Is the Inverse of this Matrix Possible?

[askor]

I tried to find the inverse of below matrix and what I get is no inverse.

$\left( \begin{array}{rrr} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right)$

Can someone please check it whether I am correct or not?

 

[fresh_42]

What have you done so far?

 

[askor]

This is what I get

$\left( \begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & \frac{4}{3} & -\frac{1}{3} & 0 \\ 0 & 0 & 0 & -1 & -2 & 1 \end{array} \right)$

 

[fresh_42]

Difficult to check something you cannot see.

What is the determinant? Maybe the matrix isn't invertible.
(You could compute the determinant along the first row. If it's zero, it isn't invertible.)

You could also try the computational method with the adjugate matrix
https://en.wikipedia.org/wiki/Adjugate_matrix#Inverses
but I would compute the determinant first.

 

[askor]

Basically, I rewrote the matrix in the below form:

$\left( \begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array} \right)$

Next, I do row operation, and what I get is this:

$\left( \begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & \frac{4}{3} & -\frac{1}{3} & 0 \\ 0 & 0 & 0 & -1 & -2 & 1 \end{array} \right)$

 

[fresh_42]

Assuming you made no mistake, what does this tell you?

 

[askor]

Assuming you made no mistake, what does this tell you?

Is this inconsistent matrix? Does this mean the matrix has no solutions/invers?

 

[fresh_42]

I don't know what an inconsistent matrix is or what you mean by a solution, since there is nowhere an equation to solve.
But it has no inverse, yes. Do you know how to compute a determinant and what it says? If not you should read about it, because it is essential when dealing with matrices.

 

[askor]

I don't know what an inconsistent matrix is or what you mean by a solution, since there is nowhere an equation to solve.
But it has no inverse, yes. Do you know how to compute a determinant and what it says? If not you should read about it, because it is essential when dealing with matrices.

OK, let me find the determinant of below matrix by row 3 expansion.

$\left( \begin{array}{rrr} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right)$

$a_{31} = 7, a_{32} = 8, a_{33} = 9$

$C_{31} = (-1)^{3+1} \left| \begin{array}{rr} 2 & 3 \\ 5 & 6 \\ \end{array} \right| = (-1)^{4}(2.6 - 3.5) = (1)(12 - 15) = (1)(-3) = -3$

$C_{32} = (-1)^{3+2} \left| \begin{array}{rr} 1 & 3 \\ 4 & 6 \\ \end{array} \right| = (-1)^{5}(1.6 - 3.4) = (-1)(6 - 12) = (-1)(-6) = 6$

$C_{33} = (-1)^{3+3} \left| \begin{array}{rr} 1 & 2 \\ 4 & 5 \\ \end{array} \right| = (-1)^{6}(1.5 - 2.4) = (1)(5 - 8) = (1)(-3) = -3$

$\det = a_{31}.C_{31} + a_{32}.C_{32} + a_{33}.C_{33}$
$= 7(-3) + 8(6) + 9(-3)$
$= -21 + 48 -27$
$= 0$

 

[mfb]

0 is correct.

There are matrices without an inverse, this is one example. And there is a nice relation between the determinant of a matrix and the existence of an inverse matrix.

 

[Stephen Tashi]

Next, I do row operation, and what I get is this:

$\left( \begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & \frac{4}{3} & -\frac{1}{3} & 0 \\ 0 & 0 & 0 & -1 & -2 & 1 \end{array} \right)$

If your text materials want you to find the inverse of the matrix by row operations, we should investigate what row operations you did.

 

[askor]

If your text materials want you to find the inverse of the matrix by row operations, we should investigate what row operations you did.

This the row operation:

$\left( \begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array} \right)$

R2 - 4R1 and R3 - 7R1

$\left( \begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & -3 & -6 & -4 & 1 & 0 \\ 0 & -6 & -12 & -7 & 0 & 1 \end{array} \right)$

R2 x $(-\frac{1}{3})$

$\left( \begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & \frac{4}{3} & -\frac{1}{3} & 0 \\ 0 & -6 & -12 & -7 & 0 & 1 \end{array} \right)$

R3 + 6R2

$\left( \begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & \frac{4}{3} & -\frac{1}{3} & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 \end{array} \right)$

As you can see in the last matrix, the 3rd row has 0 0 0. It means that this matrix is inconsistent.

 

[Stephen Tashi]

Your steps are correct.