Solveeit Logo
Login

Question

Question: Solve the following equation by matrix method \(2x+y+z=1\) , \(x-2y-3z=1\) , \(3x+2y+4z=5\) ....

Solve the following equation by matrix method 2x+y+z=12x+y+z=1 , x−2y−3z=1x-2y-3z=1 , 3x+2y+4z=53x+2y+4z=5 .

Explanation

Solution

Now we are given with three equations 2x+y+z=12x+y+z=1 , x−2y−3z=1x-2y-3z=1 , 3x+2y+4z=53x+2y+4z=5 . We will first write the equation in matrix form. Now first we will calculate the determinant of A. Now we will calculate the cofactor matrix of A. Now we know that adjoin of A is transpose of cofactor matrix of A and inverse of A is given by A−1=AdjAdet⁡A{{A}^{-1}}=\dfrac{AdjA}{\det A} . Hence we get the inverse matrix of A. Now we know that the solution to the system of linear equations is given by A−1B{{A}^{-1}}B . Hence we can easily find the solution by multiplying the matrices

Complete step-by-step answer:
Now we are given with the equations 2x+y+z=12x+y+z=1 , x−2y−3z=1x-2y-3z=1 , 3x+2y+4z=53x+2y+4z=5 .
Let us write the equations in matrix form.
(211 1−2−3 324 )(x y z )=(1 1 5 )\left( \begin{matrix} 2 & 1 & 1 \\\ 1 & -2 & -3 \\\ 3 & 2 & 4 \\\ \end{matrix} \right)\left( \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right)=\left( \begin{matrix} 1 \\\ 1 \\\ 5 \\\ \end{matrix} \right)
Now comparing with the form AX=BAX=B we get A=(211 1−2−3 324 )A=\left( \begin{matrix} 2 & 1 & 1 \\\ 1 & -2 & -3 \\\ 3 & 2 & 4 \\\ \end{matrix} \right) and B=(1 1 5 )B=\left( \begin{matrix} 1 \\\ 1 \\\ 5 \\\ \end{matrix} \right) .
Now first let us find the determinant of A.
Now we have A=(211 1−2−3 324 )A=\left( \begin{matrix} 2 & 1 & 1 \\\ 1 & -2 & -3 \\\ 3 & 2 & 4 \\\ \end{matrix} \right)
∣A∣=2(−8+6)−1(4+9)+1(2+6) ∣A∣=−4−13+8 ∣A∣=−9 \begin{aligned} & |A|=2\left( -8+6 \right)-1\left( 4+9 \right)+1\left( 2+6 \right) \\\ & |A|=-4-13+8 \\\ & |A|=-9 \\\ \end{aligned}
Now since we have determinant is not equal to zero the solution exists.
Now want to find cofactor matrix of A.
Let us calculate cofactor of each element.
C11=(−1)1+1∣−2−3 24 ∣=−2{{C}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix} -2 & -3 \\\ 2 & 4 \\\ \end{matrix} \right|=-2
C12=(−1)1+2∣1−3 34 =∣=−13{{C}_{12}}={{\left( -1 \right)}^{1+2}}\left| \begin{matrix} 1 & -3 \\\ 3 & 4 \\\ \end{matrix}= \right|=-13
C13=(−1)1+3∣1−2 32 ∣=8{{C}_{13}}={{\left( -1 \right)}^{1+3}}\left| \begin{matrix} 1 & -2 \\\ 3 & 2 \\\ \end{matrix} \right|=8
C21=(−1)2+1∣11 24 ∣=−2{{C}_{21}}={{\left( -1 \right)}^{2+1}}\left| \begin{matrix} 1 & 1 \\\ 2 & 4 \\\ \end{matrix} \right|=-2
C22=(−1)2+2∣21 34 ∣=5{{C}_{22}}={{\left( -1 \right)}^{2+2}}\left| \begin{matrix} 2 & 1 \\\ 3 & 4 \\\ \end{matrix} \right|=5
C23=(−1)2+3∣21 32 ∣=−1{{C}_{23}}={{\left( -1 \right)}^{2+3}}\left| \begin{matrix} 2 & 1 \\\ 3 & 2 \\\ \end{matrix} \right|=-1
C31=(−1)3+1∣11 −2−3 ∣=−1{{C}_{31}}={{\left( -1 \right)}^{3+1}}\left| \begin{matrix} 1 & 1 \\\ -2 & -3 \\\ \end{matrix} \right|=-1
C32=(−1)3+2∣21 1−3 ∣=7{{C}_{32}}={{\left( -1 \right)}^{3+2}}\left| \begin{matrix} 2 & 1 \\\ 1 & -3 \\\ \end{matrix} \right|=7
C33=(−1)3+3∣21 1−2 ∣=−5{{C}_{33}}={{\left( -1 \right)}^{3+3}}\left| \begin{matrix} 2 & 1 \\\ 1 & -2 \\\ \end{matrix} \right|=-5
Hence the cofactor matrix is given by C=(−2−138 −25−1 −17−5 )C=\left( \begin{matrix} -2 & -13 & 8 \\\ -2 & 5 & -1 \\\ -1 & 7 & -5 \\\ \end{matrix} \right)
Now we know that adjoin matrix is transpose of cofactor matrix.
Hence AdjA=(−2−2−1 −1357 8−1−5 )AdjA=\left( \begin{matrix} -2 & -2 & -1 \\\ -13 & 5 & 7 \\\ 8 & -1 & -5 \\\ \end{matrix} \right)
Now we know that A−1=AdjAdet⁡A{{A}^{-1}}=\dfrac{AdjA}{\det A}
Hence we get,

-2 & -2 & -1 \\\ -13 & 5 & 7 \\\ 8 & -1 & -5 \\\ \end{matrix} \right)$$ Now we know that the solution of linear equation $AX=B$ is given by $X={{A}^{-1}}B$ . Now let us calculate ${{A}^{-1}}B$ $$\begin{aligned} & {{A}^{-1}}B=\dfrac{1}{-9}\left( \begin{matrix} -2 & -2 & -1 \\\ -13 & 5 & 7 \\\ 8 & -1 & -5 \\\ \end{matrix} \right)\left( \begin{matrix} 1 \\\ 1 \\\ 5 \\\ \end{matrix} \right) \\\ & \Rightarrow {{A}^{-1}}B=\dfrac{1}{-9}\left( \begin{matrix} -2-2-5 \\\ -13+5+35 \\\ 8-1-25 \\\ \end{matrix} \right) \\\ & \Rightarrow {{A}^{-1}}B=\dfrac{1}{-9}\left( \begin{matrix} -9 \\\ 27 \\\ -18 \\\ \end{matrix} \right) \\\ & \Rightarrow {{A}^{-1}}B=\left( \begin{matrix} 1 \\\ -3 \\\ 2 \\\ \end{matrix} \right) \\\ \end{aligned}$$ Now we have $X={{A}^{-1}}B$ . Hence we can say x = 1, y = – 3 and z = 2. **Note:** Now there are a number of ways to solve systems of linear equations. We can also write the equation AX = BI where I is an identity matrix and use row and column transformation in a way that A = I. Then we will get the matrix in the form of X = BA’ . Hence we can find the solution to the linear equation.