NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4

NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4 - Determinants

Edited By Ramraj Saini | Updated on Dec 03, 2023 04:15 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.6

NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4 Determinants are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In this article, you will get NCERT solutions for Class 12 Maths chapter 4 exercise 4.6 consists of questions related to applications of matrix and determinants. There are a lot of applications of matrices and determinants in the field of research, statistics, computer graphics, business, economics, and mathematics. In exercise 4.6 Class 12 Maths you will learn the specific application of matrix and determinant i.e. solving system of linear equation using the inverse of the matrix. All the questions in Class 12 Maths ch 4 ex 4.6 are related to solving the system of linear equations having unique solutions.

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In this Class 12 Maths chapter 4 exercise 4.6 solutions you will get questions related to finding consistency of linear equations. 12th class Maths exercise 4.6 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Determinants Exercise: 4.6

Question:1 Examine the consistency of the system of equations.

$\small x+2y=2$

$\small 2x+3y=3$

Answer:

We have given the system of equations:18967

$\small x+2y=2$

$\small 2x+3y=3$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &2 \\ 2&3 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 2\\3 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(3) -2(2) = -1 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:2 Examine the consistency of the system of equations

$\small 2x-y=5$

$\small x+y=4$

Answer:

We have given the system of equations:

$\small 2x-y=5$

$\small x+y=4$

The given system of equations can be written in the form of matrix; $AX =B$

where $A= \begin{bmatrix} 2 &-1 \\ 1&1 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\4 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 2(1) -1(-1) = 3 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:3 Examine the consistency of the system of equations.

$\small x+3y=5$

$\small 2x+6y=8$

Answer:

We have given the system of equations:

$\small x+3y=5$

$\small 2x+6y=8$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &3 \\ 2&6 \end{bmatrix}$ , $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\8 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(6) -2(3) = 0$

Here A is singular matrix therefore now we will check whether the $(adjA)B$ is zero or non-zero.

$adjA= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}$

So, $(adjA)B= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}\begin{bmatrix} 5\\8 \end{bmatrix} = \begin{bmatrix} 30-24\\-10+8 \end{bmatrix}=\begin{bmatrix} 6\\-2 \end{bmatrix} \neq 0$

As, $(adjA)B \neq 0$ , the solution of the given system of equations does not exist.

Hence, the given system of equations is inconsistent.

Question:4 Examine the consistency of the system of equations.

$\small x+y+z=1$

$\small 2x+3y+2z=2$

$\small ax+ay+2az=4$

Answer:

We have given the system of equations:

$\small x+y+z=1$

$\small 2x+3y+2z=2$

$\small ax+ay+2az=4$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 1& 1&1 \\ 2& 3& 2\\ a& a &2a \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 1\\2 \\ 4 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 1(6a-2a) -1(4a-2a)+1(2a-3a)$

$= 4a -2a-a = 4a -3a =a \neq 0$

[If zero then it won't satisfy the third equation]

Here A is non- singular matrix therefore there exist $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:5 Examine the consistency of the system of equations.

$\small 3x-y-2z=2$

$\small 2y-z=-1$

$\small 3x-5y=3$

Answer:

We have given the system of equations:

$\small 3x-y-2z=2$

$\small 2y-z=-1$

$\small 3x-5y=3$

The given system of equations can be written in the form of matrix; $AX =B$

where $A = \begin{bmatrix} 3& -1&-2 \\ 0& 2& -1\\ 3& -5 &0 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 3(0-5) -(-1)(0+3)-2(0-6)$

$= -15 +3+12 = 0$

Therefore matrix A is a singular matrix.

So, we will then check $(adjA)B,$

$(adjA) = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}$

$\therefore (adjA)B = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}\begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix} = \begin{bmatrix} -10-10+15\\ -6-6+9 \\ -12-12+18 \end{bmatrix} = \begin{bmatrix} -5\\-3 \\ -6 \end{bmatrix} \neq 0$

As, $(adjA)B$ is non-zero thus the solution of the given system of the equation does not exist. Hence, the given system of equations is inconsistent.

Question:6 Examine the consistency of the system of equations.

$\small 5x-y+4z=5$

$\small 2x+3y+5z=2$

$\small 5x-2y+6z=-1$

Answer:

We have given the system of equations:

$\small 5x-y+4z=5$

$\small 2x+3y+5z=2$

$\small 5x-2y+6z=-1$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 5& -1&4 \\ 2& 3& 5\\ 5& -2 &6 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 5\\2 \\ -1 \end{bmatrix}$ .

So, we want to check for the consistency of the equations;

$|A| = 5(18+10) +1(12-25)+4(-4-15)$

$= 140-13-76 = 51 \neq 0$

Here A is non- singular matrix therefore there exist $A^{-1}$ .

