Download App

Solution of Simultaneous Linear Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsSolution of Simultaneous Linear Equations5 Marks
Question
If $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}$, find $A^{-1}.$ Using $A^{-1},$ solve the system of linear equations:
$x - 2y = 10, 2x + y + 3z = 8, -2y + z = 7$
✓

Answer

Here, $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}$ $\text{|A|}=1{(1+6)}+2{(2-0)}+0{(-4-0)}$ $=7+4+0$ $=11$
Let $C_{ij}$ be the co-factors of the elements $a_{ij}$ in $A = [a_{ij}]$,
Then, $\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}1&3\\ -2&1\end{vmatrix}=7,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}2&3\\ 0&1\end{vmatrix}=-2,\\ \text{C}_{13}={(-1)}^{1+3}\begin{vmatrix}2&1\\ 0&-2\end{vmatrix}=-4$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}-2&0\\ -2&1\end{vmatrix}=2,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}1&0\\ 0&1\end{vmatrix}=2,\\ \text{C}_{23}={(-1)}^{2+3}\begin{vmatrix}1&-2\\ 0&-2\end{vmatrix}=2$ $\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}-2&0\\ 1&3\end{vmatrix}=-6,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}1&0\\ 2&3\end{vmatrix}=-3,\\ \text{C}_{23}={(-1)}^{3+3}\begin{vmatrix}1&-2\\ 2&1\end{vmatrix}=5$
$\therefore\text{adj A}=\begin{bmatrix}7&-2&-4\\ 2&1&2\\ -6&-3&5\end{bmatrix}^\text{T}$
$=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$ $\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{11}\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$ or,
$\text{AX = B}$ where, $\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$
Now, $\therefore\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{11}\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}\begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{11}\begin{bmatrix}70+16-42\ \\ -20+8-21\\ -40+16+35\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{11}\begin{bmatrix}44\\ -33\\ 11\end{bmatrix}$ $\therefore$ $x = 4, y = -3$ and $z = 1$

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.(5 Marks)" from Solution of Simultaneous Linear Equations→Solution of Simultaneous Linear Equations — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $\text{A}=\begin{bmatrix}0&1&0\\0&0&1\\\text{p}&\text{q}&\text{r}\end{bmatrix},$ and I is the identity matrix of order $3,$ show that $A^3 = pI + qA + rA^2.$
→
Find the equation of the plane passing through the intersection of the planes $2x + 3y - z + 1 = 0$ and $x + y - 2z + 3 = 0$ and perpendicular to the plane $3x - y - 2z - 4 = 0$.
→
Solve the following equation for x:
$\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\text{n}\pi+\frac{3\pi}{4}$
→
Find the maximum and the minimum values, if any, without using derivaives of the following functions:f(x) = sin2x + 5 on R.
→
Find the absolute maximum and minimum values of the function of given by
$\text{f}(\text{x})=\cos^{2}\text{x}+\sin\text{x}, \text{x}\in[0,\pi]$
→
Find a point on the parabola $y = (x - 4)^2$​​​​​​​, where the tangent is parallel to the chord joining $(4, 0)$ and $(5, 1).$
→
A class has 15 students whose ages are 14, 17, 15, 14, 21, 19, 20, 16, 18, 17, 20, 17, 16, 19 and 20 years respectively. One student is selected in such a manner that each has the same chance to being selected and the age X of the selected student is recorded. What is the probability distribution of the random variable X.
→
Find the equations of the tangent and the normal to the following curves at the indicated points.
$y = 2x^2 − 3x − 1 at (1, −2)$
→
Find the particular solution of the differential equation $(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x +2y},$ given that when x = 1, y = 0.
→
If $\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
→