} = (2)(3) - (-1)(2) = 6 + 2 = 8$$M21​=(2)(3)−(−1)(2)=6+2=8

$$c_{21} = (-1)^{2+1} \cdot 8 = (-1)^3 \cdot 8 = -8$$c21​=(−1)2+1⋅8=(−1)3⋅8=−8

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To find $c_{33}$c33​, delete Row 3 and Column 3:

$$M_{33} = \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix}$$M33​=​10​2−3​​

$$M_{33} = (1)(-3) - (2)(0) = -3 - 0 = -3$$M33​=(1)(−3)−(2)(0)=−3−0=−3

$$c_{33} = (-1)^{3+3} \cdot (-3) = (-1)^6 \cdot (-3) = -3$$c33​=(−1)3+3⋅(−3)=(−1)6⋅(−3)=−3

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$$c_{21} \cdot c_{33} = (-8) \times (-3) = \boxed{24}$$c21​⋅c33​=(−8)×(−3)=24​