# 5.2 another Ax=B with an inverse

• MHB
• karush
In summary, having an inverse in the equation Ax=B allows us to solve for the unknown matrix x by multiplying both sides of the equation by A^-1. To determine if a matrix has an inverse, its determinant must not be equal to 0. A matrix can only have one inverse and it is possible for a matrix to have no inverse if its determinant is 0. In real-life scenarios, inverse matrices are used in solving systems of linear equations, cryptography, data encryption, and optimization problems.
karush
Gold Member
MHB
Suppose we are given
$A^{-1} = \begin{bmatrix} 1 & 4 & 0 \\ 2 & 3 & 0 \\ 4 & 2 & 2 \end{bmatrix} = \left[\begin{array}{rrr|rrr} 1&4&0&1&0&0\\ 2&3&0&0&1&0\\ 4&2&2&0&0&1 \end{array}\right]$
then
$A= \begin{bmatrix} -\frac{3}{5}&\frac{4}{5}&0\\ \frac{2}{5}&-\frac{1}{5}&0\\ \frac{4}{5}&-\frac{7}{5}&\frac{1}{2} \end{bmatrix}$
Solve the matrix equation $AX=B$ to find $x$, $y$ and $z$ where $X = \begin{bmatrix} x \\y \\z \end{bmatrix}, \textit{ and } B = \begin{bmatrix} -1 \\2 \\3 \end{bmatrix}$
$\therefore AX=B$ is
$\begin{bmatrix} -\frac{3}{5}&\frac{4}{5}&0\\ \frac{2}{5}&-\frac{1}{5}&0\\ \frac{4}{5}&-\frac{7}{5}&\frac{1}{2} \end{bmatrix} \begin{bmatrix} x \\y \\z \end{bmatrix} =\begin{bmatrix} -1 \\2 \\3 \end{bmatrix}$
$W\vert A$ returned
$x=7,y=4,z=6$ok my question is that A has 2 zero's in $C_3$
I was able to get x=7 and y=4 by a simple double simultanious equation
but not sure what the proper methed is take advantage of the 2 zeros

karush said:
Suppose we are given
$A^{-1} = \begin{bmatrix} 1 & 4 & 0 \\ 2 & 3 & 0 \\ 4 & 2 & 2 \end{bmatrix} = \left[\begin{array}{rrr|rrr} 1&4&0&1&0&0\\ 2&3&0&0&1&0\\ 4&2&2&0&0&1 \end{array}\right]$
then
$A= \begin{bmatrix} -\frac{3}{5}&\frac{4}{5}&0\\ \frac{2}{5}&-\frac{1}{5}&0\\ \frac{4}{5}&-\frac{7}{5}&\frac{1}{2} \end{bmatrix}$
Solve the matrix equation $AX=B$ to find $x$, $y$ and $z$ where $X = \begin{bmatrix} x \\y \\z \end{bmatrix}, \textit{ and } B = \begin{bmatrix} -1 \\2 \\3 \end{bmatrix}$
$\therefore AX=B$ is
$\begin{bmatrix} -\frac{3}{5}&\frac{4}{5}&0\\ \frac{2}{5}&-\frac{1}{5}&0\\ \frac{4}{5}&-\frac{7}{5}&\frac{1}{2} \end{bmatrix} \begin{bmatrix} x \\y \\z \end{bmatrix} =\begin{bmatrix} -1 \\2 \\3 \end{bmatrix}$
$W\vert A$ returned
$x=7,y=4,z=6$ok my question is that A has 2 zero's in $C_3$
I was able to get x=7 and y=4 by a simple double simultanious equation
but not sure what the proper methed is take advantage of the 2 zeros
If you are asking if it makes sense to write out the two equations in x and y that don't involve z and solving the simpler system for x and y then I think you are doing just fine.

-Dan

You are given that $$A^{-1}$$ exists and is equal to $$\begin{bmatrix}1& 4 & 0 \\ 2 & 3 & 0 \\ 4 & 2 & 2 \end{bmatrix}$$. So the solution to $$AX= B$$ is just $$X= A^{-1}B$$.

With B given as $$\begin{bmatrix}-1 \\ 2 \\ 3\end{bmatrix}$$, as here, we have immediately that $$X= \begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}1& 4 & 0 \\ 2 & 3 & 0 \\ 4 & 2 & 2 \end{bmatrix}\begin{bmatrix}-1 \\ 2 \\ 3\end{bmatrix}= \begin{bmatrix}7 \\ 4 \\ 6\end{bmatrix}$$. That is, x= 7, y= 4, and z= 6. But we can write the initial "AX= B" as the system of equstions $$-\frac{3}{5}x+ \frac{4}{5}y= -1$$, $$\frac{2}{5}x- \frac{1}{5}y= 2$$, and $$\frac{4}{5}x- \frac{7}{5}y+ \frac{1}{2}z= 3$$. As topsquark said, the fact that the first two entries in the last column are "0" mean that the first two equations have no "z" and can be solved for x and y indpendently of z. I would first multiply each equation by 5 to get $$-3x+ 4y= -5$$ and $$2x- y= 10$$. Multiply the second equation by 4 to geta $$8x- 4y= 40$$ and add that to the first equation to eliminate y and get $$5x= 35$$ so that $$x= 7$$. Setting x= 7 in $$-3x+ 4y= -5$$ gives $$-21+ 4y= -5$$ so that $$4y= 16$$ and y= 4. Finally, put x= 7, y= 4 into the last equation to get $$\frac{28}{5}- \frac{28}{5}+ \frac{1}{2}z= \frac{1}{2}z= 3$$ so that $$z= 6$$. We get x= 7, y= 4, z= 6 as before.

## What is the meaning of "5.2 another Ax=B with an inverse" in scientific terms?

In scientific terms, "5.2 another Ax=B with an inverse" refers to a mathematical equation where the matrix A is multiplied by a vector x to equal another vector B, and there exists an inverse matrix A^-1 that can be multiplied by B to get x.

## How is the inverse of a matrix represented in mathematical notation?

The inverse of a matrix A is represented as A^-1.

## What is the significance of finding the inverse of a matrix in scientific research?

Finding the inverse of a matrix is significant in scientific research as it allows for the solving of systems of linear equations and can be used to find the solution to many mathematical problems.

## How is the inverse of a matrix calculated?

The inverse of a matrix can be calculated using various methods such as Gauss-Jordan elimination, LU decomposition, or by using the adjugate matrix and the determinant of the matrix.

## What are the applications of "5.2 another Ax=B with an inverse" in scientific fields?

The equation "5.2 another Ax=B with an inverse" has various applications in scientific fields such as physics, engineering, and computer science. It is used to solve systems of linear equations, perform matrix transformations, and find the inverse of a matrix.

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