In this exercise of Determinants 4.5, we will learn how to solve a system of linear equations using matrices and determinants. The main idea is to write the system in matrix form as AX = B and then find the solution using the inverse of matrix A .
For a system to have a unique solution, the determinant of the coefficient matrix must be non-zero, and the solution is given by: X = A^{-1}B
Key Concepts
Consistent System: A system of linear equations is called consistent if it has one or more solutions.
Inconsistent System: A system of linear equations is called inconsistent if it has no solution.
In this exercise, as per NCERT, we shall mainly deal with systems having a unique solution, that is, consistent systems for which the inverse of the coefficient matrix exists.
Consistency Check (for Unique Solution)
Before solving a system of linear equations using matrices, we first check whether the system is consistent or not.
Consider the matrix equation:
AX = B
where
A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}
X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}
B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}
Key Points
Case 1: If |A| \neq 0
- A is a non-singular matrix, therefore A^{-1} exists.
- The system is consistent.
- The system has a unique solution given by:
X = A^{-1}B
Case 2: If |A| = 0
- A is a singular matrix, therefore A^{-1} does not exist.
- In this case, two possibilities may arise:
- If \operatorname{adj}(A)B \neq 0 , then the system is inconsistent and has no solution.
- If \operatorname{adj}(A)B = 0 , then the system may have infinitely many solutions.
Let us now solve all the NCERT Exercise 4.5 questions step by step.
In this exercise of determinants 4.5, we shall restrict ourselves only to systems having a unique solution.
Question 1: Determinants 4.5
1. Examine the consistency of the system of equations
x+2y=2
2x+3y=3
Solution
We can write the given system in matrix form as
AX=B
where
A=\begin{bmatrix}1 & 2 \\ 2 & 3\end{bmatrix},\quad X=\begin{bmatrix}x \\ y\end{bmatrix},\quad B=\begin{bmatrix}2 \\ 3\end{bmatrix}
Let’s examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix}
=(1)(3)-(2)(2)=3-4=-1
Since |A|=-1\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Question 2: Consistency Check
2. Examine the consistency of the system of equations
2x-y=5
x+y=4
Solution
Writing the system in matrix form,
AX=B
where
A=\begin{bmatrix}2 & -1 \\ 1 & 1\end{bmatrix},\quad X=\begin{bmatrix}x \\ y\end{bmatrix},\quad B=\begin{bmatrix}5 \\ 4\end{bmatrix}
Now, find |A| .
|A|=\begin{vmatrix}2 & -1 \\ 1 & 1\end{vmatrix}
=(2)(1)-(-1)(1)=2+1=3
Since |A|=3\neq0 , the matrix A is non-singular.
Therefore, the given system is consistent and has a unique solution.
You may already be following Maths Better for important concepts and exam-oriented topics like integration tricks, Differentiation rules and formulas and solving Linear Equations using the Matrix Method. Similarly, this NCERT Solutions series for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills.
Now, let’s move on to the next question of Determinants 4.5.
Question 3: Determinants 4.5
3. Examine the consistency of the system of equations
x+3y=5
2x+6y=8
Solution
Writing the system in matrix form,
AX=B
where
A=\begin{bmatrix}1 & 3 \\ 2 & 6\end{bmatrix},\quad X=\begin{bmatrix}x \\ y\end{bmatrix},\quad B=\begin{bmatrix}5 \\ 8\end{bmatrix}
Now, find |A| .
|A|=\begin{vmatrix}1 & 3 \\ 2 & 6\end{vmatrix}
=(1)(6)-(3)(2)=6-6=0
Since |A|=0 , the matrix A is singular and A^{-1} does not exist.
Also,
\operatorname{adj}(A)=\begin{bmatrix}6 & -3 \\ -2 & 1\end{bmatrix}
Now,
\operatorname{adj}(A)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\begin{bmatrix}0 \\ 0\end{bmatrix}
Therefore, the given system is inconsistent and has no solution.