Hence, the given system of equations is consistent.

Question:7 Solve system of linear equations, using matrix method.

$\small 5x+2y=4$

$\small 7x+3y=5$

Answer:

The given system of equations

$\small 5x+2y=4$

$\small 7x+3y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &4 \\ 7& 3 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 4\\5 \end{bmatrix}$

we have,

$|A| = 15-14=1 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = (adjA) = \begin{bmatrix} 3 &-2 \\ -7& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 3 &-2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 12-10\\ -28+25 \end{bmatrix} = \begin{bmatrix} 2\\-3 \end{bmatrix}$

Hence the solutions of the given system of equations;

x = 2 and y =-3.

Question:8 Solve system of linear equations, using matrix method.

$2x-y=-2$

$3x+4y=3$

Answer:

The given system of equations

$2x-y=-2$

$3x+4y=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &-1 \\ 3& 4 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} -2\\3 \end{bmatrix}$

we have,

$|A| = 8+3=11 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3& 2 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} -2\\3 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -8+3\\ 6+6 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -5\\12 \end{bmatrix}= \begin{bmatrix} -\frac{5}{11}\\ \\-\frac{12}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-5}{11} \ and\ y =\frac{12}{11}.$

Question:9 Solve system of linear equations, using matrix method.

$\small 4x-3y=3$

$\small 3x-5y=7$

Answer:

The given system of equations

$\small 4x-3y=3$

$\small 3x-5y=7$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 4 &-3 \\ 3& -5 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\7 \end{bmatrix}$

we have,

$|A| = -20+9=-11 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{-1}{11}\begin{bmatrix} -5 &3 \\ -3& 4 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 5 &-3 \\ 3& -4 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 5 &-3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3\\7 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 15-21\\ 9-28 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -6\\-19 \end{bmatrix}= \begin{bmatrix} -\frac{6}{11}\\ \\-\frac{19}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-6}{11} \ and\ y =\frac{-19}{11}.$

Question:10 Solve system of linear equations, using matrix method.

$\small 5x+2y=3$

$\small 3x+2y=5$

Answer:

The given system of equations

$\small 5x+2y=3$

$\small 3x+2y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &2 \\ 3& 2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\5 \end{bmatrix}$

we have,

$|A| = 10-6=4 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3 & 5 \end{bmatrix}\begin{bmatrix} 3\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 6-10\\ -9+25 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -4\\16 \end{bmatrix}= \begin{bmatrix} -1\\4 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =-1 \ and\ y =4.$

Question:11 Solve system of linear equations, using matrix method.

$\small 2x+y+z=1$

$\small x-2y-z= \frac{3}{2}$

$\small 3y-5z=9$

Answer:

The given system of equations

$\small 2x+y+z=1$

$\small x-2y-z= \frac{3}{2}$

$\small 3y-5z=9$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &1 &1 \\ 1 & -2 &-1 \\ 0& 3 &-5 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ and $B =\begin{bmatrix} 1\\ \\ \frac{3}{2} \\ \\ 9 \end{bmatrix}$

we have,

$|A| =2(10+3)-1(-5-0)+1(3-0) = 26+5+3 = 34 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(10+3) = 13$ $A_{12} =(-1)^{1+2}(-5-0) = 5$

$A_{13} =(-1)^{1+3}(3-0) = 3$ $A_{21} =(-1)^{2+1}(-5-3) = 8$

$A_{22} =(-1)^{2+2}(-10-0) = -10$ $A_{23} =(-1)^{2+3}(6-0) = -6$

$A_{31} =(-1)^{3+1}(-1+2) = 1$ $A_{32} =(-1)^{3+2}(-2-1) = 3$

$A_{33} =(-1)^{3+3}(-4-1) = -5$

$(adjA) =\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}\begin{bmatrix} 1\\\frac{3}{2} \\ 9 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 13+12+9\\5-15+27 \\ 3-9-45 \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 34\\17 \\ -51 \end{bmatrix}= \begin{bmatrix} 1\\\frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =\frac{1}{2},\ and\ \ z=-\frac{3}{2}.$

Question:12 Solve system of linear equations, using matrix method.

