Question

assignment 6: problem 12
(1 point)
find the value of ( k ) for which the matrix
( a=left\begin{array}{ccc}-6 & -3 & -3 \\ 2 & 4 & -2 \\ 9 & -2 & kend{array}
ight )
has rank 2.
( k=)
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Explanation:

Step1: Row - reduce the matrix

Given \(A=

$$\begin{bmatrix}-6&-3&-3\\2&4&-2\\9&-2&k\end{bmatrix}$$

\)
First, \(R_1 = R_1\div(- 3)\), we get \(

$$\begin{bmatrix}2&1&1\\2&4&-2\\9&-2&k\end{bmatrix}$$

\)
Then \(R_2=R_2 - R_1\), \(R_3 = R_3-\frac{9}{2}R_1\)
\(R_2=

$$\begin{bmatrix}2&1&1\\0&3&-3\\9&-2&k\end{bmatrix}$$

\), \(R_3=

$$\begin{bmatrix}2&1&1\\0&3&-3\\0&-\frac{13}{2}&k-\frac{9}{2}\end{bmatrix}$$

\)
Next, \(R_2 = R_2\div3\), we have \(

$$\begin{bmatrix}2&1&1\\0&1&-1\\0&-\frac{13}{2}&k - \frac{9}{2}\end{bmatrix}$$

\)
Then \(R_3=R_3+\frac{13}{2}R_2\)
\(R_3=

$$\begin{bmatrix}2&1&1\\0&1&-1\\0&0&k - \frac{9}{2}-\frac{13}{2}\end{bmatrix}$$

\)

Step2: Use the rank condition

Since \(\text{rank}(A) = 2\), the third - row must be all zeros.
Set \(k-\frac{9 + 13}{2}=0\)

Answer:

\(k = 11\)