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Problem #37 from pg. 61 and problem #1 from pg. 68 are graded..

#3,pg.100: Let

$A = \left[ \begin{array}{ll} 2 & -5 \\ 3 & -2 \end{array} \right]$ . Compute

$3I_2-A$ and

$(3I_2)A$ .
Solution: First we compute

$\begin{array}{ll} 3I_2-A &= \left[ \begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array} \right] - \left[ \begin{array}{ll} 2 & -5 \\ 3 & -2 \end{array} \right] \\ &= \left[ \begin{array}{ll} 1 & 5 \\ -3 & 5 \end{array} \right]. \end{array}$ Now compute

$\begin{array}{ll} (3I_2)A &= \left[ \begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array} \right] \left[ \begin{array}{ll} 2 & -5 \\ 3 & -2 \end{array} \right] \\ &= \left[ \begin{array}{ll} 6 & -15 \\ 9 & -6 \end{array} \right] \end{array}$ #18,pg.110: Solve the eqution

$AB=BC$ for

$A$ , assuming that

$A,B,C$ are square and

$B$ is invertible.
Solution: From the equation

$AB=BC$ and the fact that

$B$ is invertible, we will multiply the equaton on the right by

$B^{-1}$ to get

$ABB^{-1} = BCB^{-1}.$ By the definition of inverse, we know that

$ABB^{-1}=AI=A$ and so we have shown that

$A = BCB^{-1}.$ (note: we cannot go further and say that

$BCB^{-1}=CBB^{-1}=C$ because that would assume that

$CB^{-1}=B^{-1}C$ which is not true in general!)