By using properties of determinants, show that: \(\begin{vmatrix}-a^2&ab&ac\\ba&-b^2&bc\\ca&cb&-c^2\end{vmatrix}\)=4a2b2c2
△=\(\begin{vmatrix}-a^2&ab&ac\\ba&-b^2&bc\\ca&cb&-c^2\end{vmatrix}\)
=\(abc\begin{vmatrix}-a&b&c\\a&-b&c\\a&b&-c\end{vmatrix}\) [Taking out factors a,b,c from R1,R2and R3]
=a2b2c2 \(\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}\) [Taking out factors a,b,c from C1,C2and C3]
Applying R2 → R2 + R1 and R3 → R3 + R1, we have:
△=a2b2c2\(\begin{vmatrix}-1&1&1\\0&0&2\\0&2&0\end{vmatrix}\)
=a2b2c2(-1)\(\begin{vmatrix}0&2\\2&0\end{vmatrix}\)
=-a2b2c2(0-4)
=4a2b2c2
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:
Let I be the identity matrix of order 3 × 3 and for the matrix $ A = \begin{pmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{pmatrix} $, $ |A| = -1 $. Let B be the inverse of the matrix $ \text{adj}(A \cdot \text{adj}(A^2)) $. Then $ |(\lambda B + I)| $ is equal to _______
If $ y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\27 & 28 & 27 \\1 & 1 & 1 \end{vmatrix} $, $ x \in \mathbb{R} $, then $ \frac{d^2y}{dx^2} + y $ is equal to
Read More: Properties of Determinants