Question

2x-y+3z=1 X+2y-z=2 5y-5z=3

Answer 1

Answer:

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Step-by-step explanation:

Matrix Method

Solution.

The given system of linear equations is

\begin{lgathered}\color{red}{\begin{array}{cccccccc}2x&-&y&+&3z&=&1&\quad...(1)\\ x&+&2y&-&z&=&2&\quad...(2)\\ 0.x&+&5y&-&5z&=&3&\quad...(3)\end{array}}\end{lgathered}2xx0.x−++y2y5y+−−3zz5z===123...(1)...(2)...(3)

The system of equations can be put in matrix form,

\quad\quad AX=BAX=B ,

where \begin{lgathered}A=\left[\begin{array}{ccc}2&-1&3\\1&2&-1\\0&5&-5\end{array}\right]\end{lgathered}A=⎣⎡210−1253−1−5⎦⎤ ,

\begin{lgathered}X=\left[\begin{array}{c}x\\y\\z\end{array}\right]\end{lgathered}X=⎣⎡xyz⎦⎤ and \begin{lgathered}B=\left[\begin{array}{c}1\\2\\3\end{array}\right]\end{lgathered}B=⎣⎡123⎦⎤

Now, \begin{lgathered}|A|=\left|\begin{array}{ccc}2&-1&3\\1&2&-1\\0&5&-5\end{array}\right|\end{lgathered}∣A∣=∣∣∣∣∣∣210−1253−1−5∣∣∣∣∣∣

\quad=2(-10+5)+1(-5-0)+3(5-0)=2(−10+5)+1(−5−0)+3(5−0)

\quad ( Expanding along R_{1}R1 )

\quad=2(-5)+1(-5)+3(5)=2(−5)+1(−5)+3(5)

\quad=-10-5+15=−10−5+15

\quad=0=0

Now, \begin{lgathered}Adj\:A=\left[\begin{array}{ccc}-5&5&5\\10&-10&-10\\-5&5&5\end{array}\right]^{T}\end{lgathered}AdjA=⎣⎡−510−55−1055−105⎦⎤T

\begin{lgathered}=\left[\begin{array}{ccc}-5&10&-5\\5&-10&5\\5&-10&5\end{array}\right]\end{lgathered}=⎣⎡−55510−10−10−555⎦⎤

\begin{lgathered}\therefore (Adj\:A)B=\left[\begin{array}{ccc}-5&10&-5\\5&-10&5\\5&-10&5\end{array}\right]\left[\begin{array}{c}1\\2\\3\end{array}\right]\end{lgathered}∴(AdjA)B=⎣⎡−55510−10−10−555⎦⎤⎣⎡123⎦⎤

\begin{lgathered}=\left[\begin{array}{c}-5+20-15\\5-20+15\\5-20+15\end{array}\right]\end{lgathered}=⎣⎡−5+20−155−20+155−20+15⎦⎤

\begin{lgathered}=\left[\begin{array}{c}0\\0\\0\end{array}\right]\end{lgathered}=⎣⎡000<