Question 4: Determinants Ex 4.5
4. Examine the consistency of the system of equations
x+y+z=1;
2x+3y+2z=2;
and ax+ay+2az=4
Solution
Writing the system in matrix form,
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},\quad B=\begin{bmatrix}1 \\ 2 \\ 4\end{bmatrix}
Now, find |A| .
|A|=\begin{vmatrix}1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a\end{vmatrix}
Expanding along the first row,
|A|= a\begin{vmatrix}3 & 2 \\ 1 & 2 \end{vmatrix}-a\begin{vmatrix}2 & 2 \\ 1 & 2\end{vmatrix}+a\begin{vmatrix}2 & 3 \\ 1 & 1\end{vmatrix}
=a[(6-2)]-a[(4-2)]+a[(2-3)]
So, |A|= 4a-2a-a =a
Since |A|=a , two cases arise.
Case 1: If a\neq0
Then |A|\neq0 , so the system is consistent and has a unique solution.
Case 2: If a=0
Then |A|=0 , so the matrix is singular.
In this case, the third equation becomes
0=4
which is impossible.
Therefore, when a=0 , the system is inconsistent.
Question 5: Determinants 4.5
5. Examine the consistency of the system of equations
3x-y-2z=2;
2y-z=-1;
and 3x-5y=3
Solution
Writing the system in matrix form,
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},\quad B=\begin{bmatrix}2 \\ -1 \\ 3\end{bmatrix}
Now, find |A| .
|A|=\begin{vmatrix}3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{vmatrix}
Expanding along the first column,
|A|=3\begin{vmatrix}2 & -1 \\ -5 & 0\end{vmatrix}-0\begin{vmatrix}0 & 3 \\ -1 & 0\end{vmatrix}-2\begin{vmatrix}0 & 3 \\ 2 & -5\end{vmatrix}
=3[(2)(0)-(-5)(-1)] - 0 +3[(-1)(-1)-(2)(-2)]
=3(0-5) +3(1+4)= -15+15 = 0
Since |A|=0 , the matrix is singular.
Therefore, the given system is inconsistent and has no solution.
These types of questions are considered scoring if the basic concepts and operations are clear.
Question 6: Determinants Ex 4.5
6. Examine the consistency of the system of equations
5x-y+4z=5;
2x+3y+5z=2;
and 5x-2y+6z=-1
Solution
Writing the system in matrix form,
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},\quad B=\begin{bmatrix}5 \\ 2 \\ -1\end{bmatrix}
Now, find |A| .
|A|=\begin{vmatrix}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{vmatrix}
Expanding along the first row,
|A|=5\begin{vmatrix}3 & 5 \\ -2 & 6\end{vmatrix}-(-1)\begin{vmatrix}2 & 6 \\ 5 & 5\end{vmatrix}+4\begin{vmatrix}2 & -2 \\ 5 & 3\end{vmatrix}
=5[(3)(6)-(-2)(5)] +1[(2)(6)-(5)(5)]+4[(2)(-2)-(5)(3)]
i.e. |A| =5(18+10)+(12-25)+4(-4-15)
or |A| =140-13-76 = 51
Since |A|=51\neq0 , the matrix A is non-singular.
Therefore, the given system is consistent and has a unique solution.
Question 7: Ex 4.5 Determinants
7. Solve the following system of equations using matrix method
5x+2y=4
7x+3y=5
Solution
We can write the given system in matrix form as
AX=B
where
A=\begin{bmatrix}5 & 2 \\ 7 & 3\end{bmatrix},\quad X=\begin{bmatrix}x \\ y\end{bmatrix},\quad B=\begin{bmatrix}4 \\ 5\end{bmatrix}
First, examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}5 & 2 \\ 7 & 3\end{vmatrix}
=(5)(3)-(2)(7)=15-14=1
Since |A|=1\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Now, find the inverse of matrix A .
