#### Need solution for RD Sharma maths class 12 chapter Determinants exercise 5.1 question 1 subquestion (vii)

\begin{aligned} &\mathrm{M}_{11}=-9, \mathrm{M}_{21}=9, \mathrm{M}_{31}=-9, \mathrm{M}_{41}=0 \\ &\mathrm{C}_{11}=-9, \mathrm{C}_{21}=-9, \mathrm{C}_{31}=-9, \mathrm{C}_{41}=0 \end{aligned}

Hint:

Let Mij, Cij represents minor and cofactor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present.

Given:

\begin{aligned} &\mathrm{A}=\left[\begin{array}{cccc} 2 & -1 & 0 & 1 \\ -3 & 0 & 1 & -2 \\ 1 & 1 & -1 & 1 \\ 2 & -1 & 5 & 0 \end{array}\right] \\ \end{aligned}

Solution:

\begin{aligned} &\mathrm{A}=\left[\begin{array}{cccc} 2 & -1 & 0 & 1 \\ -3 & 0 & 1 & -2 \\ 1 & 1 & -1 & 1 \\ 2 & -1 & 5 & 0 \end{array}\right] \\ &\begin{aligned} M_{11}=\left|\begin{array}{ccc} 0 & 1 & -2 \\ 1 & -1 & 1 \\ -1 & 5 & 0 \end{array}\right| &=0(0-5)-1(0+1)-2(5-1) \\ &=-1-8 \\ &=-9 \end{aligned} \end{aligned}

\begin{aligned} \mathrm{M}_{21}=\left|\begin{array}{ccc} -1 & 0 & 1 \\ 1 & -1 & 1 \\ -1 & 5 & 0 \end{array}\right| &=-1(0-5)+1(5-1) \\ &=5+4 \\ &=9 \end{aligned}

\begin{aligned} &\mathrm{M}_{31}=\left|\begin{array}{ccc} -1 & 0 & 1 \\ 0 & 1 & -2 \\ -1 & 5 & 0 \end{array}\right| \\ & =-1(0+10)+1(0+1) \\ & =-10+1 \\ & =-9 \\ & \begin{array}{rll} M_{41}=\left|\begin{array}{ccc} -1 & 0 & 1 \\ 0 & 1 & -2 \\ 1 & -1 & 1 \end{array}\right| & =-1(1-2)+1(0-1) \\ & =1-1 \\ & =0 \end{array} \end{aligned}

\begin{aligned} &C_{i j}=(-1)^{\mathrm{i}+j} M_{i j} \\ &C_{11}=(-1)^{2}(-9)=-9 \\ &C_{21}=(-1)^{3}(9)=-9 \\ &C_{31}=(-1)^{4}(-9)=-9 \\ &C_{41}=(-1)^{5}(0)=0 \end{aligned}

\begin{aligned} \mathrm{D} &=2\left|\begin{array}{ccc} 0 & 1 & -2 \\ 1 & -1 & 1 \\ -1 & 5 & 0 \end{array}\right|+1\left|\begin{array}{ccc} -3 & 1 & -2 \\ 1 & -1 & 1 \\ 2 & 5 & 0 \end{array}\right|-1\left|\begin{array}{ccc} -3 & 0 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 5 \end{array}\right| \\ &=-18-27+15 \\ &=-30 \end{aligned}

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