Consider the following matrix $A$, and let $B=A^{-1}$. Find $b_{31}, b_{32}$, and $b_{33}$. (i.e., find the entries in the third row of $A^{-1}$ ).
$$
A=\left[\begin{array}{ccc}
1 & 1 & -2 \
-4 & -1 & 0 \
0 & -1 & 2
\end{array}\right]
$$

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the inverse of matrix $A$, we first need to write down the matrix $A$: $$ A = \begin{pmatrix} 1 & 1 & -2 \ -4 & -1 & 0 \ 0 & -1 & 2 \end{pmatrix} $$ ‖ ‖ step 2 ⋮ Calculate the determinant of matrix $A$. The determinant of a 3x3 matrix $A$ is given by: $$ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) $$ For our matrix $A$: $$ \text{det}(A) = 1((-1)(2) - (0)(-1)) - 1((-4)(2) - (0)(0)) + (-2)((-4)(-1) - (-1)(0)) $$ $$ \text{det}(A) = 1(-2) - 1(-8) + (-2)(4) $$ $$ \text{det}(A) = -2 + 8 - 8 = -2 $$ ‖ ‖ step 3 ⋮ Find the matrix of minors for $A$. The minor of an element is the determinant of the 2x2 matrix that remains after removing the row and column of that element: $$ \text{Minors}(A) = \begin{pmatrix} \text{det}\begin{pmatrix} -1 & 0 \ -1 & 2 \end{pmatrix} & \text{det}\begin{pmatrix} -4 & 0 \ 0 & 2 \end{pmatrix} & \text{det}\begin{pmatrix} -4 & -1 \ 0 & -1 \end{pmatrix} \ \text{det}\begin{pmatrix} 1 & -2 \ -1 & 2 \end{pmatrix} & \text{det}\begin{pmatrix} 1 & -2 \ 0 & 2 \end{pmatrix} & \text{det}\begin{pmatrix} 1 & 1 \ 0 & -1 \end{pmatrix} \ \text{det}\begin{pmatrix} 1 & -2 \ -4 & 0 \end{pmatrix} & \text{det}\begin{pmatrix} 1 & 1 \ -4 & 0 \end{pmatrix} & \text{det}\begin{pmatrix} 1 & 1 \ -4 & -1 \end{pmatrix} \end{pmatrix} $$ $$ = \begin{pmatrix} (-1)(2) - (0)(-1) & (-4)(2) - (0)(0) & (-4)(-1) - (-1)(0) \ (1)(2) - (-2)(-1) & (1)(2) - (-2)(0) & (1)(-1) - (1)(0) \ (1)(0) - (-2)(-4) & (1)(0) - (1)(-4) & (1)(-1) - (1)(-4) \end{pmatrix} $$ $$ = \begin{pmatrix} -2 & -8 & 4 \ 2 - 2 & 2 & -1 \ 0 - 8 & 0 + 4 & -1 + 4 \end{pmatrix} $$ $$ = \begin{pmatrix} -2 & -8 & 4 \ 0 & 2 & -1 \ -8 & 4 & 3 \end{pmatrix} $$ ‖ ‖ step 4 ⋮ Find the matrix of cofactors by applying the checkerboard pattern of signs to the matrix of minors: $$ \text{Cofactors}(A) = \begin{pmatrix} -2 & 8 & 4 \ 0 & 2 & 1 \ -8 & -4 & 3 \end{pmatrix} $$ ‖ ‖ step 5 ⋮ Transpose the matrix of cofactors to get the adjugate matrix: $$ \text{Adj}(A) = \begin{pmatrix} -2 & 0 & -8 \ 8 & 2 & -4 \ 4 & 1 & 3 \end{pmatrix} $$ ‖ ‖ step 6 ⋮ Finally, divide the adjugate matrix by the determinant to get the inverse of $A$: $$ A^{-1} = \frac{1}{\text{det}(A)} \text{Adj}(A) = \frac{1}{-2} \begin{pmatrix} -2 & 0 & -8 \ 8 & 2 & -4 \ 4 & 1 & 3 \end{pmatrix} $$ $$ = \begin{pmatrix} 1 & 0 & 4 \ -4 & -1 & 2 \ -2 & -\frac{1}{2} & -\frac{3}{2} \end{pmatrix} $$ ‖ ‖ step 7 ⋮ The entries in the third row of $A^{-1}$ are: $$ b_{31} = -2, \quad b_{32} = -\frac{1}{2}, \quad b_{33} = -\frac{3}{2} $$ ‖ ∻Answer∻ ⚹ The entries in the third row of $A^{-1}$ are $b_{31} = -2$, $b_{32} = -\frac{1}{2}$, and $b_{33} = -\frac{3}{2}$. ⚹ ∻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.⚹◊What is the formula to find the determinant of a 3x3 matrix $A$?⍭ Generate me a similar question◊

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