Solving system of linear equations involving 3x3 matrix using adjoint matrix

Question

Solve the following system using the adjoint matrix. $$2x+4y-10z=-2$$ $$3x+9y-21z=0$$ $$x+5y-12z=1$$

Now, I have tried to solve it, and I got the determinant of matrix $A$ is equal to zero. What is the solution of the above system? Please help me. Thank you.

Answer 1

The determinant of the matrix is $-6$: $$ \det\begin{bmatrix} 2 & 4 & -10 \\ 3 & 9 & -21 \\ 1 & 5 & -12 \end{bmatrix} = -216-84-150+90+210+144=-6 $$ Thus the system has a single solution. Write it as $AX=B$, where $$ A=\begin{bmatrix} 2 & 4 & -10 \\ 3 & 9 & -21 \\ 1 & 5 & -12 \end{bmatrix} \quad X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \quad B=\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} $$ and use that $$ \operatorname{adj}A\cdot A=(\det A)I_3 $$ so you get $$ (\det A)X=\operatorname{adj}A\cdot B $$

Answer 2

Hint : Since the determinant of the coefficient matrix is zero, the system of linear equations has no unique solution. Try to find the row echelon form of the augmented matrix $(A|B)$ , where $A$ is the coefficient matrix, and $B$ is the matrix of the right side. This will help.

Answer 3

The system of equation can be expressed in the matrix form as: $$\begin{pmatrix}2&4&-10\\ 3&9&-21\\1&5&-12\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-2\\0\\1\end{pmatrix} \iff AX=B.$$ The formula to find $X$: $$AX=B \iff A^{-1}AX=A^{-1}B \iff X=A^{-1}B=\frac{\text{adj}A}{|A|}B.$$ The determinant of $A$ (performing row operations): $$|A|=\begin{vmatrix}2&4&-10\\ 3&9&-21\\1&5&-12\end{vmatrix}=\begin{vmatrix}0&-6&14\\ 0&-6&15\\1&5&-12\end{vmatrix}=\begin{vmatrix}0&0&-1\\ 0&-6&15\\1&5&-12\end{vmatrix}=-6.$$ The adjoint of $A$: $$\text{adj}A=(C_{ij})^{\text{T}}=\begin{pmatrix}9(-12)-(-21)5&-(4(-12)-(-10)5)&4(-21)-(-10)9\\ -(3(-12)-(-21)1)&2(-12)-(-10)1&-(2(-21)-(-10)3)\\3\cdot 5-9\cdot 1&-(2\cdot 5-4\cdot 1)&2\cdot 9-4\cdot 3\end{pmatrix}=\\ =\begin{pmatrix}-3&-2&6\\ 15&-14&12\\6&-6&6\end{pmatrix}$$ Can you calculate $X$?