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boya's Question
Math
Posted 2 months ago

Consider the following matrix AA, and let B=A1B=A^{-1}. Find b31,b32b_{31}, b_{32}, and b33b_{33}. (i.e., find the entries in the third row of A1A^{-1} ).
A=[112410012]
A=\left[\begin{array}{ccc}
1 & 1 & -2 \\
-4 & -1 & 0 \\
0 & -1 & 2
\end{array}\right]
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
To find the inverse of matrix AA, we first need to write down the matrix AA: A=(112410012) A = \begin{pmatrix} 1 & 1 & -2 \\ -4 & -1 & 0 \\ 0 & -1 & 2 \end{pmatrix}
step 2
Calculate the determinant of matrix AA. The determinant of a 3x3 matrix AA is given by: det(A)=a(eifh)b(difg)+c(dheg) \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) For our matrix AA: det(A)=1((1)(2)(0)(1))1((4)(2)(0)(0))+(2)((4)(1)(1)(0)) \text{det}(A) = 1((-1)(2) - (0)(-1)) - 1((-4)(2) - (0)(0)) + (-2)((-4)(-1) - (-1)(0)) det(A)=1(2)1(8)+(2)(4) \text{det}(A) = 1(-2) - 1(-8) + (-2)(4) det(A)=2+88=2 \text{det}(A) = -2 + 8 - 8 = -2
step 3
Find the matrix of minors for AA. The minor of an element is the determinant of the 2x2 matrix that remains after removing the row and column of that element: Minors(A)=(det(1012)det(4002)det(4101)det(1212)det(1202)det(1101)det(1240)det(1140)det(1141)) \text{Minors}(A) = \begin{pmatrix} \text{det}\begin{pmatrix} -1 & 0 \\ -1 & 2 \end{pmatrix} & \text{det}\begin{pmatrix} -4 & 0 \\ 0 & 2 \end{pmatrix} & \text{det}\begin{pmatrix} -4 & -1 \\ 0 & -1 \end{pmatrix} \\ \text{det}\begin{pmatrix} 1 & -2 \\ -1 & 2 \end{pmatrix} & \text{det}\begin{pmatrix} 1 & -2 \\ 0 & 2 \end{pmatrix} & \text{det}\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} \\ \text{det}\begin{pmatrix} 1 & -2 \\ -4 & 0 \end{pmatrix} & \text{det}\begin{pmatrix} 1 & 1 \\ -4 & 0 \end{pmatrix} & \text{det}\begin{pmatrix} 1 & 1 \\ -4 & -1 \end{pmatrix} \end{pmatrix} =((1)(2)(0)(1)(4)(2)(0)(0)(4)(1)(1)(0)(1)(2)(2)(1)(1)(2)(2)(0)(1)(1)(1)(0)(1)(0)(2)(4)(1)(0)(1)(4)(1)(1)(1)(4)) = \begin{pmatrix} (-1)(2) - (0)(-1) & (-4)(2) - (0)(0) & (-4)(-1) - (-1)(0) \\ (1)(2) - (-2)(-1) & (1)(2) - (-2)(0) & (1)(-1) - (1)(0) \\ (1)(0) - (-2)(-4) & (1)(0) - (1)(-4) & (1)(-1) - (1)(-4) \end{pmatrix} =(2842221080+41+4) = \begin{pmatrix} -2 & -8 & 4 \\ 2 - 2 & 2 & -1 \\ 0 - 8 & 0 + 4 & -1 + 4 \end{pmatrix} =(284021843) = \begin{pmatrix} -2 & -8 & 4 \\ 0 & 2 & -1 \\ -8 & 4 & 3 \end{pmatrix}
step 4
Find the matrix of cofactors by applying the checkerboard pattern of signs to the matrix of minors: Cofactors(A)=(284021843) \text{Cofactors}(A) = \begin{pmatrix} -2 & 8 & 4 \\ 0 & 2 & 1 \\ -8 & -4 & 3 \end{pmatrix}
step 5
Transpose the matrix of cofactors to get the adjugate matrix: Adj(A)=(208824413) \text{Adj}(A) = \begin{pmatrix} -2 & 0 & -8 \\ 8 & 2 & -4 \\ 4 & 1 & 3 \end{pmatrix}
step 6
Finally, divide the adjugate matrix by the determinant to get the inverse of AA: A1=1det(A)Adj(A)=12(208824413) A^{-1} = \frac{1}{\text{det}(A)} \text{Adj}(A) = \frac{1}{-2} \begin{pmatrix} -2 & 0 & -8 \\ 8 & 2 & -4 \\ 4 & 1 & 3 \end{pmatrix} =(10441221232) = \begin{pmatrix} 1 & 0 & 4 \\ -4 & -1 & 2 \\ -2 & -\frac{1}{2} & -\frac{3}{2} \end{pmatrix}
step 7
The entries in the third row of A1A^{-1} are: b31=2,b32=12,b33=32 b_{31} = -2, \quad b_{32} = -\frac{1}{2}, \quad b_{33} = -\frac{3}{2}
Answer
The entries in the third row of A1A^{-1} are b31=2b_{31} = -2, b32=12b_{32} = -\frac{1}{2}, and b33=32b_{33} = -\frac{3}{2}.
Key Concept
Matrix Inversion
Explanation
To find the inverse of a 3x3 matrix, we use the determinant and the adjugate matrix. The inverse is given by the adjugate matrix divided by the determinant.

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