Chapter 11.3, Problem 14AYU

$\left[\begin{array}{l}\begin{array}{ccc}3& -9& 4\end{array}\\ \begin{array}{ccc}1& 4& 0\end{array}\\ \begin{array}{ccc}8& -3& 1\end{array}\end{array}\right]$

To determine

To find: The value of the determinant $\left|\begin{array}{ccc}3& -9& 4\\ 1& 4& 0\\ 8& -3& 1\end{array}\right|$.

Answer to Problem 14AYU

$-119$

Explanation of Solution

Given:

$\left|\begin{array}{ccc}3& -9& 4\\ 1& 4& 0\\ 8& -3& 1\end{array}\right|$

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:

$D=\left|\begin{array}{ccc}3& -9& 4\\ 1& 4& 0\\ 8& -3& 1\end{array}\right|$

$D=3\times \left|\begin{array}{cc}4& 0\\ -3& 1\end{array}\right|-\left(-9\right)\times \left|\begin{array}{cc}1& 0\\ 8& 1\end{array}\right|+4\times \left|\begin{array}{cc}1& 4\\ 8& -3\end{array}\right|$

$D=3.\left(4+0\right)+9.\left(1-0\right)+4.\left(-3-32\right)$

$D=3\left(4\right)+9\left(1\right)+4\left(-35\right)$

$D=12+9-140$

$D=-119$