Using the property of determinants and without expanding, prove that: \(\begin{vmatrix}x&a&x+a\\y&b&y+b\\z&C&z+c\end{vmatrix}=0\)
\(\begin{vmatrix}x&a&x+a\\y&b&y+b\\z&C&z+c\end{vmatrix}\)
=\(\begin{vmatrix}x&a&x\\y&b&y\\z&c&z\end{vmatrix}+\begin{vmatrix}x&a&a\\y&b&b\\z&c&c\end{vmatrix}=0+0=0\)
[Here the two columns of the determinants are identical]
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