Download App

DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS3 Marks
Question
Using the proprties of determinants in Exercise:
$\begin{vmatrix}\text{a}^2+2\text{a}&2\text{a}+1&1\\2\text{a}+1& \text{a}+2&1\\3&3&1\end{vmatrix}=(\text{a}-1)^3$
✓

Answer

We have to prove,
$\begin{vmatrix}\text{a}^2+2\text{a}&2\text{a}+1&1\\2\text{a}+1& \text{a}+2&1\\3&3&1\end{vmatrix}=(\text{a}-1)^3$
$\therefore\ \text{L.H.S}=\begin{vmatrix}\text{a}^2+2\text{a}&2\text{a}+1&1\\2\text{a}+1& \text{a}+2&1\\3&3&1\end{vmatrix}$
$=\begin{vmatrix}\text{a}^2+2\text{a}-2\text{a}-1&2\text{a}+1-\text{a}-2&0\\2\text{a}+1-3&\text{a}+2-3&0\\3&3&1\end{vmatrix}$
$[\because\ \text{R}_1\rightarrow\text{R}_1-\text{R}_2\text{ and R}_2\rightarrow\text{R}_2-\text{R}_3]$
$=\begin{vmatrix}(\text{a}-1)(\text{a}+1)&(\text{a}-1)&0\\2(\text{a}-1)&(\text{a}-1)&0\\3&3&1\end{vmatrix}$
$=(\text{a}-1)^2\begin{vmatrix}(\text{a}+1)&1&0\\2&1&0\\3&3&1\end{vmatrix}$
$[\text{Taking (a}-1)\text{ common from R}_1\text{ and R}_2\text{ each}]$
$=(\text{a}-1)^2\big[1(\text{a}+1)-2\big]=(\text{a}-1)^3$
$=\text{R.H.S}$
Hence proved.

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 "3 Marks Question" from DETERMINANTS→DETERMINANTS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the value of $\lambda$ so that the following vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\vec{\text{c}}=\lambda\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$
→
Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=7$ and $\vec{\text{r}}\cdot(\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})=26$
→
Find $\int e^{-x} \sin 2 x d dx$. Hence show that $\int_{-\pi / 4}^{\pi / 4} e^{-x}|\sin 2 x| d x=\frac{1}{5}\left(4+e^{\pi / 4}-e^{-\pi / 4}\right)$
→
Find the volume of the parallelopiped whose coterminous edges are represented by the vectors:
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$
→
If $\vec{a}=(\hat{i}-\hat{j}), \vec{b}=(3 \hat{j}-\hat{k})$ and $\vec{c}=(7 \hat{i}-\hat{k})$, find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and for which $\vec{c} \cdot \vec{d}=1$
→
Show that $\text{y}=\text{Ae}^{\text{bx}}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{y}}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2.$
→
Evaluate the following integrals:
$\int\tan^{-1}\Big(\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
→
Find the coordinates of the point on the curve $y^2 = 3 - 4x$ where tangent is parallel to the line $2x + y - 2 = 0$.
→
Find the unit vector in the direction of sum of vectors $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{j}}+\hat{\text{k}}.$
→
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}(\text{x})=\text{x}^{3}-6\text{x}^{2}+9\text{x}+15$
→