Download App

Algebra of Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices5 Marks
Question
If $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix},$ prove that $\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}$
✓

Answer

Given: $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
Now,
$\text{P}\text{Q}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
$ =\begin{bmatrix}\text{xa}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{y}\text{b}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{zc}\end{bmatrix}$
$ =\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}\ \dots(4)$
Also,
$\text{Q}\text{P}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
$=\begin{bmatrix}\text{ax}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{by}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{cz}\end{bmatrix}$
$=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}\ \dots(5)$
From (4) and (5), we get
$\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}$

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 Algebra of Matrices→Algebra of Matrices — 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 foot of the perpendicular from (1, 2, -3) on the line $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{-2}=\frac{\text{z}}{-1}.$
→
Show that the points A, B, C with position vectors $2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}, \hat{\text{i}} - 3\hat{\text{j}} - 5\hat{\text{k}} \text{ and } 3\hat{\text{i}} - 4\hat{\text{j}} - 4\hat{\text{k}}$ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle.
→
Find the values of x and y if the vectors $\vec{\text{a}}=3\hat{\text{i}}+\text{x}\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+\text{y}\hat{\text{k}}$ are mutually perpendicular vectors of equal magnitude.
→
A wire of length $28m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined areas of the square and the circle is minimum?
→
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{e}^{3\text{x}}\sin4\text{x}\times2^\text{x}$
→
Evaluate the following integrals as limit of sum:
$\int\limits^2_0(\text{x}+3)\text{dx}$
→
$\text{If } x^{\text{m }} \text{y}^{\text{n}} = ({x + \text{y)}^{\text{m + n}}}, \text{prove that} \frac{\text{d}^{2}\text{y}}{\text{d}x^{2}} = 0.$
→
Three schools A, B and C want to award their selected students for the values of Honesty, Regularity and Hard work. Each school decided to award a sum of $\text{₹ 2,500, ₹ 3,100, ₹ 5,100}$ per student for the respective values. The number of students to be awarded by the three schools is given below in the table:
School A B C
Values
Honesty 3 4 6
Regularity 4 5 2
Hard work 6 3 2
Find the total money given in awards by the three schools separately, using matrices. Apart from the above-given values, suggest one more value which should be considered for giving award.
→
Differentiate the following functions with respect to x:
$\text{x}^{\frac{1}{\text{x}}}$
→
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
→