For a 2\times2 matrix,
\begin{bmatrix}a & b \\ c & d\end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}
Thus,
A^{-1}=\dfrac{1}{1}\begin{bmatrix}3 & -2 \\ -7 & 5\end{bmatrix}
=\begin{bmatrix}3 & -2 \\ -7 & 5\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}3 & -2 \\ -7 & 5\end{bmatrix}\begin{bmatrix}4 \\ 5\end{bmatrix}
=\begin{bmatrix}(3)(4)+(-2)(5) \\ (-7)(4)+(5)(5)\end{bmatrix}
i.e. LHS
=\begin{bmatrix}12-10 \\ -28+25\end{bmatrix}
=\begin{bmatrix}2 \\ -3\end{bmatrix}
Hence,
x=2,\quad y=-3
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter, to help you understand the concepts more visually.
Now, let’s move to the next question.
Question 8: Determinants 4.5
8. Solve the following system of equations using matrix method
2x-y=-2
3x+4y=3
Solution
We can write the given system in matrix form as
AX=B
where
A=\begin{bmatrix}2 & -1 \\ 3 & 4\end{bmatrix},\quad X=\begin{bmatrix}x \\ y\end{bmatrix},\quad B=\begin{bmatrix}-2 \\ 3\end{bmatrix}
First, examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}2 & -1 \\ 3 & 4\end{vmatrix}
=(2)(4)-(-1)(3)=8+3=11
Since |A|=11\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Now, find the inverse of matrix A .
For a 2\times2 matrix,
\begin{bmatrix}a & b \\ c & d\end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}
Thus,
A^{-1}=\dfrac{1}{11}\begin{bmatrix}4 & 1 \\ -3 & 2\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y\end{bmatrix}=\dfrac{1}{11}\begin{bmatrix}4 & 1 \\ -3 & 2\end{bmatrix}\begin{bmatrix}-2 \\ 3\end{bmatrix}
=\dfrac{1}{11}\begin{bmatrix}(4)(-2)+(1)(3) \\ (-3)(-2)+(2)(3)\end{bmatrix}
or LHS
=\dfrac{1}{11}\begin{bmatrix}-8+3 \\ 6+6\end{bmatrix}
=\dfrac{1}{11}\begin{bmatrix}-5 \\ 12\end{bmatrix}
or it is
=\begin{bmatrix}-\dfrac{5}{11} \\ \dfrac{12}{11}\end{bmatrix}
Hence,
x=-\dfrac{5}{11},\quad y=\dfrac{12}{11}
Question 9: Inverse of a Matrix
9. Solve the following system of equations using matrix method
4x-3y=3
3x-5y=7
Solution
We can write the given system in matrix form as
AX=B
where
A=\begin{bmatrix}4 & -3 \\ 3 & -5\end{bmatrix},\quad X=\begin{bmatrix}x \\ y\end{bmatrix},\quad B=\begin{bmatrix}3 \\ 7\end{bmatrix}
First, examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}4 & -3 \\ 3 & -5\end{vmatrix}
=(4)(-5)-(-3)(3)=-20+9=-11
Since |A|=-11\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Now, find the inverse of matrix A .
For a 2\times2 matrix,
\begin{bmatrix}a & b \\ c & d\end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}
Thus,
A^{-1}=\dfrac{1}{-11}\begin{bmatrix}-5 & 3 \\ -3 & 4\end{bmatrix}
=\dfrac{1}{11}\begin{bmatrix}5 & -3 \\ 3 & -4\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y\end{bmatrix}=\dfrac{1}{11}\begin{bmatrix}5 & -3 \\ 3 & -4\end{bmatrix}\begin{bmatrix}3 \\ 7\end{bmatrix}
=\dfrac{1}{11}\begin{bmatrix}(5)(3)+(-3)(7) \\ (3)(3)+(-4)(7)\end{bmatrix}
or it is
=\dfrac{1}{11}\begin{bmatrix}15-21 \\ 9-28\end{bmatrix}
=\dfrac{1}{11}\begin{bmatrix}-6 \\ -19\end{bmatrix}
So, LHS
=\begin{bmatrix}-\dfrac{6}{11} \\ -\dfrac{19}{11}\end{bmatrix}
Hence,
x=-\dfrac{6}{11},\quad y=-\dfrac{19}{11}
Question 10: Determinants 4.5 Exercise
10. Solve the following system of equations using matrix method
5x+2y=3
3x+2y=5
Solution
We can write the given system in matrix form as
AX=B
where
A=\begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix},\quad X=\begin{bmatrix}x \\ y\end{bmatrix},\quad B=\begin{bmatrix}3 \\ 5\end{bmatrix}
First, examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}5 & 2 \\ 3 & 2\end{vmatrix}
=(5)(2)-(2)(3)=10-6=4
Since |A|=4\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Now, find the inverse of matrix A .
