Question

If |A|=7 where A=[abc/def/gbc] then det (2A)-1=​

Answer 1

Given : |A|=7 ,  $A=\left[\begin{array}{ccc}a&b&b\\d&e&f\\g&b&c\end{array}\right]$

To find : Det ( 2A)⁻¹

Solution:

| A | = 7

| kA|  = kⁿ|A|

where n is the order of matrix

$\left[\begin{array}{ccc}a&b&b\\d&e&f\\g&b&c\end{array}\right]$

Matrix is of order 3 x 3

=> n = 3

=> | 2 A |  = 2³ | A |

=> | 2 A |  = 8 * 7

=> | 2 A |  = 56

|B⁻¹ |  = 1/| B |

=> Det ( (2A)⁻¹ )  = 1/56

| (2A)⁻¹ | = 1/56

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