\displaystyle \text{Solve the following system of homogeneous linear equations by the matrix method:}
\displaystyle \textbf{Question 1: } \qquad \begin{aligned}2x-y+z&=0\\3x+2y-z&=0\\x+4y+3z&=0\end{aligned}
\displaystyle \text{Answer:}
\displaystyle \text{The given system of homogeneous linear equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{or, }AX=O,
\displaystyle \text{where }A=\begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }O=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&-1&1\\3&2&-1\\1&4&3\end{vmatrix}
\displaystyle =2\begin{vmatrix}2&-1\\4&3\end{vmatrix}+\begin{vmatrix}3&-1\\1&3\end{vmatrix}+\begin{vmatrix}3&2\\1&4\end{vmatrix}
\displaystyle =2(6+4)+(9+1)+(12-2)=40\ne0
\displaystyle \therefore A\text{ is non-singular and }A^{-1}\text{ exists.}
\displaystyle AX=O
\displaystyle \Rightarrow A^{-1}(AX)=A^{-1}O
\displaystyle \Rightarrow (A^{-1}A)X=O
\displaystyle \Rightarrow I_3X=O
\displaystyle \Rightarrow X=O
\displaystyle \therefore x=y=z=0
\displaystyle \text{Hence, the given system has only the trivial solution }x=y=z=0.
\displaystyle \\

\displaystyle \textbf{Question 2: } \qquad \begin{aligned}2x-y+2z&=0\\5x+3y-z&=0\\x+5y-5z&=0\end{aligned}
\displaystyle \text{Answer:}
\displaystyle \text{The given system of homogeneous linear equations is}
\displaystyle \begin{aligned}2x-y+2z&=0\qquad\ldots(1)\\5x+3y-z&=0\qquad\ldots(2)\\x+5y-5z&=0\qquad\ldots(3)\end{aligned}
\displaystyle \text{It can be written in matrix form as}
\displaystyle \begin{bmatrix}2&-1&2\\5&3&-1\\1&5&-5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{or, }AX=O,
\displaystyle \text{where }A=\begin{bmatrix}2&-1&2\\5&3&-1\\1&5&-5\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }O=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&-1&2\\5&3&-1\\1&5&-5\end{vmatrix}
\displaystyle =2(-15+5)+(-25+1)+2(25-3)
\displaystyle =-20-24+44=0
\displaystyle \therefore |A|=0
\displaystyle \text{Hence, }A\text{ is singular and the given homogeneous system has infinitely many solutions.}
\displaystyle \text{Put }z=k,\text{ where }k\text{ is any real number.}
\displaystyle \text{Then, from equations (1) and (2), we get}
\displaystyle 2x-y=-2k\qquad\ldots(4)
\displaystyle 5x+3y=k\qquad\ldots(5)
\displaystyle \text{Equations (4) and (5) can be written as}
\displaystyle \begin{bmatrix}2&-1\\5&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2k\\k\end{bmatrix}
\displaystyle \text{or, }A_1X_1=B,
\displaystyle \text{where }A_1=\begin{bmatrix}2&-1\\5&3\end{bmatrix},\quad X_1=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}-2k\\k\end{bmatrix}
\displaystyle |A_1|=\begin{vmatrix}2&-1\\5&3\end{vmatrix}=6+5=11\ne0
\displaystyle \therefore A_1^{-1}\text{ exists.}
\displaystyle \text{Now, }\text{adj }A_1=\begin{bmatrix}3&1\\-5&2\end{bmatrix}
\displaystyle \therefore A_1^{-1}=\frac{1}{|A_1|}\text{adj }A_1=\frac1{11}\begin{bmatrix}3&1\\-5&2\end{bmatrix}
\displaystyle X_1=A_1^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\end{bmatrix}=\frac1{11}\begin{bmatrix}3&1\\-5&2\end{bmatrix}\begin{bmatrix}-2k\\k\end{bmatrix}
\displaystyle =\frac1{11}\begin{bmatrix}-6k+k\\10k+2k\end{bmatrix}=\begin{bmatrix}-\frac{5k}{11}\\\frac{12k}{11}\end{bmatrix}
\displaystyle \therefore x=-\frac{5k}{11},\qquad y=\frac{12k}{11},\qquad z=k,
\displaystyle \text{where }k\text{ is any real number.}
\displaystyle \text{These values also satisfy equation (3).}
\displaystyle \text{Hence, }x=-\frac{5k}{11},\ y=\frac{12k}{11}\text{ and }z=k,\text{ where }k\in\mathbb{R},\text{ are the solutions.}
\displaystyle \\