For a 2\times2 matrix,
\begin{bmatrix}a & b \\ c & d\end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}
Thus,
A^{-1}=\dfrac{1}{4}\begin{bmatrix}2 & -2 \\ -3 & 5\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y\end{bmatrix}=\dfrac{1}{4}\begin{bmatrix}2 & -2 \\ -3 & 5\end{bmatrix}\begin{bmatrix}3 \\ 5\end{bmatrix}
=\dfrac{1}{4}\begin{bmatrix}(2)(3)+(-2)(5) \\ (-3)(3)+(5)(5)\end{bmatrix}
or it is
=\dfrac{1}{4}\begin{bmatrix}6-10 \\ -9+25\end{bmatrix}
=\dfrac{1}{4}\begin{bmatrix}-4 \\ 16\end{bmatrix}
i.e. LHS
=\begin{bmatrix}-1 \\ 4\end{bmatrix}
Hence,
x=-1,\quad y=4
The questions in Determinants 4.5 are mostly based on adjoint and inverse of matrices. Once the concepts are clear, these questions become quite manageable and scoring.
Mathematics becomes easier with regular practice, so try to solve each question step by step and understand the logic behind every operation. If you have any doubt, feel free to leave a comment, I’ll be happy to help.
Here’s the next question.
Question 11: Determinants Exercise 4.5
11. Solve the following system of equations using matrix method
2x+y+z=1;
x-2y-z=\dfrac{3}{2};
and 3y-5z=9
Solution
We can write the given system in matrix form as
AX=B
where
A=\begin{bmatrix}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}1 \\ \dfrac{3}{2} \\ 9\end{bmatrix}
First, examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{vmatrix}
Expanding along the first row,
|A|=2\begin{vmatrix}-2 & -1 \\ 3 & -5\end{vmatrix}-1\begin{vmatrix}1 & -1 \\ 0 & -5\end{vmatrix}+1\begin{vmatrix}1 & -2 \\ 0 & 3\end{vmatrix}
=2[(10+3)]-1[(-5-0)]+1[(3-0)]
i.e. |A|=26+5+3=34
Since |A|=34\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Now, find the cofactors of matrix A .
Cofactors of first row:
A_{11}=\begin{vmatrix}-2 & -1 \\ 3 & -5\end{vmatrix}=10+3=13
A_{12}=-\begin{vmatrix}1 & -1 \\ 0 & -5\end{vmatrix}=5
and A_{13}=\begin{vmatrix}1 & -2 \\ 0 & 3\end{vmatrix}=3
Cofactors of second row:
A_{21}=-\begin{vmatrix}1 & 1 \\ 3 & -5\end{vmatrix}=-(-5-3)=8
A_{22}=\begin{vmatrix}2 & 1 \\ 0 & -5\end{vmatrix}=-10
and A_{23}=-\begin{vmatrix}2 & 1 \\ 0 & 3\end{vmatrix}=-6
Cofactors of third row:
A_{31}=\begin{vmatrix}1 & 1 \\ -2 & -1\end{vmatrix}=-1+2=1
A_{32}=-\begin{vmatrix}2 & 1 \\ 1 & -1\end{vmatrix}=-(-2-1)=3
and A_{33}=\begin{vmatrix}2 & 1 \\ 1 & -2\end{vmatrix}=-4-1=-5
Thus, the cofactor matrix is
\begin{bmatrix}13 & 5 & 3 \\ 8 & -10 & -6 \\ 1 & 3 & -5\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{34}\begin{bmatrix}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\dfrac{1}{34}\begin{bmatrix}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{bmatrix}\begin{bmatrix}1 \\ \dfrac{3}{2} \\ 9\end{bmatrix}
=\dfrac{1}{34}\begin{bmatrix}13+12+9 \\ 5-15+27 \\ 3-9-45\end{bmatrix}
