The value of, arraycc_5 10 & 2 \\[4pt] 2 & _10 5array, is Option 1: -3 Option 2: -7/2 Option 3: 1 Option 4: 0

Answer: -3. Solution: For a 2 x 2 matrix vmatrix a & b \\ c & d vmatrix, the determinant = ad - bc So our determinant: = (_5 10)(_10 5) - (2)(2) = (_5 10)(_10 5) - 4 Using the change of base identity: _a b x _b a = 1 This is because _a b = 1/(_b a), so multiplying them gives 1. Here, a = 5 and b = 10: _5 10 x _10 5 = 1 Substituting back: = 1 - 4 = -3

Algebra

Relations & Functions

Medium

applied

The value of $\left|\begin{array}{cc}\log_5 10 & 2 \\[4pt] 2 & \log_{10} 5\end{array}\right|$ is

$-3$

$-\frac{7}{2}$

$1$

$0$

✅ Correct Option: 1

For a $2 \times 2$ matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ , the determinant $= ad - bc$

So our determinant:

$= (\log_5 10)(\log_{10} 5) - (2)(2)$

$= (\log_5 10)(\log_{10} 5) - 4$

Using the change of base identity:

$\log_a b \times \log_b a = 1$

This is because $\log_a b = \dfrac{1}{\log_b a}$ , so multiplying them gives $1$ .

Here, $a = 5$ and $b = 10$ :

$\log_5 10 \times \log_{10} 5 = 1$

Substituting back:

$= 1 - 4$

$= -3$

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Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The value of $\left|\begin{array}{cc}\log_5 10 & 2 \\[4pt] 2 & \log_{10} 5\end{array}\right|$ is