Chapter E, Problem 1E

To determine

To find:

The value of the determinant:

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

Answer to Problem 1E

Solution:

$26$

Explanation of Solution

Concept:

To find the value of a second order $2\times 2$ determinant:

$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=a\times d-b\times c=ad-bc$

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:

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

The above is a second order $2\times 2$ determinant.

So by the given concept, it can be expanded as:

$\left|\begin{array}{cc}3& 5\\ -1& 7\end{array}\right|=3\times 7-5\times \left(-1\right)$

Simplify further:

$\left|\begin{array}{cc}3& 5\\ -1& 7\end{array}\right|=21+5$

$\left|\begin{array}{cc}3& 5\\ -1& 7\end{array}\right|=26$

Thus the value of the given determinant is $26.$

Conclusion:

Therefore, the required value of the determinant is $26.$