i.e. X =\dfrac{1}{34}\begin{bmatrix}34 \\ 17 \\ -51\end{bmatrix}
or X =\begin{bmatrix}1 \\ \dfrac{1}{2} \\ -\dfrac{3}{2}\end{bmatrix}
Hence,
x=1,\quad y=\dfrac{1}{2},\quad z=-\dfrac{3}{2}
Question 12: Determinants 4.5
12. Solve the following system of equations using matrix method
x-y+z=4;
2x+y-3z=0;
and x+y+z=2
Solution
We can write the given system in matrix form as
AX=B
where
A=\begin{bmatrix}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}4 \\ 0 \\ 2\end{bmatrix}
First, examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{vmatrix}
Expanding along the first row,
We have, |A|=1\begin{vmatrix}1 & -3 \\ 1 & 1\end{vmatrix}-(-1)\begin{vmatrix}2 & -3 \\ 1 & 1\end{vmatrix}+1\begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix}
=(1+3)+(2+3)+(2-1)
=4+5+1=10
Since |A|=10\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Now, find the cofactors of matrix A .
Cofactors of first row:
A_{11}=\begin{vmatrix}1 & -3 \\ 1 & 1\end{vmatrix}=1+3=4
A_{12}=-\begin{vmatrix}2 & -3 \\ 1 & 1\end{vmatrix}=-(2+3)=-5
and A_{13}=\begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix}=2-1=1
Cofactors of second row:
A_{21}=-\begin{vmatrix}-1 & 1 \\ 1 & 1\end{vmatrix}=-(-1-1)=2
A_{22}=\begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix}=1-1=0
and A_{23}=-\begin{vmatrix}1 & -1 \\ 1 & 1\end{vmatrix}=-(1+1)=-2
Cofactors of third row:
A_{31}=\begin{vmatrix}-1 & 1 \\ 1 & -3\end{vmatrix}=3-1=2
A_{32}=-\begin{vmatrix}1 & 1 \\ 2 & -3\end{vmatrix}=-(-3-2)=5
and A_{33}=\begin{vmatrix}1 & -1 \\ 2 & 1\end{vmatrix}=1+2=3
Thus, the cofactor matrix is
\begin{bmatrix}4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{10}\begin{bmatrix}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\dfrac{1}{10}\begin{bmatrix}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{bmatrix}\begin{bmatrix}4 \\ 0 \\ 2\end{bmatrix}
=\dfrac{1}{10}\begin{bmatrix}16+0+4 \\ -20+0+10 \\ 4+0+6\end{bmatrix}
i.e. X =\dfrac{1}{10}\begin{bmatrix}20 \\ -10 \\ 10\end{bmatrix}
=\begin{bmatrix}2 \\ -1 \\ 1\end{bmatrix}
Hence,
x=2,\quad y=-1,\quad z=1
Now, let’s move on to the next question of Determinants 4.5
Question 13: Ex. 4.5 Determinants
13. Solve the following system of equations using matrix method
2x+3y+3z=5;
x-2y+z=-4;
and 3x-y-2z=3
Solution
We can write the given system in matrix form as
AX=B
where
A=\begin{bmatrix}2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}5 \\ -4 \\ 3\end{bmatrix}
First, examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2\end{vmatrix}
Expanding along the first row,
|A|=2\begin{vmatrix}-2 & 1 \\ -1 & -2\end{vmatrix}-3\begin{vmatrix}1 & 1 \\ 3 & -2\end{vmatrix}+3\begin{vmatrix}1 & -2 \\ 3 & -1\end{vmatrix}
=2(4+1)-3(-2-3)+3(-1+6)
So, |A|=10+15+15=40
Since |A|=40\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Now, find the cofactors of matrix A .
