Compute $A - 4I_3$ and $(4I_3)A$, where $A = \begin{bmatrix} 4 & -2 & 4 \ -4 & 4 & -7 \ -4 & 1 & 1 \end{bmatrix}$ \newline $A - 4I_3 = $
Added by Larry R.
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Let's denote the column vector as v: v = \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} Now, let's find the product Av: A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} Av = \begin{bmatrix} a_{11} & Show more…
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