Linear Algebra, Find a matrix C st CA = B

In summary, the matrix C for CA = B can be found by multiplying the corresponding rows of A by a nonzero number c for part (a), swapping the rows of A for part (b), and adding c times row i of A to row j of A for part (c). However, it is important to note that matrix multiplication is not commutative and the order of the matrices matters. It is also important to review the rules and properties of matrix multiplication before attempting to solve for C.
  • #1
Poke

Homework Statement


Let A be an arbitrary m× n matrix. Find a matrix C such that CA = B for each of the following matrices B.
a. B is the matrix that results from multiplying row i of A by a nonzero number c.
b. B is the matrix that results from swapping rows i and j of A.
c. B is the matrix that results from adding c times row i of A to row j of A.

Homework Equations


Multiply the two matrices

The Attempt at a Solution


(a)
Let A = a11 a12 ... a1n
a21 a22 ... a2n
...
am1 am2 ... amn
.
And C = c11 c12 ... c1n
c21 c22 ... c2n
...
cm1 cm2 ... cmn

.
AC = a11c11 a12c12 ... a1nc1n
a21c21 a22c22 ... a2nc2n
...
am1cm1 am2cm2 ... amncmn

.
Because B = cA, c11, coefficients in matrix C are all constant c
So AC = c*a11 c*a12 ... c*a1n
c*a21 c*a22 ... c*a2n
...
c*am1 c*am2 ... c*amn

.
which is equal to cA. Then B = AC =cA.. Part (a) solved
For part b, I am not sure if it is actually asking to switch columns, if so..
C = 0 1
1 0

.
will do.
And part c will be,
C = 0 c
c 0
...
I have no idea if it is actually asking to switch ROWS instead :(
 
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  • #2
Poke said:

Homework Statement


Let A be an arbitrary m× n matrix. Find a matrix C such that CA = B for each of the following matrices B.
a. B is the matrix that results from multiplying row i of A by a nonzero number c.
b. B is the matrix that results from swapping rows i and j of A.
c. B is the matrix that results from adding c times row i of A to row j of A.

Homework Equations


Multiply the two matrices

The Attempt at a Solution


(a)
Let A = a11 a12 ... a1n
a21 a22 ... a2n
...
am1 am2 ... amn
.
And C = c11 c12 ... c1n
c21 c22 ... c2n
...
cm1 cm2 ... cmn

.
AC = a11c11 a12c12 ... a1nc1n
a21c21 a22c22 ... a2nc2n
...
am1cm1 am2cm2 ... amncmn

.
Because B = cA, c11, coefficients in matrix C are all constant c
So AC = c*a11 c*a12 ... c*a1n
c*a21 c*a22 ... c*a2n
...
c*am1 c*am2 ... c*amn

.
which is equal to cA. Then B = AC =cA.. Part (a) solved
For part b, I am not sure if it is actually asking to switch columns, if so..
C = 0 1
1 0

.
will do.
And part c will be,
C = 0 c
c 0
...
I have no idea if it is actually asking to switch ROWS instead :(

You need to review matrix multiplication. You wrote that
$$\begin{bmatrix} a_{11}& a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}
\cdot
\begin{bmatrix} c_{11} & c_{12} & \cdots & c_{1n} \\
c_{21} & c_{22} & \cdots & c_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
c_{m1} & c_{m2} & \cdots & c_{mn} \end{bmatrix}
$$
is equal to
$$\begin{bmatrix} a_{11} c_{11} & a_{12} c_{12} & \cdots & a_{1n} c_{1n} \\
a_{21} c_{21} & a_{22} c_{22} & \cdots & a_{2n} c_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} c_{m1} & a_{m2} c_{m2}& \cdots & a_{mn} c_{mn} \end{bmatrix}
$$
That is absolutely false.

Besides. you were asked about CA, not AC. For matrices, these two are generally different; in fact, if ##m \neq n## only one of the two matrix products can exist at all! You could have CA with C being and ##m \times m## matrix or you could have AD, with D being an ##n \times n## matrix, but when ##m \neq n## you cannot have both CA and AC with exactly the same C.

If you have a textbook I suggest you go back and read the sections on matrix multiplication; if you do not have a textbook, Google "matrix multiplication" to see on-line sources.
 
Last edited:
  • #3
Another idea is to test with the case ##n=m=2 ## which should give you an idea of the solution.
 

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations, vector spaces, and linear transformations. It involves the use of matrices and their operations to solve problems related to lines, planes, and higher-dimensional objects.

2. What does it mean to find a matrix C st CA = B?

Finding a matrix C such that CA = B means finding a matrix C that, when multiplied with another matrix A, results in the matrix B. In other words, it is finding the solution to a system of linear equations represented by the matrix equation CA = B.

3. What is the importance of finding a matrix C st CA = B?

Finding a matrix C st CA = B is important because it allows us to solve systems of linear equations efficiently. It is also essential in applications such as data analysis, computer graphics, and machine learning.

4. How do I find a matrix C st CA = B?

To find a matrix C st CA = B, you can use the method of Gaussian elimination or the method of inverse matrices. These methods involve performing a series of elementary row operations on the augmented matrix [A|B] to transform it into the form [I|C], where I is the identity matrix. The resulting matrix C will be the solution to the equation CA = B.

5. Are there any special cases to consider when finding a matrix C st CA = B?

Yes, there are a few special cases to consider when finding a matrix C st CA = B. If the matrix A is not invertible, then there will be no unique solution for C, and the system may have either infinitely many solutions or no solutions at all. Additionally, if the dimensions of A and B do not match, then it is not possible to find a matrix C st CA = B.

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