Download App

JUNE 2024 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsJUNE 20244 Marks
Question
Solve system of linear equation, using matrix method :
$\begin{aligned}
x+y+z & =6 \\
y+3 z & =11 \\
x+z & =2 y
\end{aligned}$
✓

Answer

Let first, second and third numbers be denoted by $x, y$ and $z$, respectively.
Then, according to given conditions, we have
$\begin{array}{l}
x+y+z=6 \\
y+3 z=11 \\
x+z=2 y \text { or } x-2 y+z=0
\end{array}$
This system can be written as $AX = B$, where,
$A =\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{c}6 \\ 11 \\ 0\end{array}\right]$
Hence, $| A |=1(1+6)-1(0-3)+1(0-1)=9 \neq 0$.
Now we find adj A.
$\begin{array}{l}
A_{11}=1(1+6)=7, \\
A_{12}=-(0-3)=3, \\
A_{13}=-1 \\
A_{21}=-(1+2)=-3,
\end{array}$
$\begin{array}{l}A_{22}=0 \\ A_{23}=-(-2-1)=3 \\ A_{31}=(3-1)=2, A_{32}=-(3-0)=-3, \\ A_{33}=(1-0)=1\end{array}$
Hence, adj $A =\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
Thus, $\quad A ^{-1}=\frac{1}{|A|} \operatorname{adj}( A )$
$=\frac{1}{9}\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
Since $\quad X=A^{-1} B$
$X =\frac{1}{9}\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]\left[\begin{array}{c}6 \\ 11 \\ 0\end{array}\right]$
or $\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{9}\left[\begin{array}{c}42-33+0 \\ 18+0+0 \\ -6+33+0\end{array}\right]$
$=\frac{1}{9}\left[\begin{array}{c}9 \\ 18 \\ 27\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
Thus, $\quad x=1, y=2, z=3$.

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

Similar questions

Show that the points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.
→
Prove that the line through A(0, –1, –1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(–4, 4, 4).
→
A particle moves along the curve y = x2 + 2x. At what point(s) on the curve are the x and y coordinates of the particle changing at the same rate?
→
Solve the following linear programming problem graphically.Minimise $\text{z = 3x + 5y}$
subject to the constraints
$\text{x + 2y}\geq 10$
$\text{x + y}\geq 6$
$\text{3x + y}\geq 8$
$\text{x, y}\geq 0.$
→
Find the image of the point with position vector $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ in the plane $\vec{\text{r}}. (2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=4.$ Also, find the position vectors of the foot of the prependicular and the equation of the perpendicular line through $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}.$
→
Evaluate the following integrals as limit of sum:
$\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$
→
Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
→
If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.
→
Find A-1, If $\text{A}=\begin{bmatrix}1&2&5\\ 1&-1&-1\\ 2&3&-1\end{bmatrix}$. Hence solve the follwing system of linear equations:
x + 2y +5z = 10, x- y - z = - 2, 2x + 3y - z = - 11
→
$\begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&-3\\-2&4\end{bmatrix}=\begin{bmatrix}-4&6\\-9&\text{x}\end{bmatrix}$ find x.
→