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Question: Using matrices, solve the following system of equations: \(2x+3y+3z=5,x-2y+z=-4,3x-y-2z=3.\)...

Using matrices, solve the following system of equations:
2x+3y+3z=5,x−2y+z=−4,3x−y−2z=3.2x+3y+3z=5,x-2y+z=-4,3x-y-2z=3.

Explanation

Solution

We first consider the given system of linear equations and we represent them in the matrix form AX=BAX=B. We have to find the solution that is, matrix X=A−1BX={{A}^{-1}}B. Before finding the solution, we make sure that the matrix A−1{{A}^{-1}} exists by checking whether the determinant of matrix AA is zero or not. After making sure that the determinant of matrix AA is not equal to zero, we find the matrix A−1{{A}^{-1}} using the formula 1∣A∣Adj(A)\dfrac{1}{\left| A \right|}Adj\left( A \right). Then we obtain the final result XXusing the formula X=A−1BX={{A}^{-1}}B.

Complete step by step answer:
Let us consider the given system of linear equations,
2x+3y+3z=5 x−2y+z=−4 3x−y−2z=3 \begin{aligned} & 2x+3y+3z=5 \\\ & x-2y+z=-4 \\\ & 3x-y-2z=3 \\\ \end{aligned}
Now, we represent this system of linear equations in matrix form i.e., AX=BAX=B where
A=[233 1−21 3−1−2 ]A=\left[ \begin{matrix} 2 & 3 & 3 \\\ 1 & -2 & 1 \\\ 3 & -1 & -2 \\\ \end{matrix} \right], X=[x y z ]X=\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right] and B=[5 −4 3 ]B=\left[ \begin{matrix} 5 \\\ -4 \\\ 3 \\\ \end{matrix} \right]
So, let us check whether the determinant is equal to zero or not equal to zero.
Now, let us consider the formula,
∣a11a12a13 a21a22a23 a31a32a33 ∣=a11(a22a33−a23a32)−a21(a12a33−a13a32)+a31(a12a23−a13a22)\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)-{{a}_{21}}\left( {{a}_{12}}{{a}_{33}}-{{a}_{13}}{{a}_{32}} \right)+{{a}_{31}}\left( {{a}_{12}}{{a}_{23}}-{{a}_{13}}{{a}_{22}} \right)
By using the above formula,
∣A∣=∣233 1−21 3−1−2 ∣=2(4+1)−1(−6+3)+3(3+6)\left| A \right|=\left| \begin{matrix} 2 & 3 & 3 \\\ 1 & -2 & 1 \\\ 3 & -1 & -2 \\\ \end{matrix} \right|=2\left( 4+1 \right)-1\left( -6+3 \right)+3\left( 3+6 \right)
∣A∣=2(5)−1(−3)+3(9) ∣A∣=10+3+27 ∣A∣=40 \begin{aligned} & \left| A \right|=2\left( 5 \right)-1\left( -3 \right)+3\left( 9 \right) \\\ & \left| A \right|=10+3+27 \\\ & \left| A \right|=40 \\\ \end{aligned}
So, we got
∣A∣≠0\left| A \right|\ne 0
So, we can say that A−1{{A}^{-1}} exists.
Let us consider the formula,
A−1=1∣A∣AdjA{{A}^{-1}}=\dfrac{1}{\left| A \right|}AdjA
So now, we find the matrix AdjAAdjA.
To find the adjoint of a matrix, first we find the cofactor matrix of the given matrix. Then we find the transpose of the cofactor matrix.
We have, A=[233 1−21 3−1−2 ]A=\left[ \begin{matrix} 2 & 3 & 3 \\\ 1 & -2 & 1 \\\ 3 & -1 & -2 \\\ \end{matrix} \right]
For finding the inverse we have to find the cofactor matrix of A. So, first let us find the cofactors of every element in the matrix.
Cofactor of 2 = ∣−21 −1−2 ∣=−2(−2)−1(−1)=5\left| \begin{matrix} -2 & 1 \\\ -1 & -2 \\\ \end{matrix} \right|=-2\left( -2 \right)-1\left( -1 \right)=5
Cofactor of 3 = −∣11 3−2 ∣=−(1(−2)−1(3))=5-\left| \begin{matrix} 1 & 1 \\\ 3 & -2 \\\ \end{matrix} \right|=-\left( 1\left( -2 \right)-1\left( 3 \right) \right)=5
Cofactor of 3 = ∣1−2 3−1 ∣=1(−1)−(−2)(3)=5\left| \begin{matrix} 1 & -2 \\\ 3 & -1 \\\ \end{matrix} \right|=1\left( -1 \right)-\left( -2 \right)\left( 3 \right)=5