\displaystyle \textbf{Question 3: } \qquad \begin{aligned}3x-y+2z&=0\\4x+3y+3z&=0\\5x+7y+4z&=0\end{aligned}
\displaystyle \text{Answer:}
\displaystyle \text{The given system of homogeneous linear equations is}
\displaystyle \begin{aligned}3x-y+2z&=0\qquad\ldots(1)\\4x+3y+3z&=0\qquad\ldots(2)\\5x+7y+4z&=0\qquad\ldots(3)\end{aligned}
\displaystyle \text{It can be written in matrix form as}
\displaystyle \begin{bmatrix}3&-1&2\\4&3&3\\5&7&4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{or, }AX=O,
\displaystyle \text{where }A=\begin{bmatrix}3&-1&2\\4&3&3\\5&7&4\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }O=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&-1&2\\4&3&3\\5&7&4\end{vmatrix}
\displaystyle =3(12-21)+(16-15)+2(28-15)
\displaystyle =-27+1+26=0
\displaystyle \therefore |A|=0
\displaystyle \text{Hence, }A\text{ is singular and the given homogeneous system has infinitely many solutions.}
\displaystyle \text{Put }z=k,\text{ where }k\text{ is any real number.}
\displaystyle \text{Then, from equations (1) and (2), we get}
\displaystyle 3x-y=-2k\qquad\ldots(4)
\displaystyle 4x+3y=-3k\qquad\ldots(5)
\displaystyle \text{Equations (4) and (5) can be written as}
\displaystyle \begin{bmatrix}3&-1\\4&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2k\\-3k\end{bmatrix}
\displaystyle \text{or, }A_1X_1=B,
\displaystyle \text{where }A_1=\begin{bmatrix}3&-1\\4&3\end{bmatrix},\quad X_1=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}-2k\\-3k\end{bmatrix}
\displaystyle |A_1|=\begin{vmatrix}3&-1\\4&3\end{vmatrix}=9+4=13\ne0
\displaystyle \therefore A_1^{-1}\text{ exists.}
\displaystyle \text{Now, }\text{adj }A_1=\begin{bmatrix}3&1\\-4&3\end{bmatrix}
\displaystyle \therefore A_1^{-1}=\frac{1}{|A_1|}\text{adj }A_1=\frac1{13}\begin{bmatrix}3&1\\-4&3\end{bmatrix}
\displaystyle X_1=A_1^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\end{bmatrix}=\frac1{13}\begin{bmatrix}3&1\\-4&3\end{bmatrix}\begin{bmatrix}-2k\\-3k\end{bmatrix}
\displaystyle =\frac1{13}\begin{bmatrix}-6k-3k\\8k-9k\end{bmatrix}=\begin{bmatrix}-\frac{9k}{13}\\-\frac{k}{13}\end{bmatrix}
\displaystyle \therefore x=-\frac{9k}{13},\qquad y=-\frac{k}{13},\qquad z=k,
\displaystyle \text{where }k\text{ is any real number.}
\displaystyle \text{These values also satisfy equation (3).}
\displaystyle \text{Hence, }x=-\frac{9k}{13},\ y=-\frac{k}{13}\text{ and }z=k,\text{ where }k\in\mathbb{R},\text{ are the solutions.}
\displaystyle \\

\displaystyle \textbf{Question 4: } \qquad \begin{aligned}x+y-6z&=0\\x-y+2z&=0\\-3x+y+2z&=0\end{aligned}
\displaystyle \text{Answer:}
\displaystyle \text{The given system of homogeneous linear equations is}
\displaystyle \begin{aligned}x+y-6z&=0\qquad\ldots(1)\\x-y+2z&=0\qquad\ldots(2)\\-3x+y+2z&=0\qquad\ldots(3)\end{aligned}
\displaystyle \text{It can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&-6\\1&-1&2\\-3&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{or, }AX=O,
\displaystyle \text{where }A=\begin{bmatrix}1&1&-6\\1&-1&2\\-3&1&2\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }O=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&-6\\1&-1&2\\-3&1&2\end{vmatrix}
\displaystyle =1(-2-2)-(2+6)-6(1-3)
\displaystyle =-4-8+12=0
\displaystyle \therefore |A|=0
\displaystyle \text{Hence, }A\text{ is singular and the given homogeneous system has infinitely many solutions.}
\displaystyle \text{Put }z=k,\text{ where }k\text{ is any real number.}
\displaystyle \text{Then, from equations (1) and (2), we get}
\displaystyle x+y=6k\qquad\ldots(4)
\displaystyle x-y=-2k\qquad\ldots(5)
\displaystyle \text{Equations (4) and (5) can be written as}
\displaystyle \begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}6k\\-2k\end{bmatrix}
\displaystyle \text{or, }A_1X_1=B,
\displaystyle \text{where }A_1=\begin{bmatrix}1&1\\1&-1\end{bmatrix},\quad X_1=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}6k\\-2k\end{bmatrix}
\displaystyle |A_1|=\begin{vmatrix}1&1\\1&-1\end{vmatrix}=-1-1=-2\ne0
\displaystyle \therefore A_1^{-1}\text{ exists.}
\displaystyle \text{Now, }\text{adj }A_1=\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}
\displaystyle \therefore A_1^{-1}=\frac{1}{|A_1|}\text{adj }A_1=-\frac12\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}
\displaystyle X_1=A_1^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\end{bmatrix}=-\frac12\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}\begin{bmatrix}6k\\-2k\end{bmatrix}
\displaystyle =-\frac12\begin{bmatrix}-6k+2k\\-6k-2k\end{bmatrix}=\begin{bmatrix}2k\\4k\end{bmatrix}
\displaystyle \therefore x=2k,\qquad y=4k,\qquad z=k,
\displaystyle \text{where }k\text{ is any real number.}
\displaystyle \text{These values also satisfy equation (3).}
\displaystyle \text{Hence, }x=2k,\ y=4k\text{ and }z=k,\text{ where }k\in\mathbb{R},\text{ are the solutions.}
\displaystyle \\

