Determine the product −444−7185−3−11−111−2−2213 and use it to solve the system ofequations x−y+z=4,x−2y−2z=9,2x+y+3z=1

let A=−444−7135−31 and B=1−111−2−2213A⋅B=−444−7135−311−111−2−2213=−4+4+84−8+4−4−8+12−7+1+67−2+3−7−2+95−3−2−5+6−15+6−3AB=800080008=8100010001⇒18AB=I3∵AB=BI3⇒18AB⋅B−1=I3B−1⇒18A=B−1⇒B−1=18A ⇒B−1=18−444−7135−31…(i)Matrix from of equations1−111−2−2213xyz=491BX=C⇒X=B−1C⇒xyz=18−444−7135−31491⇒xyz=18−16+36+4−25+9+320−27−1=1824−16−8=3−2−1 Hence, x=3, y=−2, z=-1

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Determine the product $[\begin{array}{ccc} - 4 & 4 & 4 \\ - 7 & 1 & 8 \\ 5 & - 3 & - 1 \end{array}] [\begin{array}{ccc} 1 & - 1 & 1 \\ 1 & - 2 & - 2 \\ 2 & 1 & 3 \end{array}]$ and use it to solve the system of
equations $x - y + z = 4, x - 2 y - 2 z = 9, 2 x + y + 3 z = 1$

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Detailed Solution

let $A = [\begin{array}{ccc} - 4 & 4 & 4 \\ - 7 & 1 & 3 \\ 5 & - 3 & 1 \end{array}] and B = [\begin{array}{ccc} 1 & - 1 & 1 \\ 1 & - 2 & - 2 \\ 2 & 1 & 3 \end{array}]$
$\begin{array}{l} A \cdot B = [\begin{array}{ccc} - 4 & 4 & 4 \\ - 7 & 1 & 3 \\ 5 & - 3 & 1 \end{array}] [\begin{array}{lcc} 1 & - 1 & 1 \\ 1 & - 2 & - 2 \\ 2 & 1 & 3 \end{array}] \\ = [\begin{array}{ccc} - 4 + 4 + 8 & 4 - 8 + 4 & - 4 - 8 + 12 \\ - 7 + 1 + 6 & 7 - 2 + 3 & - 7 - 2 + 9 \\ 5 - 3 - 2 & - 5 + 6 - 1 & 5 + 6 - 3 \end{array}] \\ AB = [\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}] = 8 [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow (\frac{1}{8} A) B = I_{3} (∵ AB = {BI}_{3}) \\ \Rightarrow (\frac{1}{8} A) B \cdot B^{- 1} = I_{3} B^{- 1} \\ \Rightarrow \frac{1}{8} A = B^{- 1} \Rightarrow B^{- 1} = \frac{1}{8} A \end{array}$
$\Rightarrow B^{- 1} = \frac{1}{8} [\begin{array}{ccc} - 4 & 4 & 4 \\ - 7 & 1 & 3 \\ 5 & - 3 & 1 \end{array}] \dots (i)$
Matrix from of equations
$[\begin{array}{ccc} 1 & - 1 & 1 \\ 1 & - 2 & - 2 \\ 2 & 1 & 3 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 4 \\ 9 \\ 1 \end{array}]$
$\begin{array}{l} BX = C \\ \Rightarrow X = B^{- 1} C \end{array}$
$\begin{array}{l} \Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{8} [\begin{array}{ccc} - 4 & 4 & 4 \\ - 7 & 1 & 3 \\ 5 & - 3 & 1 \end{array}] [\begin{matrix} 4 \\ 9 \\ 1 \end{matrix}] \\ \Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{8} [\begin{array}{c} - 16 + 36 + 4 \\ - 25 + 9 + 3 \\ 20 - 27 - 1 \end{array}] \\ = \frac{1}{8} [\begin{array}{c} 24 \\ - 16 \\ - 8 \end{array}] = [\begin{array}{c} 3 \\ - 2 \\ - 1 \end{array}] \end{array}$
$Hence, x = 3, y = - 2, z = - 1$