Download App

Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics4 Marks
Question
If three numbers are added, their sum is $2.$ If two times the second number is subtracted from the sum of first and third numbers we get $8$ and if three times the first number is added to the sum of second and third numbers we get $4.$ Find the numbers using matrices.
✓

Answer

Let the three numbers $\text{x , y , z.}$
From given condition, we have
$x + y + z = 2 .......(1)$
$x + z - 2y = 8$
$x - 2y + z = 8 ......(2)$
And
$3x + y + z = 4 .....(3)$
Given all equation can be written in matrix form as ,
$\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & -2 & 1 \\ 3 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}2 \\ 8 \\ 4\end{array}\right]$
Consider , $AX = B$
On multiplying $A^{-1}$ both sides , we get
$X = A^{-1} . B ......(4)$
Now
$| A |=\left|\begin{array}{ccc}1 & 1 & 1 \\ 0 & -3 & 0 \\ 0 & -2 & -2\end{array}\right|\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}2 \\ 6 \\ -2\end{array}\right]$
$\left|\begin{array}{ccc}x+ & y+ & z \\ 0- & 3 y & +0 \\ 0 & -2 & -2\end{array}\right|=\left[\begin{array}{c}2 \\ 6 \\ -2\end{array}\right]$
By equality of matrices,
$x + y + z = 2 ……(1)$
$-3y = 6 ……(2)$
$– 2y – 2z = -2 ……..(3)$
From $(2), y = -2$
Substituting $y = -2$ in $(3),$ we get,
$-2(-2) – 2z$
$= -2$
$\therefore -2z = -6$
$\therefore z = 3$
Substituting $y = -2, z = 3$ in $(1),$ we get,
$x – 2 + 3$
$= 2$
$\therefore x = 1$
Hence, the required numbers are $1, -2$ and $3.$

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 "Solve the Following Question.(4 Marks)" from Matrics→Matrics — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

A point source of light is hung 30 feet directly above a straight horizontal path on which a man of 6 feet in height is walking. How fast will the man's shadow lengthen and how fast the tip of shadow is move when he is walking away from the light at the rate of $100 ft / min$
→
A box with a square base is to have an open top. The surface area of box is 147 sq. cm. What should be its dimensions in order that the volume is largest?
→
Diffrentiate the following w. r. t. x.$\cot ^{-1}\left(\frac{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}+\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}{\sqrt{1+\sin \left(\frac{4 x}{3}\right)}-\sqrt{1-\sin \left(\frac{4 x}{3}\right)}}\right)$
→
Find the derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ w.r.t. $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$.
→
$\int \frac{5 e ^x}{\left( e ^x+1\right)\left( e ^{2 x}+9\right)} d x$
→
Find the second order derivatives of the following:

$2 x^5-4 x^3-\frac{2}{x^2}-9$

→
Evaluate : $\int \sqrt{a^2+x^2} \cdot d x=\frac{x}{2} \cdot \sqrt{x^2+a^2}+\frac{a^2}{2} \cdot \log \left(x+\sqrt{x^2+a^2}\right)+c$
→
Solve the following equations by the method of reduction. $x + y + z = 1, 2x + 3y + 2z = 2$ and $x + y + 2z = 4.$
→
Solve the following differential equation $y^2dx + (xy + x^2) dy = 0$
→
Differentiate $\tan ^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$ w.r. to $\tan ^{-1}\left[\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right]$
→