Solveeit Logo

Question

Mathematics Question on Applications of Determinants and Matrices

Solve the system of equations x+2y+z=4x + 2y + z = 4, x+y+z=0- x + y + z = 0 and x3y+z=4x - 3y + z = 4.

A

x=2x = 2 , y=0y = 0, z=2z = 2

B

x=2x = 2,y=0y = 0, z=2z = -2

C

x=2x = -2, y=2y = 2, z=0z = 0

D

x=2x = -2, y0y - 0, z=2z = 2

Answer

x=2x = 2 , y=0y = 0, z=2z = 2

Explanation

Solution

Given system of equations is x+2y+z=4x + 2y + z = 4, x+y+z=0- x + y + z = 0 and x3y+z=4x - 3y + z = 4 In matrix form, it can be written as [121 111 131][x y z]=[4 0 4]\left[\begin{matrix}1&2&1\\\ -1&1&1\\\ 1&-3&1\end{matrix}\right]\left[\begin{matrix}x\\\ y\\\ z\end{matrix}\right]=\left[\begin{matrix}4\\\ 0\\\ 4\end{matrix}\right] or AX=BAX = B where, A=[121 111 131],B=[4 0 4]andX=[x y z]A =\left[\begin{matrix}1&2&1\\\ -1&1&1\\\ 1&-3&1\end{matrix}\right], B=\left[\begin{matrix}4\\\ 0\\\ 4\end{matrix}\right] and X=\left[\begin{matrix}x\\\ y\\\ z\end{matrix}\right] Now, A=1(1+3)2(11)+1(31)=4+4+2=10|A| = 1\left(1 + 3\right) - 2 \left(-1 - 1\right) + 1\left(3 - 1\right)= 4 + 4 + 2 = 10 A0\because\quad\left|A\right|\ne0, hence unique solution exists. Cofactor matrix of A=[422 505 123]A =\left[\begin{matrix}4&2&2\\\ -5&0&5\\\ 1&-2&3\end{matrix}\right] adj(A)=[422 505 123]T=[451 202 253]\therefore\quad adj\left(A\right) =\left[\begin{matrix}4&2&2\\\ -5&0&5\\\ 1&-2&3\end{matrix}\right]^{T}= \left[\begin{matrix}4&-5&1\\\ 2&0&-2\\\ 2&5&3\end{matrix}\right] A1=1Aadj(A)=110[451 202 253]Now,X=A1B=110[451 202 253][4 0 4]A^{-1}=\frac{1}{\left|A\right|} adj \left(A\right)=\frac{1}{10}\left[\begin{matrix}4&-5&1\\\ 2&0&-2\\\ 2&5&3\end{matrix}\right] Now, X=A^{-1}B=\frac{1}{10}\left[\begin{matrix}4&-5&1\\\ 2&0&-2\\\ 2&5&3\end{matrix}\right]\left[\begin{matrix}4\\\ 0\\\ 4\end{matrix}\right] [x y z]=110[16+0+4 8+08 8+0+12]=110[20 0 20]=[2 0 2]\left[\begin{matrix}x\\\ y\\\ z\end{matrix}\right]=\frac{1}{10} \left[\begin{matrix}16+0+4\\\ 8+0-8\\\ 8+0+12\end{matrix}\right]=\frac{1}{10} \left[\begin{matrix}20\\\ 0\\\ 20\end{matrix}\right]=\left[\begin{matrix}2\\\ 0\\\ 2\end{matrix}\right] On comparing the corresponding elements, we get x=2x = 2, y=0y = 0 and z=2z = 2