Download App

MARCH 2024 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMARCH 20244 Marks
Question
Solve system of linear equations, using matrix method :
$\begin{array}{l}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6, \quad \frac{1}{y}+\frac{3}{z}=11, \\
\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=0
\end{array}$
✓

Answer


$\begin{array}{l}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6 \\ \frac{1}{y}+\frac{3}{z}=11 \\ \frac{1}{x}-\frac{2}{y}+\frac{1}{z}=0\end{array}$
$\Rightarrow$ The equation can be written as matrix form,
$\begin{array}{l}
{\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & -2 & 1
\end{array}\right]\left[\begin{array}{c}
\frac{1}{x} \\
\frac{1}{y} \\
\frac{1}{z}
\end{array}\right]=\left[\begin{array}{c}
6 \\
11 \\
0
\end{array}\right]} \\
\therefore A=\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & -2 & 1
\end{array}\right], X=\left[\begin{array}{c}
\frac{1}{x} \\
\frac{1}{y} \\
\frac{1}{z}
\end{array}\right], B=\left[\begin{array}{c}
6 \\
11 \\
0
\end{array}\right]
\end{array}$
$\begin{array}{l}
|A|=\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & -2 & 1
\end{array}\right] \\
=1(1+6)-1(0-3)+1(0-1) \\
=7+3-1 \\
=9 \neq 0 \\
\therefore \quad A^{-1} \text { exists. }
\end{array}$
$\operatorname{adj} A=\left[\begin{array}{rrr}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
$\therefore \quad A^{-1}=\frac{1}{|A|}$ adj A
$=\frac{1}{9}\left[\begin{array}{rrr}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
$\therefore$ $AX = B$
$\therefore$ $A ^{-1} AX = A ^{-1} B$
$\therefore$ $IX = A ^{-1} B$
$\therefore$ $X = A ^{-1} B$
$\begin{array}{l}=\frac{1}{9}\left[\begin{array}{rrr}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]\left[\begin{array}{c}6 \\ 11 \\ 0\end{array}\right] \\ =\frac{1}{9}\left[\begin{array}{rrr}42 & -33 & +0 \\ 18 & +0 & -0 \\ -6 & +33+0\end{array}\right] \\ =\frac{1}{9}\left[\begin{array}{c}9 \\ 18 \\ 27\end{array}\right]\end{array}$
$\left[\begin{array}{c}\frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z}\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
$\begin{array}{l}\therefore \quad \text { Solution : } \frac{1}{x}=1, \frac{1}{y}=2, \frac{1}{z}=3 \\ \quad \therefore x=1, y=\frac{1}{z}, z=\frac{1}{3}\end{array}$

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 "4 Marks" from MARCH 2024→MARCH 2024 — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Sketch the graph y = |x + 1|. Evaluate $\int\limits_{-4}^{2}|\text{x}+1|\text{dx} $ . What does this value of the integral represent on the graph.
→
$\int\frac{\text{x}+1}{\sqrt{2\text{x}+3}}\text{dx}$
→
There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommended daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grams of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using matrices. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the 2 families. What awareness can you create among people about the balanced diet from this question?
→
A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 units of calories. Two foods A and B are available at a cost of Rs. 5 and Rs. 4 per unit respectively. One unit of the food A contains 200 units of vitamins, 1 unit of minerals and 40 units of calories, while one unit of the food B contains 100 units of vitamins, 2 units of minerals and 40 units of calories. Find what combination of the foods A and B should be used to have least cost, but it must satisfy the requirements of the sick person. Form the question as LPP and solve it graphically.
→
Show that the function defined by f(x) = cos (x2 ) is a continuous function.
→
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)

$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}+1)\text{e}^{-\text{x}}$

→
Show that the lines $\frac{\text{x}-1}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-1}{5}$ and $\frac{\text{x}+2}{4}=\frac{\text{y}-1}{3}=\frac{\text{z}+1}{-2}$ do not intersect.
→
A factory owner purchases two types of machines, A and B, for his factory. The requirements and limitations for the machines are as follows:
 
Area occupied by the
machine
Labour force for each
machine
Daliy outputin
units
Machines
1000 sp.m
12 mem
60
Machines
1200 sp.m
8 mem
40
He has an area of 7600 sq. m available and 72 skilled men who can operate the machines.
How many machines of each type should he buy to maximize the daily output?
→
Differentiate the following functions with respect to x:
$\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2}$
→
For each of the differential equations given in find the general solution:

 

$\cos^2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\tan\text{x}\Big(0\leq\text{x}<\frac{\pi}{2}\Big)$

→