\displaystyle \textbf{Question 5: } \qquad \begin{aligned}x+y+z&=0\\x-y-5z&=0\\x+2y+4z&=0\end{aligned}
\displaystyle \text{Answer:}
\displaystyle \text{The given system of homogeneous linear equations is}
\displaystyle \begin{aligned}x+y+z&=0\qquad\ldots(1)\\x-y-5z&=0\qquad\ldots(2)\\x+2y+4z&=0\qquad\ldots(3)\end{aligned}
\displaystyle \text{It can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&1\\1&-1&-5\\1&2&4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{or, }AX=O,
\displaystyle \text{where }A=\begin{bmatrix}1&1&1\\1&-1&-5\\1&2&4\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }O=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&1\\1&-1&-5\\1&2&4\end{vmatrix}
\displaystyle =1(-4+10)-1(4+5)+1(2+1)
\displaystyle =6-9+3=0
\displaystyle \therefore |A|=0
\displaystyle \text{Hence, }A\text{ is singular and the given homogeneous system has infinitely many solutions.}
\displaystyle \text{Put }z=k,\text{ where }k\text{ is any real number.}
\displaystyle \text{Then, from equations (1) and (2), we get}
\displaystyle x+y=-k\qquad\ldots(4)
\displaystyle x-y=5k\qquad\ldots(5)
\displaystyle \text{Equations (4) and (5) can be written as}
\displaystyle \begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-k\\5k\end{bmatrix}
\displaystyle \text{or, }A_1X_1=B,
\displaystyle \text{where }A_1=\begin{bmatrix}1&1\\1&-1\end{bmatrix},\quad X_1=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}-k\\5k\end{bmatrix}
\displaystyle |A_1|=\begin{vmatrix}1&1\\1&-1\end{vmatrix}=-1-1=-2\ne0
\displaystyle \therefore A_1^{-1}\text{ exists.}
\displaystyle \text{Now, }\text{adj }A_1=\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}
\displaystyle \therefore A_1^{-1}=\frac{1}{|A_1|}\text{adj }A_1=-\frac12\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}
\displaystyle X_1=A_1^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\end{bmatrix}=-\frac12\begin{bmatrix}-1&-1\\-1&1\end{bmatrix}\begin{bmatrix}-k\\5k\end{bmatrix}
\displaystyle =-\frac12\begin{bmatrix}k-5k\\k+5k\end{bmatrix}=\begin{bmatrix}2k\\-3k\end{bmatrix}
\displaystyle \therefore x=2k,\qquad y=-3k,\qquad z=k,
\displaystyle \text{where }k\text{ is any real number.}
\displaystyle \text{These values also satisfy equation (3).}
\displaystyle \text{Hence, }x=2k,\ y=-3k\text{ and }z=k,\text{ where }k\in\mathbb{R},\text{ are the solutions.}
\displaystyle \\

