Question

The value of \(K\),If \(\begin{bmatrix}    1 & K & 3          \\[0.3em]        3 & K   & -2 \\[0.3em]  2  & 3 & -1      \end{bmatrix}=33\),is :

• A. \(-1\)
• B. \(0\)
• C. \(1\)
• D. \(2\)

Answer 1

The Correct Option is B

To find the value of \(K\) for which the determinant of the matrix \(\begin{bmatrix}1 & K & 3 \\ 3 & K & -2 \\ 2 & 3 & -1\end{bmatrix}\) is 33, we calculate the determinant using the formula for a 3x3 matrix:

\[\text{Determinant} = a(ei-fh) - b(di-fg) + c(dh-eg)\]

Given the matrix:

1	K	3
3	K	-2
2	3	-1

Substituting into the determinant formula, we have:

\[1((K \times -1) - (-2 \times 3)) - K((3 \times -1) - (-2 \times 2)) + 3((3 \times 3) - (K \times 2))\]

Simplifying each term:

\[1(-K + 6) - K(-3 + 4) + 3(9 - 2K)\]

\[= 1(-K + 6) + K + 3(9 - 2K)\]

\[= -K + 6 + K + 27 - 6K\]

Simplifying further:

\[= 33 - 6K\]

We set the determinant equal to 33:

\[33 - 6K = 33\]

Solving for \(K\), we subtract 33 from both sides:

\[-6K = 0\]

Thus, dividing both sides by -6 gives:

\[K = 0\]

Therefore, the value of \(K\) is \(0\).