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  • Please solve RD Sharma class 12 chapter Determinants exercise 5.1 question 8 maths textbook solution

Please solve RD Sharma class 12 chapter Determinants exercise 5.1 question 8 maths textbook solution

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Answer:

\begin{aligned} |A B|=|A||B| \\ \end{aligned}

Given:

 

\begin{aligned} & A=\left[\begin{array}{ll} 2 & 5 \\ 2 & 1 \end{array}\right], B=\left[\begin{array}{cc} 4 & -3 \\ 2 & 5 \end{array}\right] \\ & \text { To prove: }|A B|=|A||B| \end{aligned}

 

Solution:

Consider AB in LHS:

\begin{aligned} \mathrm{AB} &=\left[\begin{array}{ll} 2 & 5 \\ 2 & 1 \end{array}\right]\left[\begin{array}{cc} 4 & -3 \\ 2 & 5 \end{array}\right] \\ &=\left[\begin{array}{cc} 8+10 & -6+25 \\ 8+2 & -6+5 \end{array}\right] \\ &=\left[\begin{array}{cc} 18 & 19 \\ 10 & -1 \end{array}\right] \\ |A B| &=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21} \\ &=-18-190 \\ |A B| &=-208 \end{aligned}

Consider RHS:

\begin{aligned} &|A|=2-10=-8 \\ &|B|=20+6=26 \\ &\begin{array}{l} |A||B|=|-8||26| \\ =-208 \\ \therefore \text { LHS }=\text { RHS } \\ |A B|=|A||B| \text { is proved. } \end{array} \end{aligned}

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