The correct answer is 5376
Sum of all diagonal elements is equal to sum of square of each element of the matrix.
\(i.e., A =\) \(\begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{bmatrix}\)
then \(t_r(A.A^T)\)
\(= a^{2}_{1}+a^{2}_{2}+a^{2}_{3}+b^{2}_{1}+b^{2}_{2}+b^{2}_{3}+c^{2}_{1}+c^{2}_{2}+c^{2}_{3}\)
\(∵ a_i, b_i, c_i ∈{-1,0,1} \)for \(i = 1,2,3\)
∴ Exactly three of them are zero and rest are 1 or – 1.
Total number of possible matrices
\(^9C_3×2^6\)
\(= \frac{9×8×7}{6}×64\)
= 5376
Consider the following cell: $ \text{Pt}(s) \, \text{H}_2 (1 \, \text{atm}) | \text{H}^+ (1 \, \text{M}) | \text{Cr}_2\text{O}_7^{2-}, \, \text{Cr}^{3+} | \text{H}^+ (1 \, \text{M}) | \text{Pt}(s) $
Given: $ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
At equilibrium: $ \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
Objective: $ \text{Determine the pH at the cathode where } E_{\text{cell}} = 0. $
A matrix is a rectangular array of numbers, variables, symbols, or expressions that are defined for the operations like subtraction, addition, and multiplications. The size of a matrix is determined by the number of rows and columns in the matrix.