Question

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

StartLayout left-bracket 1st Row 1st Column StartLayout 1st Row 1st Column 2 2nd Column 2 EndLayout 2nd Column k EndLayout right-bracket times Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 3 3rd Column 0 2nd Row 1st Column 3 2nd Column 0 3rd Column 5 3rd Row 1st Column 0 2nd Column 5 3rd Column -1 EndMatrix StartBinomialOrMatrix StartLayout 1st Row 1st Column 2 2nd Row 1st Column 2 EndLayout Choose k EndBinomialOrMatrix equals 0



Note: Enter "NA" if no answers. If multiple answers enter as a row matrix, in ascending order of size.

k equals

Answer 1

There are two such values of
`k`,
`k = 6-4√(3)` and
`k = 6+4√(3).`

Explanation:

The question is to find all values of
`k`, if any, that satisfy the following matrix equation.

`\begin{bmatrix}2 & 2 & k\end{bmatrix} \cdot \begin{bmatrix}1 & 1 & 0\\1 & 0 & 3 \\0 & 3 & -1\end{bmatrix} \cdot \begin{bmatrix}2 \\2\\k\\\end{bmatrix} = 0`

Let's multiply the first two matrices. We can do that, since the number of columns in the first matrix equals the number of rows in the second matrix, which means that their product is defined.

`\begin{bmatrix}2 \cdot 1 + 2 \cdot 1 + k \cdot 0 & 2 \cdot 1 + 2 \cdot 0+ k \cdot 3 & 2 \cdot 0 + 2 \cdot 3 + k \cdot (-1)\end{bmatrix} \cdot \begin{bmatrix}2\\2\\k\end{bmatrix}= 0`

Next, we need to solve the matrix equation

`\begin{bmatrix} 4 & 2+3k & 6-k\end{bmatrix} \cdot \begin{bmatrix}2\\2\\k\end{bmatrix}= 0`

Again, the number of columns in the first matrix equals the number of rows in the second matrix, which means that their product is defined and we can multiply them. The result will be
`1`×
`1` matrix, and that will be the dimension of the zero matrix, as well.

`\begin{bmatrix}4 \cdot 2 + (2+3k) \cdot2 + (6+k)\cdot k\end{bmatrix} = 0\\\begin{bmatrix}8 + 4 + 6 \cdotk + 6 \cdot k - k^2 \end{bmatrix} = 0 & \iff 8 + 4 + 6 \cdot k + 6 \cdot k - k^2 = 0`

Now, all we need to do is to solve the quadratic equation

`-k^2 +12k +12 = 0`

By using the well known formula, we obtain

`k_(1/2) = (-12 \pm √(12^2 - 4 \cdot (-1) \cdot 12))/(-2) = (-12 \pm √(192))/(-2) = (-12 \pm √(64 \cdot 3))/(-2) = (-12 \pm 8√(3))/(-2)`

Therefore, we obtain two values for
`\mathbf{k}`,
`k = 6-4√(3)` and
`k = 6+4√(3)`.