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 \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) 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
Standard electrode potential for \( \text{Sn}^{4+}/\text{Sn}^{2+} \) couple is +0.15 V and that for the \( \text{Cr}^{3+}/\text{Cr} \) couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be:
To calculate the cell potential (\( E^\circ_{\text{cell}} \)), we use the standard electrode potentials of the given redox couples.
Given data:
\( E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15V \)
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74V \)
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Read More: Properties of Determinants