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Find the value of $X$ if

$$\begin{pmatrix}2& 0& 7\\ 0 & 1& 0 \\ 1& -2& 1 \\ \end{pmatrix} \begin{pmatrix}-X& 14X& 7X\\ 0 & 1& 6 \\ X& -4X& -2X \\ \end{pmatrix} = \begin{pmatrix}1& 0& 0\\ 0 & 1& 0 \\ 0& 0& 1 \\ \end{pmatrix}$$

I tried to solve the question by finding the inverse of the matrix on the left side, then multiplying both sides by that inverse matrix to be able to solve for X. I got X=1/5, but the problem is I got 6 = 0 which means inconsistent system. Please help!!

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  • $\begingroup$ Another way is do the multiplication and identify coefficients, it is linear you know. But you got same solution x=1/5 and 6=0 so yes, the system IS inconsistant, there are no solution. You got it right. By the way since AB=I, the inverse matrice of A should be B. $\endgroup$ – zwim Mar 4 '17 at 7:43
  • $\begingroup$ Your solution is the answer to your own question, what you have done is shown that there is no $x$ such that the equation is true. $\endgroup$ – Prince M Mar 4 '17 at 7:48
  • $\begingroup$ Can anyone try to solve it by inverse of matrices?? $\endgroup$ – user421354 Mar 4 '17 at 8:22
  • $\begingroup$ If you multiply out the two matrices you will see that the $(2,3)$-entry of the result is a constant $6$, which independently of$~X$ gives a contradiction with the equation. So there is no point in even trying to solve anything here. $\endgroup$ – Marc van Leeuwen Mar 4 '17 at 9:59
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Let $A=\begin{pmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{pmatrix}$, note that det(A)=-5, so A is nonsingular.

Since $AA^{-1}=I$, according to the equation, you have: $A^{-1}=\begin{pmatrix} -X & 14X & 7X \\ 0 & 1 & 6 \\ X & -4X & -2X \end{pmatrix}$

After computation, $A^{-1}= \begin{pmatrix} -0.2 & 2.8 & 1.4\\ 0 & 1 & 0 \\ 0.2 & -0.8 & -0.4 \end{pmatrix}$

$\underline{Edit}$: Notice that $6\neq 0$, referring to $A^{-1}_{(2,3)}$, so there is no solution for X.

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    $\begingroup$ there is a little problem with coefficient on row 2 column 3: $6 \neq 0$... $\endgroup$ – Jean Marie Mar 4 '17 at 8:25
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    $\begingroup$ This is false since the (2,3)-element of $AA^{-1}$ using your $A^{-1}$ is not $0$. There is no solution. $\endgroup$ – ancientmathematician Mar 4 '17 at 8:26
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It is impossible.

Let $M$ be the initial matrix, and $N$ its inverse.

Use the formula of the inverse as being the transpose of the matrix of cofactors, multiplied by $1/det(M)=-1/5$.

Coefficient $N_{23}$ should be $(-1/5) \times \ \underbrace{-\begin{vmatrix} 2 & 7 \\ 0 & 0 \end{vmatrix}}_{\text{cofactor of } \ M_{32}}=0$, instead of $6$.

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    $\begingroup$ @ancient mathematician It is not necessary to downvote an answer that I was changing, being conscious of the problem (see the similar remark I have done to the OP) $\endgroup$ – Jean Marie Mar 4 '17 at 8:36
  • $\begingroup$ @ancient mathematician Nothing is irredeemably wrong. I show it by my new answer. Besides, if you are an ancient mathematician, you are maybe an ancient teacher, and I whish you have never said that remark to your students. $\endgroup$ – Jean Marie Mar 4 '17 at 8:45
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Your matrix has full rank, $\rho = 3$. Therefore, we just need to find the inverse of the matrix.

The following is based on the assumption that the input matrix is valid and the inverse matrix exists. The implies that there there is a typo in the original question.

The solution follows.


Gauss-Jordan elimination

Use augmented reduction.

Clear column 1 $$ % clear \left[ \begin{array}{rcc} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % A I \left[ \begin{array}{crc|ccc} 2 & 0 & 7 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 1 & -2 & 1 & 0 & 0 & 1 \\ \end{array} \right] % out = \left[ \begin{array}{crr|rcc} \boxed{1} & 0 & \frac{7}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & -2 & -\frac{5}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % $$

Clear column 2 $$ % clear \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \\ \end{array} \right] % A I \left[ \begin{array}{crr|rcc} \boxed{1} & 0 & \frac{7}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & -2 & -\frac{5}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % out = \left[ \begin{array}{ccr|rcc} \boxed{1} & 0 & \frac{7}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \boxed{1} & 0 & 0 & 1 & 0 \\ 0 & 0 & -\frac{5}{2} & -\frac{1}{2} & 2 & 1 \\ \end{array} \right] % $$

Clear column 3 $$ % clear \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \\ \end{array} \right] % A I \left[ \begin{array}{ccr|rcc} \boxed{1} & 0 & \frac{7}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \boxed{1} & 0 & 0 & 1 & 0 \\ 0 & 0 & -\frac{5}{2} & -\frac{1}{2} & 2 & 1 \\ \end{array} \right] % out = \left[ \begin{array}{ccc|rrr} \boxed{1} & 0 & 0 & -\frac{1}{5} & \frac{14}{5} & \frac{7}{5} \\ 0 & \boxed{1} & 0 & 0 & 1 & 0 \\ 0 & 0 & \boxed{1} & \frac{1}{5} & -\frac{4}{5} & -\frac{2}{5} \\ \end{array} \right] % $$

Harvest inverse matrix $$ \mathbb{A}^{-1} = \left( \begin{array}{rrr} -\frac{1}{5} & \frac{14}{5} & \frac{7}{5} \\ 0 & 1 & 0 \\ \frac{1}{5} & -\frac{4}{5} & -\frac{2}{5} \\ \end{array} \right) $$


Conclusion

$$ \boxed{ X = \frac{1}{5} } $$


Verify answer

$$ \mathbf{A} \mathbf{A}^{-1} = % \left[ \begin{array}{crc} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \\ \end{array} \right] % \left[ \begin{array}{rrr} -\frac{1}{5} & \frac{14}{5} & \frac{7}{5} \\ 0 & 1 & 0 \\ \frac{1}{5} & -\frac{4}{5} & -\frac{2}{5} \\ \end{array} \right] % = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] % $$

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