Chapter 11.3, Problem 12AYU

To determine

To find: The value of the determinant $\left|\begin{array}{ccc}1& 3& -2\\ 6& 1& -5\\ 8& 2& 3\end{array}\right|$.

Answer to Problem 12AYU

$-169$

Explanation of Solution

Given:

$\left|\begin{array}{ccc}1& 3& -2\\ 6& 1& -5\\ 8& 2& 3\end{array}\right|$

Formula used:

To find the value of a third order $3\times 3$ determinant:

Step 1: Expand along the first row with signs alternatively $+$ and $-$:

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=a_1\left|\begin{array}{cc}{b}_2& c_2\\ b_3& c_3\end{array}\right|-b_1\left|\begin{array}{cc}{a}_2& c_2\\ a_3& c_3\end{array}\right|+c_1\left|\begin{array}{cc}{a}_2& b_2\\ a_3& b_3\end{array}\right|$

Step 2: Expand the second order minors to get the product:

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=a_1\left(b_2{c}_3-b_3{c}_2\right)-b_1\left(a_2{c}_3-a_3{c}_2\right)+c_1\left(a_2{b}_3-a_3{b}_2\right)$

Step 3: Simplify further:

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=a_1{b}_2{c}_3-a_1{b}_3{c}_2-b_1{a}_2{c}_3+b_1{a}_3{c}_2+c_1{a}_2{b}_3-c_1{a}_3{b}_2$

Calculation:

$D=\left|\begin{array}{ccc}1& 3& -2\\ 6& 1& -5\\ 8& 2& 3\end{array}\right|$

$D=1\times \left|\begin{array}{cc}1& -5\\ 2& 3\end{array}\right|-3\times \left|\begin{array}{cc}6& -5\\ 8& 3\end{array}\right|-2\times \left|\begin{array}{cc}6& 1\\ 8& 2\end{array}\right|$

$D=1.\left(3+10\right)-3.\left(18+40\right)-2.\left(12-8\right)$

$D=1\left(13\right)-3\left(58\right)-2\left(4\right)$

$D=13-174-8$

$D=-169$