#### Provide solution for RD Sharma maths class 12 chapter Determinants exercise 5.1 question 5

$-140$

Hint:

Determinant matrix must be square (i.e. same number of rows and columns)

Given:

\begin{aligned} \left|\begin{array}{ccc} 2 & 3 & -5 \\ 7 & 1 & -2 \\ -3 & 4 & 1 \end{array}\right|\\ \end{aligned}

Solution:

\begin{aligned} &\Delta=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21+} \mathrm{a}_{31} \mathrm{C}_{31} \end{aligned}

\begin{aligned} &=(-1)^{1+1} 2(1+8)+(-1)^{1+2} 3(7-6)+(-1)^{1+3}(-5)(28+3) \\ &=2(1+8)-3(7-6)-5(28+3) \\ &=18-3-155 \\ &\Delta=-140 \end{aligned}

Second method is the Sarus method where we adjoin the first two columns to the right to get:

$\left|\begin{array}{ccccc} 2 & 3 & -5 & 2 & 3 \\ 7 & 1 & -2 & 7 & 1 \\ -3 & 4 & 1 & -3 & 4 \end{array}\right|$

\begin{aligned} \Delta &=\{((2 \times 1 \times 1)+(3 \times-2 \times-3)-(5 \times 7 \times 4))-((-5 \times 1 \times-3)+(2 \times-2 \times 4)+(3 \times 7 \times 1))\} \\ &=(2+18-140)-(15-16+21) \\ &=-120-20 \\ \Delta &=-140 \end{aligned}