Question

Without expanding, evaluate the following determinants:
(a) $$\left|\begin{array}{lll}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|$$
(b) $$\left|\begin{array}{lll}2& 3& 4\\ 5& 6& 8\\ 6x& 9x& 12x\end{array}\right|$$ (c) $$\left|\begin{array}{lll}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|$$

Answer 1

Hint:
Here, we need to find the values of the given determinants. We will use the properties of determinants to evaluate the value of the determinants without expanding.

Complete step by step solution:
We will use the properties of determinants to evaluate the given determinants.
(a)
We know that if any two rows or columns of the determinant are added, then the value of the determinant remains the same.
Rewriting column 3 as a sum of column 2 and column 3, we get
$$\Rightarrow \left|\begin{array}{lll}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|=\left|\begin{array}{lll}1& a& a+b+c\\ 1& b& a+b+c\\ 1& c& a+b+c\end{array}\right|$$
Taking $$a+b+c$$ common from column 3, we get
$$\Rightarrow \left|\begin{array}{lll}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|=\left(a+b+c\right)\left|\begin{array}{lll}1& a& 1\\ 1& b& 1\\ 1& c& 1\end{array}\right|$$
If any two rows or columns of a determinant are equal, then its value is zero.
Since column 1 and column 3 are equal, we get
$$\Rightarrow \left|\begin{array}{lll}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|=\left(a+b+c\right)\left(0\right)$$
Multiplying the terms, we get
$$\Rightarrow \left|\begin{array}{lll}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|=0$$
Thus, we get the value of the determinant $$\left|\begin{array}{lll}1& a& b+c\\ 1& b& c+a\\ 1& c& a+b\end{array}\right|$$ as 0.
(b)
If any row or column of a determinant is multiplied by a scalar quantity, then the value of the determinant is also multiplied by the same scalar quantity.
Multiplying column 2 by the scalar quantity $${\displaystyle \frac{4}{3}}$$, we get
$$\Rightarrow \left|\begin{array}{lll}2& 3\times {\displaystyle \frac{4}{3}}& 4\\ 5& 6\times {\displaystyle \frac{4}{3}}& 8\\ 6x& 9x\times {\displaystyle \frac{4}{3}}& 12x\end{array}\right|={\displaystyle \frac{4}{3}}\left|\begin{array}{lll}2& 3& 4\\ 5& 6& 8\\ 6x& 9x& 12x\end{array}\right|$$
Multiplying the terms, we get
$$\Rightarrow \left|\begin{array}{lll}2& 4& 4\\ 5& 8& 8\\ 6x& 12x& 12x\end{array}\right|={\displaystyle \frac{4}{3}}\left|\begin{array}{lll}2& 3& 4\\ 5& 6& 8\\ 6x& 9x& 12x\end{array}\right|$$
We know that if any two columns or rows of a determinant are equal, then its value is zero.
Since column 2 and column 3 are equal, we get
$$\Rightarrow 0={\displaystyle \frac{4}{3}}\left|\begin{array}{lll}2& 3& 4\\ 5& 6& 8\\ 6x& 9x& 12x\end{array}\right|$$
Dividing both sides by $${\displaystyle \frac{4}{3}}$$, we get
$$\begin{array}{l}\Rightarrow {\displaystyle \frac{0}{{\displaystyle \frac{4}{3}}}}={\displaystyle \frac{{\displaystyle \frac{4}{3}}\left|\begin{array}{ccc}2& 3& 4\\ 5& 6& 8\\ 6x& 9x& 12x\end{array}\right|}{{\displaystyle \frac{4}{3}}}}\\ \Rightarrow \left|\begin{array}{ccc}2& 3& 4\\ 5& 6& 8\\ 6x& 9x& 12x\end{array}\right|=0\end{array}$$
Thus, we get the value of the determinant $$\left|\begin{array}{lll}2& 3& 4\\ 5& 6& 8\\ 6x& 9x& 12x\end{array}\right|$$ as 0.
(c)
Rewriting column 3 as a difference of column 3 and 9 times column 2, that is $$C_3\to C_3-9C_2$$, we get
$$\Rightarrow \left|\begin{array}{lll}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|=\left|\begin{array}{lll}2& 7& 65-9\left(7\right)\\ 3& 8& 75-9\left(8\right)\\ 5& 9& 86-9\left(9\right)\end{array}\right|$$
Multiplying the terms, we get
$$\Rightarrow \left|\begin{array}{lll}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|=\left|\begin{array}{lll}2& 7& 65-63\\ 3& 8& 75-72\\ 5& 9& 86-81\end{array}\right|$$
Subtracting the terms, we get
$$\Rightarrow \left|\begin{array}{lll}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|=\left|\begin{array}{lll}2& 7& 2\\ 3& 8& 3\\ 5& 9& 5\end{array}\right|$$
We know that if any two columns or rows of a determinant are equal, then its value is zero.Since column 1 and column 3 are equal, we get
$$\Rightarrow \left|\begin{array}{lll}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|=0$$

Thus, we get the value of the determinant $$\left|\begin{array}{lll}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|$$ as 0.

Note:
We took $$a+b+c$$ common from column 3 of $$\left|\begin{array}{lll}1& a& a+b+c\\ 1& b& a+b+c\\ 1& c& a+b+c\end{array}\right|$$ to get $$\left(a+b+c\right)\left|\begin{array}{lll}1& a& 1\\ 1& b& 1\\ 1& c& 1\end{array}\right|$$. This is because of the property of scalar multiplication of a determinant. Multiplying any row or column by a scalar multiplies the value of the determinant by the same scalar quantity.
We need to also remember the properties of determinants:
1) If any two rows or columns of the determinant are added, then the value of the determinant remains the same.
2) If any two columns or rows of a determinant are equal, then its value is zero.
3) If any row or column of a determinant is multiplied by a scalar quantity, then the value of the determinant is also multiplied by the same scalar quantity.