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  • Please solve RD Sharma Class 12 Chapter 5 Determinants Exercise Very Short Answer Question 24 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 5 Determinants Exercise Very Short Answer Question 24 maths textbook solution.

Answers (1)

Answer:-4

Hint: Here we use basic concept of determinant of matrix

Given:

\rightarrow \mathrm{A} \text { is } 3 \times 3 \text { matrix }

So, n=3

and \left | A \right |=4

Solution :

\left | -A \right | for what we use below formula |k A|=k^{n}|A|

when k is constant
n is order of matrix

\rightarrow In |-A|,-1 is constant

So,

\begin{array}{ll} |k A|=k^{n}|A| \\ |-1 A|=(-1)^{n}|A| & {[k=-1]} \\ |-1 A|=(-1)^{3}|A| & {[n=3]} \end{array}

\begin{aligned} &|-1 A|=(-1)^{3} \times 4 \quad[|A|=4] \\ &|-1 A|=-4 \end{aligned}
 

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