Evaluate: 1&a&bc 1&b&ca 1&c&ab - Vidyadip

Question

Evaluate:
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$

✓

Answer

$\triangle=\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
When a = b, the first two rows become identical. Hence, a - b is a factor. Similarly, when b = c and c = a, the second and third and third and first rows become indetical. Hence, b - c and c - a are also factors. The degree of product of the diagonal elements is 3. Hence, there are no other factors.
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=\lambda(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$ [Where $\lambda$ is a constant]
$\begin{vmatrix}1&0&2\\1&1&0\\1&2&0\end{vmatrix}=2\lambda$ $[$Putting a = 0, b = 1 and c = 2 to find $\lambda]$
$\Rightarrow2=2\lambda$
$\Rightarrow\lambda=1$
Hence,
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from DETERMINANTS→DETERMINANTS — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Find graphically, the maximum value of $\text{z = 2x + 5y},$ subject to constraints given below:
$2x + 4y \leq 8$
$3x + y \leq 6$
$x + y \leq 4$
$x\geq 0, y\geq 0$

The function $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{\text{a}},&\text{if }0\leq\text{ x}<1\\\text{a},&\text{if }1\leq\text{x}<\sqrt{2}\\\frac{2\text{b}^2-4\text{b}}{\text{x}^2},&\text{if }\sqrt{2}\leq\text{x}<\infty\end{cases}$ is continuous on $(0,\infty),$ then find the most suitable value of a and b.

Evalute the following integrals:
$\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$

Find the area lying above the $x-$axis and under the paraola $y^2 = 4x - x^2.$

Sketch the region bounded by the curves $y = x^2 + 2, y = x, x = 0$ and $x = 1.$ Also find the area of this region.

Evaluate the following intregals:
$\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$

Find the distance of the point (-1, -5, -10) from the point of intersection of the line $\vec{\text{r}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}+\lambda\Big(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}}\Big)$ and the plane $\vec{\text{r}}.\Big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\Big)=5.$

If $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos^2\text{t}}},\text{y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$

Show that the points whose position vectors are$\vec{\text{a}}=4\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}, \vec{\text{b}}=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}-\hat{\text{j}}$ from a right triangle.

If $xy^2 = 1,$ prove that $2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$