Find X , using the inverse of matrices

Question

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!!

Answer 1

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.

Answer 2

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 $6$.

Answer 3

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.

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

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Conclusion

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

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Verify answer

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