Linear Algebra, Find a matrix C st CA = B

[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 :(​

 

[Ray Vickson]

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.

 

[fresh_42]

Another idea is to test with the case $n=m=2$ which should give you an idea of the solution.