$\small x-y+z=4$

$\small 2x+y-3z=0$

$\small x+y+z=2$

Answer:

The given system of equations

$\small x-y+z=4$

$\small 2x+y-3z=0$

$\small x+y+z=2$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &1 \\ 2 & 1 &-3 \\ 1& 1 &1 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 4\\ 0 \\ 2 \end{bmatrix}.$

we have,

$|A| =1(1+3)+1(2+3)+1(2-1) = 4+5+1= 10 \neq 0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(1+3) = 4$ $A_{12} =(-1)^{1+2}(2+3) = -5$

$A_{13} =(-1)^{1+3}(2-1) = 1$ $A_{21} =(-1)^{2+1}(-1-1) = 2$

$A_{22} =(-1)^{2+2}(1-1) = 0$ $A_{23} =(-1)^{2+3}(1+1) = -2$

$A_{31} =(-1)^{3+1}(3-1) = 2$ $A_{32} =(-1)^{3+2}(-3-2) = 5$

$A_{33} =(-1)^{3+3}(1+2) = 3$

$(adjA) =\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}\begin{bmatrix} 4\\0 \\ 2 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 16+0+4\\-20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 20\\-10 \\ 10 \end{bmatrix}= \begin{bmatrix} 2\\-1 \\ 1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =-1,\ and\ \ z=1.$

Question:13 Solve system of linear equations, using matrix method.

$\small 2x+3y+3z=5$

$\small x-2y+z=-4$

$\small 3x-y-2z=3$

Answer:

The given system of equations

$\small 2x+3y+3z=5$

$\small x-2y+z=-4$

$\small 3x-y-2z=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &3 &3 \\ 1 & -2 &1 \\ 3& -1 &-2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 5\\ -4 \\ 3 \end{bmatrix}.$

we have,

$|A| =2(4+1) -3(-2-3)+3(-1+6) = 10+15+15 = 40$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(4+1) = 5$ $A_{12} =(-1)^{1+2}(-2-3) = 5$

$A_{13} =(-1)^{1+3}(-1+6) = 5$ $A_{21} =(-1)^{2+1}(-6+3) = 3$

$A_{22} =(-1)^{2+2}(-4-9) = -13$ $A_{23} =(-1)^{2+3}(-2-9) = 11$

$A_{31} =(-1)^{3+1}(3+6) = 9$ $A_{32} =(-1)^{3+2}(2-3) = 1$

$A_{33} =(-1)^{3+3}(-4-3) = -7$

$(adjA) =\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5&11 & -7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}\begin{bmatrix} 5\\-4 \\ 3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 25-12+27\\25+52+3 \\ 25-44-21 \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 40\\80 \\ -40 \end{bmatrix}= \begin{bmatrix} 1\\2 \\ -1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=-1.$

Question:14 Solve system of linear equations, using matrix method.

$\small x-y+2z=7$

$\small 3x+4y-5z=-5$

$\small 2x-y+3z=12$

Answer:

The given system of equations

$\small x-y+2z=7$

$\small 3x+4y-5z=-5$

$\small 2x-y+3z=12$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &2 \\ 3 & 4 &-5 \\ 2& -1 &3 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 7\\ -5 \\ 12 \end{bmatrix}.$

we have,

$|A| =1(12-5) +1(9+10)+2(-3-8) = 7+19-22 = 4 \neq0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{12-5} = 7$ $A_{12} =(-1)^{1+2}(9+10) = -19$

$A_{13} =(-1)^{1+3}(-3-8) = -11$ $A_{21} =(-1)^{2+1}(-3+2) = 1$

$A_{22} =(-1)^{2+2}(3-4) = -1$ $A_{23} =(-1)^{2+3}(-1+2) = -1$

$A_{31} =(-1)^{3+1}(5-8) = -3$ $A_{32} =(-1)^{3+2}(-5-6) = 11$

$A_{33} =(-1)^{3+3}(4+3) = 7$

$(adjA) =\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}\begin{bmatrix} 7\\-5 \\ 12 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 49-5-36\\-133+5+132 \\ -77+5+84 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8\\4 \\ 12 \end{bmatrix}= \begin{bmatrix} 2\\1 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =1,\ and\ \ z=3.$

Question:15 If $A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}$ , find $A^-^1$ . Using $A^-^1$ solve the system of equations

$2x-3y+5z=11$

$3x+2y-4z=-5$

$x+y-2z=-3$

Answer:

The given system of equations

$2x-3y+5z=11$

$3x+2y-4z=-5$

$x+y-2z=-3$

can be written in the matrix form of AX =B, where

$A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}$ , $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 11\\ -5 \\ -3 \end{bmatrix}.$

we have,

$|A| =2(-4+4) +3(-6+4)+5(3-2) = 0-6+5 = -1 \neq0$ .

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{-4+4} = 0$ $A_{12} =(-1)^{1+2}(-6+4) = 2$

$A_{13} =(-1)^{1+3}(3-2) = 1$ $A_{21} =(-1)^{2+1}(6-5) = -1$

$A_{22} =(-1)^{2+2}(-4-5) = -9$ $A_{23} =(-1)^{2+3}(2+3) = -5$

$A_{31} =(-1)^{3+1}(12-10) = 2$ $A_{32} =(-1)^{3+2}(-8-15) = 23$

$A_{33} =(-1)^{3+3}(4+9) = 13$

$(adjA) =\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = -1\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix} = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}\begin{bmatrix} 11\\-5 \\ -3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-5+6\\-22-45+69 \\ -11-25+39 \end{bmatrix} = \begin{bmatrix} 1\\2 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=3.$

Question:16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

Answer:

So, let us assume the cost of onion, wheat, and rice be x, y and z respectively.

