Question

Solve the following equation by matrix method
x+y+z=1
x+2y+3z=6
x+3y+4z=6

Answer 1

Answer:

i dont know ............

Answer 2

Answer:

$x=1, y=-5 \text{ and } z=5$

Step-by-step explanation:

Given equations are

$x+y+z=1\\x+2y+3z=6 \\x+3y+4z=6$

Let given equations written in matrix be

$\left[\begin{array}{ccc}1&1&1\\1&2&3\\1&3&4\end{array}\right] \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}1&6&6\end{array}\right]$

Let A X = B

$\therefore A = \left[\begin{array}{ccc}1&1&1\\1&2&3\\1&3&4\end{array}\right], B=\left[\begin{array}{c}1&6&6\end{array}\right] \text{ and } X=\left[\begin{array}{c}x&y&z\end{array}\right]$

$|A| = 1 \left|\begin{array}{cc}2&3\\3&4\end{array}\right| -1 \left|\begin{array}{cc}1&3\\1&4\end{array}\right| + 1 \left|\begin{array}{cc}1&2\\1&3\end{array}\right|$

$=1(8-9)-1(4-3)+1(3-2)$

$=-1-1(1)+1(1)$

$=-1-1+1$

$=-1$

Since $|A| \neq 0$

$\therefore \text{ The system of equation is consistent and has a unique solution. }$

$\text{Now, } AX = B$

$\Rightarrow X = A^{-1} B$

Now, calculating $A^{-1}$

$A^{-1}=\cfrac{1}{| A |}\text{ adj.} A$

$\text{adj. }A = \left[\begin{array}{ccc}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{array}\right] ^\prime = \left[\begin{array}{ccc}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{array}\right]$

Now,

$A=\left[\begin{array}{ccc}1&1&1\\1&2&3\\1&3&4\end{array}\right]$

$M_{11} = \left|\begin{array}{cc}2&3\\3&4\end{array}\right| = 8 - 9 = -1$

$M_{12} = \left|\begin{array}{cc}1&3\\1&4\end{array}\right| = 4-3 = 1$

$M_{13} = \left|\begin{array}{cc}1&2\\1&3\end{array}\right| = 3 - 2 = 1$

$M_{21} = \left|\begin{array}{cc}1&1\\3&4\end{array}\right| = 4 - 3 = 1$

$M_{22} = \left|\begin{array}{cc}1&1\\1&4\end{array}\right| = 4 - 1 = 3$

$M_{23} = \left|\begin{array}{cc}1&1\\1&3\end{array}\right| = 3 - 1 = 2$

$M_{31} = \left|\begin{array}{cc}1&1\\2&3\end{array}\right| = 3 -2 = 1$

$M_{32} = \left|\begin{array}{cc}1&1\\1&3\end{array}\right| = 3 - 1 = 2$

$M_{33} = \left|\begin{array}{cc}1&1\\1&2\end{array}\right| = 2 -1 = 1$

Now,

$A_{11} = (-1)^{1+1} \cdot M_{11} = (-1)^2 \cdot (-1) = -1$

$A_{12} = (-1)^{1+2} \cdot M_{12} = (-1)^3 \cdot (1) = -1$

$A_{13} = (-1)^{1+3} \cdot M_{13} = (-1)^4 \cdot (1) = 1$

$A_{21} = (-1)^{2+1} \cdot M_{21} = (-1)^3 \cdot (1) = -1$

$A_{22} = (-1)^{2+2} \cdot M_{22} = (-1)^4 \cdot (3) = 3$

$A_{23} = (-1)^{2+3} \cdot M_{23} = (-1)^5 \cdot (2) = -2$

$A_{31} = (-1)^{3+1} \cdot M_{31} = (-1)^4 \cdot (1) = 1$

$A_{32} = (-1)^{3+2} \cdot M_{32} = (-1)^5 \cdot (2) = -2$

$A_{33} = (-1)^{3+3} \cdot M_{33} = (-1)^6 \cdot (1) = 1$

Thus,

$\text{adj. }A = \left[\begin{array}{ccc}-1&-1&1\\-1&3&-2\\1&-2&1\end{array}\right]$

Now,

$A^{-1} =\cfrac{1}{|A|} \cdot \text{ adj. }A$

$= \cfrac{1}{-1} \left[\begin{array}{ccc}-1&-1&1\\-1&3&-2\\1&-2&1\end{array}\right]$

$=\left[\begin{array}{ccc}1&1&-1\\1&-3&2\\-1&2&-1\end{array}\right]$

Now,

$X=A^{-1} B$

$\Rightarrow \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{ccc}1&1&-1\\1&-3&2\\-1&2&-1\end{array}\right] \left[\begin{array}{c}1&6&6\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}(1\times1) + (1\times 6) +(-1\times 6)\\(1\times 1) +(-3\times 6) + (2\times 6)\\(-1\times 1) + (2\times 6) + (-1 \times 6)\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}1+6-6\\1-18+12\\-1+12-6\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x&y&z\end{array}\right]=\left[\begin{array}{c}1\\-5\\5\end{array}\right]$

$\therefore x=1, y=-5 \text{ and } z = 5$

Please mark my answer as BRAINLIEST.