\displaystyle \textbf{Question 6: } \qquad \begin{aligned}x+y-z&=0\\x-2y+z&=0\\3x+6y-5z&=0\end{aligned}
\displaystyle \text{Answer:}
\displaystyle \text{The given system of homogeneous linear equations is}
\displaystyle \begin{aligned}x+y-z&=0\qquad\ldots(1)\\x-2y+z&=0\qquad\ldots(2)\\3x+6y-5z&=0\qquad\ldots(3)\end{aligned}
\displaystyle \text{It can be written in matrix form as}
\displaystyle \begin{bmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{or, }AX=O,
\displaystyle \text{where }A=\begin{bmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }O=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{vmatrix}
\displaystyle =1(10-6)-1(-5-3)-1(6+6)
\displaystyle =4+8-12=0
\displaystyle \therefore |A|=0
\displaystyle \text{Hence, }A\text{ is singular and the given homogeneous system has infinitely many solutions.}
\displaystyle \text{Put }z=k,\text{ where }k\text{ is any real number.}
\displaystyle \text{Then, from equations (1) and (2), we get}
\displaystyle x+y=k\qquad\ldots(4)
\displaystyle x-2y=-k\qquad\ldots(5)
\displaystyle \text{Equations (4) and (5) can be written as}
\displaystyle \begin{bmatrix}1&1\\1&-2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}k\\-k\end{bmatrix}
\displaystyle \text{or, }A_1X_1=B,
\displaystyle \text{where }A_1=\begin{bmatrix}1&1\\1&-2\end{bmatrix},\quad X_1=\begin{bmatrix}x\\y\end{bmatrix}\text{ and }B=\begin{bmatrix}k\\-k\end{bmatrix}
\displaystyle |A_1|=\begin{vmatrix}1&1\\1&-2\end{vmatrix}=-2-1=-3\ne0
\displaystyle \therefore A_1^{-1}\text{ exists.}
\displaystyle \text{Now, }\text{adj }A_1=\begin{bmatrix}-2&-1\\-1&1\end{bmatrix}
\displaystyle \therefore A_1^{-1}=\frac{1}{|A_1|}\text{adj }A_1=-\frac13\begin{bmatrix}-2&-1\\-1&1\end{bmatrix}
\displaystyle X_1=A_1^{-1}B
\displaystyle \Rightarrow \begin{bmatrix}x\\y\end{bmatrix}=-\frac13\begin{bmatrix}-2&-1\\-1&1\end{bmatrix}\begin{bmatrix}k\\-k\end{bmatrix}
\displaystyle =-\frac13\begin{bmatrix}-2k+k\\-k-k\end{bmatrix}=\begin{bmatrix}\frac{k}{3}\\\frac{2k}{3}\end{bmatrix}
\displaystyle \therefore x=\frac{k}{3},\qquad y=\frac{2k}{3},\qquad z=k,
\displaystyle \text{where }k\text{ is any real number.}
\displaystyle \text{These values also satisfy equation (3).}
\displaystyle \text{Hence, }x=\frac{k}{3},\ y=\frac{2k}{3}\text{ and }z=k,\text{ where }k\in\mathbb{R},\text{ are the solutions.}
\displaystyle \\

\displaystyle \textbf{Question 7: } \qquad \begin{aligned}3x+y-2z&=0\\x+y+z&=0\\x-2y+z&=0\end{aligned}
\displaystyle \text{Answer:}
\displaystyle \text{The given system of homogeneous linear equations can be written in matrix form as}
\displaystyle \begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{or, }AX=O,
\displaystyle \text{where }A=\begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }O=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&1&-2\\1&1&1\\1&-2&1\end{vmatrix}
\displaystyle =3(1+2)-(1-1)-2(-2-1)
\displaystyle =9-0+6=15\ne0
\displaystyle \therefore A\text{ is non-singular and }A^{-1}\text{ exists.}
\displaystyle AX=O
\displaystyle \Rightarrow A^{-1}(AX)=A^{-1}O
\displaystyle \Rightarrow (A^{-1}A)X=O
\displaystyle \Rightarrow I_3X=O
\displaystyle \Rightarrow X=O
\displaystyle \therefore x=y=z=0
\displaystyle \text{Hence, the given system has only the trivial solution }x=y=z=0.
\displaystyle \\
\displaystyle \textbf{Question 8: } \qquad \begin{aligned}2x+3y-z&=0\\x-y-2z&=0\\3x+y+3z&=0\end{aligned}
\displaystyle \text{Answer:}
\displaystyle \text{The given system of homogeneous linear equations can be written in matrix form as}
\displaystyle \begin{bmatrix}2&3&-1\\1&-1&-2\\3&1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle \text{or, }AX=O,
\displaystyle \text{where }A=\begin{bmatrix}2&3&-1\\1&-1&-2\\3&1&3\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{ and }O=\begin{bmatrix}0\\0\\0\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}2&3&-1\\1&-1&-2\\3&1&3\end{vmatrix}
\displaystyle =2(-3+2)-3(3+6)-(1+3)
\displaystyle =-2-27-4=-33\ne0
\displaystyle \therefore A\text{ is non-singular and }A^{-1}\text{ exists.}
\displaystyle AX=O
\displaystyle \Rightarrow A^{-1}(AX)=A^{-1}O
\displaystyle \Rightarrow (A^{-1}A)X=O
\displaystyle \Rightarrow I_3X=O
\displaystyle \Rightarrow X=O
\displaystyle \therefore x=y=z=0
\displaystyle \text{Hence, the given system has only the trivial solution }x=y=z=0.
\displaystyle \\

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