Cofactors of first row:
A_{11}=\begin{vmatrix}-2 & 1 \\ -1 & -2\end{vmatrix}=4+1=5
A_{12}=-\begin{vmatrix}1 & 1 \\ 3 & -2\end{vmatrix}=-(-2-3)=5
and A_{13}=\begin{vmatrix}1 & -2 \\ 3 & -1\end{vmatrix}=-1+6=5
Cofactors of second row:
A_{21}=-\begin{vmatrix}3 & 3 \\ -1 & -2\end{vmatrix}=-(-6+3)=3
A_{22}=\begin{vmatrix}2 & 3 \\ 3 & -2\end{vmatrix}=-4-9=-13
and A_{23}=-\begin{vmatrix}2 & 3 \\ 3 & -1\end{vmatrix}=-(-2-9)=11
Cofactors of third row:
A_{31}=\begin{vmatrix}3 & 3 \\ -2 & 1\end{vmatrix}=3+6=9
A_{32}=-\begin{vmatrix}2 & 3 \\ 1 & 1\end{vmatrix}=-(2-3)=1
and A_{33}=\begin{vmatrix}2 & 3 \\ 1 & -2\end{vmatrix}=-4-3=-7
Thus, the cofactor matrix is
\begin{bmatrix}5 & 5 & 5 \\ 3 & -13 & 11 \\ 9 & 1 & -7\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{40}\begin{bmatrix}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\dfrac{1}{40}\begin{bmatrix}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{bmatrix}\begin{bmatrix}5 \\ -4 \\ 3\end{bmatrix}
=\dfrac{1}{40}\begin{bmatrix}25-12+27 \\ 25+52+3 \\ 25-44-21\end{bmatrix}
or X =\dfrac{1}{40}\begin{bmatrix}40 \\ 80 \\ -40\end{bmatrix}
=\begin{bmatrix}1 \\ 2 \\ -1\end{bmatrix}
Hence,
x=1,\quad y=2,\quad z=-1
💡 Exam Tip: Questions of the form AX = B are often solved using the inverse matrix method, as we have seen in the above questions. However, there are some more types of such questions as well. You can refer to Example 19 of the NCERT textbook.
🎥 In fact, I would also suggest you watch this YouTube video for different types of questions which are not given in NCERT but have appeared in previous years’ examinations.
Question 14: Determinants 4.5
14. Solve the following system of equations using matrix method
x-y+2z=7;
3x+4y-5z=-5;
and 2x-y+3z=12
Solution
We can write the given system in matrix form as
AX=B
where
A=\begin{bmatrix}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}7 \\ -5 \\ 12\end{bmatrix}
First, examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{vmatrix}
Expanding along the first row,
|A|=1\begin{vmatrix}4 & -5 \\ -1 & 3\end{vmatrix}-(-1)\begin{vmatrix}3 & -5 \\ 2 & 3\end{vmatrix}+2\begin{vmatrix}3 & 4 \\ 2 & -1\end{vmatrix}
=(12-5)+(9+10)+2(-3-8)
So, |A|=7+19-22=4
Since |A|=4\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Now, find the cofactors of matrix A .
Cofactors of first row:
A_{11}=\begin{vmatrix}4 & -5 \\ -1 & 3\end{vmatrix}=12-5=7
A_{12}=-\begin{vmatrix}3 & -5 \\ 2 & 3\end{vmatrix}=-(9+10)=-19
and A_{13}=\begin{vmatrix}3 & 4 \\ 2 & -1\end{vmatrix}=-3-8=-11
Cofactors of second row:
A_{21}=-\begin{vmatrix}-1 & 2 \\ -1 & 3\end{vmatrix}=-(-3+2)=1
A_{22}=\begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix}=3-4=-1
and A_{23}=-\begin{vmatrix}1 & -1 \\ 2 & -1\end{vmatrix}=-(-1+2)=-1
Cofactors of third row:
A_{31}=\begin{vmatrix}-1 & 2 \\ 4 & -5\end{vmatrix}=5-8=-3
A_{32}=-\begin{vmatrix}1 & 2 \\ 3 & -5\end{vmatrix}=-(-5-6)=11
and A_{33}=\begin{vmatrix}1 & -1 \\ 3 & 4\end{vmatrix}=4+3=7
Thus, the cofactor matrix is
\begin{bmatrix}7 & -19 & -11 \\ 1 & -1 & -1 \\ -3 & 11 & 7\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{4}\begin{bmatrix}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\dfrac{1}{4}\begin{bmatrix}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{bmatrix}\begin{bmatrix}7 \\ -5 \\ 12\end{bmatrix}
=\dfrac{1}{4}\begin{bmatrix}49-5-36 \\ -133+5+132 \\ -77+5+84\end{bmatrix}
or X =\dfrac{1}{4}\begin{bmatrix}8 \\ 4 \\ 12\end{bmatrix}
=\begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}
Hence,
x=2,\quad y=1,\quad z=3
Question 15: Inverse Matrix
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
and x+y-2z=-3
Solution
Given,
A=\begin{bmatrix}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{bmatrix}
First, find the determinant of A .