Then we have the equations for the given situation :

$4x+3y+2z = 60$

$2x+4y+6z = 90$

$6x+2y+3y = 70$

We can find the cost of each item per Kg by the matrix method as follows;

Taking the coefficients of x, y, and z as a matrix $A$ .

We have;

$A = \begin{bmatrix} 4 &3 &2 \\ 2& 4 &6 \\ 6 & 2 & 3 \end{bmatrix},$ $X= \begin{bmatrix} x\\y \\ z \end{bmatrix}$ $and\ B = \begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}.$

$|A| = 4(12-12) -3(6-36)+2(4-24) = 0 +90-40 = 50 \neq 0$

Now, we will find the cofactors of A;

$A_{11} = (-1)^{1+1}(12-12) = 0$ $A_{12} = (-1)^{1+2}(6-36) = 30$

$A_{13} = (-1)^{1+3}(4-24) = -20$ $A_{21} = (-1)^{2+1}(9-4) = -5$

$A_{22} = (-1)^{2+2}(12-12) = 0$ $A_{23} = (-1)^{2+3}(8-18) = 10$

$A_{31} = (-1)^{3+1}(18-8) = 10$ $A_{32} = (-1)^{3+2}(24-4) = -20$

$A_{33} = (-1)^{3+3}(16-6) = 10$

Now we have adjA;

$adjA = \begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$ s

So, the solutions can be found by $X = A^{-1}B = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-450+700\\1800+0-1400 \\ -1200+900+700 \end{bmatrix} =\frac{1}{50} \begin{bmatrix} 250\\400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5\\8 \\ 8 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =5,\ y =8,\ and\ \ z=8.$

Therefore, we have the cost of onions is Rs. 5 per Kg, the cost of wheat is Rs. 8 per Kg, and the cost of rice is Rs. 8 per kg.

Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6:-

This chapter is very important for board exams and competitive exams like JEE, SRMJEE, etc.
Class 12 Maths chapter 4 exercise 4.6 solutions are helpful for you in solving the system of linear equations and linear equations consistency problems.
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 are helpful for the students who are not able to solve NCERT problems by themself.
In the Class 12th Maths chapter 4 exercise 4.6 solutions, you will learn the art of answering in board exams to get good marks.
You can take these Class 12 Maths chapter 4 exercise 4.6 solutions for reference.

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Key Features Of NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4

Comprehensive Coverage: The solutions encompass all the topics covered in ex 4.6 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 4.6, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 4.6 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 4.6 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 4.6 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 4.6 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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Frequently Asked Questions (FAQs)

1. If A is a square matrix of order 2 and |A| = 2 then det( transpose(A)) ?

|A^T| = |A| = 2

2. If three-point are collinear then find area of triangle formed by these points ?

If three-point are collinear then the area of triangle formed by these points is zero.

3. What is the definition of consistent system ?

If solutions of a system of equations exist then it is called a consistent system.

4. What is the definition of inconsistent system ?

If the solution of a system of equations doesn't exist then it is called a inconsistent system.

5. How many questions are there in the exercise 4.6 Class 12 Maths?

There are 16 questions in the exercise 4.6 Class 12 Maths. The questions are solved with all the necessary steps. Students can follow the NCERT syllabus to get a good score in the board exam.

6. The transpose of the row matrix is ?

The transpose of the row matrix is the column matrix.

7. The transpose of the column matrix is ?

The transpose of the column matrix is the row matrix.

8. If the determinant of matrix A is zero then matrix A is called ?

If the determinant of matrix A is zero then matrix A is called singular matrix.

Also Read

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

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NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

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NCERT Class 12 Physics Chapter 5 Notes Magnetism and Matter - Download PDF

NCERT Class 12 Physics Chapter 4 Notes Moving Charges and Magnetism - Download PDF

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

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NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Read Complete Answer

Yuvan 01 September,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
You have to appear for the 2025 12th board exams.
Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
Aim to register before late October to avoid extra fees.
Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

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kumardurgesh1802 10 July,2024

I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

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satyamercedes0626 23 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4 - Determinants

NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.6

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Assess NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6

Determinants Exercise: 4.6

More About NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6:-

Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6:-

Key Features Of NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4

JEE Main Important Mathematics Formulas

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NCERT Solutions of Class 12 Subject Wise

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