By using properties of determinants ,show that: \(\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}\)=0
We have,
△= \(\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}\)
Applying R1 \(\to\) cR1,we have
△=\(\frac{1}{c}\begin{vmatrix}0&ac&-bc\\-a&0&-c\\b&c&0\end{vmatrix}\)
Applying R11\(\to\) R1-bR2,we have
△=\(\frac{1}{c}\begin{vmatrix}ab&ac&0\\-a&0&-c\\b&c&0\end{vmatrix}\)
=\(\frac{1}{c}\begin{vmatrix}b&c&0\\-a&0&-c\\b&c&0\end{vmatrix}\)
Here, the two rows R1 and R3 are identical.
∴∆ = 0.
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 _______
Read More: Properties of Determinants