|A|=\begin{vmatrix}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{vmatrix}
Expanding along the first row,
|A|=2\begin{vmatrix}2 & -4 \\ 1 & -2\end{vmatrix}-(-3)\begin{vmatrix}3 & -4 \\ 1 & -2\end{vmatrix}+5\begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix}
=2(-4+4)+3(-6+4)+5(3-2)
So, |A| =0-6+5=-1
Since |A|=-1\neq0 , the matrix is non-singular and inverse exists.
Now, find the cofactors of A .
Cofactors of first row:
A_{11}=\begin{vmatrix}2 & -4 \\ 1 & -2\end{vmatrix}=-4+4=0
A_{12}=-\begin{vmatrix}3 & -4 \\ 1 & -2\end{vmatrix}=-(-6+4)=2
and A_{13}=\begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix}=3-2=1
Cofactors of second row:
A_{21}=-\begin{vmatrix}-3 & 5 \\ 1 & -2\end{vmatrix}=- (6-5)=-1
A_{22}=\begin{vmatrix}2 & 5 \\ 1 & -2\end{vmatrix}=-4-5=-9
and A_{23}=-\begin{vmatrix}2 & -3 \\ 1 & 1\end{vmatrix}=-(2+3)=-5
Cofactors of third row:
A_{31}=\begin{vmatrix}-3 & 5 \\ 2 & -4\end{vmatrix}=12-10=2
A_{32}=-\begin{vmatrix}2 & 5 \\ 3 & -4\end{vmatrix}=-(-8-15)=23
and A_{33}=\begin{vmatrix}2 & -3 \\ 3 & 2\end{vmatrix}=4+9=13
Thus, the cofactor matrix is
\begin{bmatrix}0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-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}
Now, write the given system in matrix form:
AX=B
where
X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}11 \\ -5 \\ -3\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{bmatrix}\begin{bmatrix}11 \\ -5 \\ -3\end{bmatrix}
=\begin{bmatrix}0-5+6 \\ -22-45+69 \\ -11-25+39\end{bmatrix}
i.e. X =\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}
Hence,
x=1,\quad y=2,\quad z=3
Question 16: Determinants 4.5
16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. And the cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.
Solution
Let the cost per kg of onion, wheat and rice be x , y and z respectively.
Then, the given information can be written as
4x+3y+2z=60;
2x+4y+6z=90;
and 6x+2y+3z=70
We can write the system in matrix form as
AX=B
where
A=\begin{bmatrix}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{bmatrix},\quad X=\begin{bmatrix}x \\ y \\ z\end{bmatrix},\quad B=\begin{bmatrix}60 \\ 90 \\ 70\end{bmatrix}
First, examine the consistency of the system by finding |A| .
|A|=\begin{vmatrix}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{vmatrix}
Expanding along the first row,
|A|=4\begin{vmatrix}4 & 6 \\ 2 & 3\end{vmatrix}-3\begin{vmatrix}2 & 6 \\ 6 & 3\end{vmatrix}+2\begin{vmatrix}2 & 4 \\ 6 & 2\end{vmatrix}
=4(12-12)-3(6-36)+2(4-24)
i.e. |A|=0+90-40=50
Since |A|=50\neq0 , the matrix A is non-singular.
