Given that $\mathbf{A}=\left(\begin{array}{ccc}1 & 1 & 2 \ 4 & -1 & -3 \ 4 & 0 & 2\end{array}\right)$ and $\mathbf{B}=\left(\begin{array}{ccc}1 & 1 & -3 \ 2 & -4 & 0 \ -1 & -2 & 3\end{array}\right)$, let $((\mathbf{A}-\mathbf{B})-\mathbf{B})-\mathbf{B}=\mathbf{C}$,
where $\mathbf{C}=\left(\begin{array}{lll}c_{11} & c_{12} & c_{13} \ c_{21} & c_{22} & c_{23} \ c_{31} & c_{32} & c_{33}\end{array}\right)$. What is $c_{12}$ ?

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The Answer of Given that $\mathbf{A}=\left(\begin{array}{ccc}1 & 1 & 2 \ 4 & -1 & -3 \ 4 & 0 & 2\end{array}\right)$ and $\mathbf{B}=\left(\begin{array}{ccc}1 & 1 & -3 \ 2 & -4 & 0 \ -1 & -2 & 3\end{array}\right)$, let $((\mathbf{A}-\mathbf{B})-\mathbf{B})-\mathbf{B}=\mathbf{C}$,
where $\mathbf{C}=\left(\begin{array}{lll}c_{11} & c_{12} & c_{13} \ c_{21} & c_{22} & c_{23} \ c_{31} & c_{32} & c_{33}\end{array}\right)$. What is $c_{12}$ ?

Given That A=(1124−1−3402)\mathbf{A}=\left(\begin{array}{ccc}1 & 1 & 2 \\ 4 & -1 & -3 \\ 4 & 0 & 2\end{array}\right)A=​144​1−10​2−32​​

Given That $\mathbf{A}=\left(\begin{array}{ccc}1 & 1 & 2 \\ 4 & -1 & -3 \\ 4 & 0 & 2\end{array}\right)$