Cofactor of 1 = −∣33 −1−2 ∣=−(3(−2)−3(−1))=3-\left| \begin{matrix} 3 & 3 \\\ -1 & -2 \\\ \end{matrix} \right|=-\left( 3\left( -2 \right)-3\left( -1 \right) \right)=3
Cofactor of -2 = ∣23 3−2 ∣=2(−2)−3(3)=−13\left| \begin{matrix} 2 & 3 \\\ 3 & -2 \\\ \end{matrix} \right|=2\left( -2 \right)-3\left( 3 \right)=-13
Cofactor of 1 = −∣23 3−1 ∣=−(2(−1)−3(3))=11-\left| \begin{matrix} 2 & 3 \\\ 3 & -1 \\\ \end{matrix} \right|=-\left( 2\left( -1 \right)-3\left( 3 \right) \right)=11
Cofactor of 3 = ∣33 −21 ∣=3(1)−3(−2)=9\left| \begin{matrix} 3 & 3 \\\ -2 & 1 \\\ \end{matrix} \right|=3\left( 1 \right)-3\left( -2 \right)=9
Cofactor of -1 = −∣23 11 ∣=−(2(1)−3(1))=1-\left| \begin{matrix} 2 & 3 \\\ 1 & 1 \\\ \end{matrix} \right|=-\left( 2\left( 1 \right)-3\left( 1 \right) \right)=1
Cofactor of -2 = ∣23 1−2 ∣=2(−2)−3(1)=−7\left| \begin{matrix} 2 & 3 \\\ 1 & -2 \\\ \end{matrix} \right|=2\left( -2 \right)-3\left( 1 \right)=-7
We get the cofactor matrix = [555 3−1311 91−7 ]\left[ \begin{matrix} 5 & 5 & 5 \\\ 3 & -13 & 11 \\\ 9 & 1 & -7 \\\ \end{matrix} \right]
Now, we get AdjAAdjA by taking the transpose of cofactor matrix.
AdjA=[539 5−131 511−7 ]AdjA=\left[ \begin{matrix} 5 & 3 & 9 \\\ 5 & -13 & 1 \\\ 5 & 11 & -7 \\\ \end{matrix} \right]
Now, let us consider the formula, A−1=1∣A∣AdjA{{A}^{-1}}=\dfrac{1}{\left| A \right|}AdjA
By using the above formula, we get
AdjA=140[539 5−131 511−7 ]AdjA=\dfrac{1}{40}\left[ \begin{matrix} 5 & 3 & 9 \\\ 5 & -13 & 1 \\\ 5 & 11 & -7 \\\ \end{matrix} \right]
Let us consider the formula X=A−1BX={{A}^{-1}}B
By using the above formula, we get
X=1∣40∣[539 5−131 511−7 ][5 −4 3 ]  X=1∣40∣[5(5)+3(−4)+9(3) 5(5)−13(−4)+1(3) 5(5)+11(−4)−7(3) ]  X=1∣40∣[25−12+27 25+52+3 25−44−21 ]  X=1∣40∣[40 80 −40 ]  X=[4040 8040 −4040 ]  X=[1 2 −1 ] \begin{aligned} & X=\dfrac{1}{\left| 40 \right|}\left[ \begin{matrix} 5 & 3 & 9 \\\ 5 & -13 & 1 \\\ 5 & 11 & -7 \\\ \end{matrix} \right]\left[ \begin{matrix} 5 \\\ -4 \\\ 3 \\\ \end{matrix} \right] \\\ & \\\ & X=\dfrac{1}{\left| 40 \right|}\left[ \begin{matrix} 5\left( 5 \right)+3\left( -4 \right)+9\left( 3 \right) \\\ 5\left( 5 \right)-13\left( -4 \right)+1\left( 3 \right) \\\ 5\left( 5 \right)+11\left( -4 \right)-7\left( 3 \right) \\\ \end{matrix} \right] \\\ & \\\ & X=\dfrac{1}{\left| 40 \right|}\left[ \begin{matrix} 25-12+27 \\\ 25+52+3 \\\ 25-44-21 \\\ \end{matrix} \right] \\\ & \\\ & X=\dfrac{1}{\left| 40 \right|}\left[ \begin{matrix} 40 \\\ 80 \\\ -40 \\\ \end{matrix} \right] \\\ & \\\ & X=\left[ \begin{matrix} \dfrac{40}{40} \\\ \dfrac{80}{40} \\\ \dfrac{-40}{40} \\\ \end{matrix} \right] \\\ & \\\ & X=\left[ \begin{matrix} 1 \\\ 2 \\\ -1 \\\ \end{matrix} \right] \\\ \end{aligned}

Therefore, we get x=1,y=2x=1,y=2 and z=−1z=-1.
Therefore, the solution for the system of equations2x+3y+3z=5,x−2y+z=−4,3x−y−2z=32x+3y+3z=5,x-2y+z=-4,3x-y-2z=3 is x=1,y=2,z=−1x=1,y=2,z=-1

Hence, the answer is x=1,y=2,z=−1x=1,y=2,z=-1

Note: While solving this question one might make a mistake by taking adjoint matrix as cofactor matrix directly and finding out the inverse of the matrix without converting it into the transpose of cofactor matrix. But Adjoint Matrix is the transpose of a cofactor matrix, so one needs to remember it while doing the question.