Therefore, the system is consistent and has a unique solution.
Now, find the cofactors of matrix A .
Cofactors of first row:
A_{11}=\begin{vmatrix}4 & 6 \\ 2 & 3\end{vmatrix}=12-12=0
A_{12}=-\begin{vmatrix}2 & 6 \\ 6 & 3\end{vmatrix}=-(6-36)=30
and A_{13}=\begin{vmatrix}2 & 4 \\ 6 & 2\end{vmatrix}=4-24=-20
Cofactors of second row:
A_{21}=-\begin{vmatrix}3 & 2 \\ 2 & 3\end{vmatrix}=-(9-4)=-5
A_{22}=\begin{vmatrix}4 & 2 \\ 6 & 3\end{vmatrix}=12-12=0
and A_{23}=-\begin{vmatrix}4 & 3 \\ 6 & 2\end{vmatrix}=-(8-18)=10
Cofactors of third row:
A_{31}=\begin{vmatrix}3 & 2 \\ 4 & 6\end{vmatrix}=18-8=10
A_{32}=-\begin{vmatrix}4 & 2 \\ 2 & 6\end{vmatrix}=-(24-4)=-20
and A_{33}=\begin{vmatrix}4 & 3 \\ 2 & 4\end{vmatrix}=16-6=10
Thus, the cofactor matrix is
\begin{bmatrix}0 & 30 & -20 \\ -5 & 0 & 10 \\ 10 & -20 & 10\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{50}\begin{bmatrix}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{bmatrix}
Using
X=A^{-1}B
we get
\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\dfrac{1}{50}\begin{bmatrix}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{bmatrix}\begin{bmatrix}60 \\ 90 \\ 70\end{bmatrix}
=\dfrac{1}{50}\begin{bmatrix}0-450+700 \\ 1800+0-1400 \\ -1200+900+700\end{bmatrix}
i.e. X =\dfrac{1}{50}\begin{bmatrix}250 \\ 400 \\ 400\end{bmatrix}
=\begin{bmatrix}5 \\ 8 \\ 8\end{bmatrix}
Hence, the cost per kg of onion, wheat and rice are
\text{Onion}=₹5 \text{ per kg}
\text{Wheat}=₹8 \text{ per kg}
and \text{Rice}=₹8 \text{ per kg}
Common Mistakes to Avoid
- Writing matrix incorrectly: While forming matrix A , ensure that coefficients of variables are written in the correct order.
- Forgetting to check consistency: Before finding inverse, always check whether |A|\neq0 . Only then the inverse exists and the system has a unique solution.
- Sign mistakes in cofactors: Cofactors follow alternating signs: \begin{bmatrix}+ & - & + \\ - & + & - \\ + & - & +\end{bmatrix}
- Confusing adjoint with cofactor matrix: \operatorname{adj}(A) is the transpose of the cofactor matrix, not the cofactor matrix itself.
- Using wrong formula for inverse: Remember, A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
- Errors during matrix multiplication: While calculating X=A^{-1}B , multiply corresponding row and column elements carefully.
- Ignoring determinant value: If |A|=0 , then inverse does not exist and the matrix method using A^{-1} cannot be applied directly.
- Not writing final answer clearly: Always mention the values of variables separately at the end, such as x=2,\ y=1,\ z=3 .
Continue Learning
- Understand the difference between consistent and inconsistent systems of linear equations.
- Revise the concepts of determinant, adjoint and inverse of a matrix.
- Practice solving systems of equations using X=A^{-1}B .
- Try solving the same systems using elimination method and compare the steps.
- Explore what happens when |A|=0 and why inverse does not exist in that case.
- Learn how matrices are used in real-life applications such as economics, computer graphics, coding theory and data science.
Explore More
You can also explore other Class 12 Maths chapters, MCQs, concept notes and shortcut methods available on Maths Better. Many of these questions are also explained in video format on the Mathsbetter YouTube channel.
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