Find all values of k, if any, that satisfy the equation

Thanks, what would the matrix multiplication look like in this case?

Does the matrix expand..
 
frctl said:
View attachment 13492

What should I do here?
Click to expand...
Where did you get this problem if you don't know how to do matrix multiplication. One way to do a matrix multiplication is to think of the "ith" row of the first matrix and the "jth" column of the second matrix as vectors. Then the "ith row, jth column" of the product is the dot product of those two vectors. Here the matrix on the left has only one row, \(\displaystyle \begin{bmatrix}k & 1 & 1 \end{bmatrix}\). The first column of the next matrix is \(\displaystyle \begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}\). The dot product of those vectors is k(1)+ 1(1)+ 1(0)= k+ 1. Similarly the second column is \(\displaystyle \begin{bmatrix}1 \\ 0 \\ 2 \end{bmatrix}\) and the dot product of that with \(\displaystyle \begin{bmatrix}k & 1 & 1 \end{bmatrix}\) is k(1)+ 1(0)+ 1(2)= k+ 2. Finally the third column is \(\displaystyle \begin{bmatrix}0 \\ 2 \\ -3 \end{bmatrix}\) and the dot product of that with \(\displaystyle \begin{bmatrix}k & 1 & 1 \end{bmatrix}\) is k(0)+ 1(2)+ 1(-3)= -1.

So the first matrix multiplication is \(\displaystyle \begin{bmatrix}k & 1 & 1 \end{bmatrix}\begin{bmatrix}1 & 1 & 0 \\ 1 & 0 & 2\\ 0 & 2 & -3 \end{bmatrix}= \begin{bmatrix} k+ 1 & k+ 2 & -1 \end{bmatrix}\).

So now the problem is reduced to \(\displaystyle \begin{bmatrix}k+ 1 & k+ 2 & -1 \end{bmatrix}\begin{bmatrix}k \\ 1 \\ 1 \end{bmatrix}= k(k+1)+ (k+2)(1)+ (-1)(1)= k^2+ 2k+ 1= 0\). Can you solve that equation for k?.................... edited
 
frctl said:
Thanks, what would the matrix multiplication look like in this case?
Does the matrix expand..
Click to expand...
Look at the link. By removing Solve & =0 I got the program do the multiplication.
Don't be so lazy. Learn to use the resources available to you.
 
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