Anand Classes provides detailed and exam-oriented NCERT Solutions for Determinants Exercise 4.5 Class 12 Math Chapter-4, prepared strictly according to the latest CBSE syllabus and NCERT guidelines. These solutions are designed to help students understand important determinant concepts, step-by-step problem solving, and scoring techniques for board examinations. The PDF notes are ideal for quick revision, concept clarity, and practice before exams. Click the print button to download study material and notes.
NCERT Question.1 : Examine the consistency of the system of equations
$x+2y=2$
$2x+3y=3$
Solution
Matrix form of the given equations is
$$
AX=B
$$
where
$$
A=\begin{bmatrix}1 & 2\\ 2 & 3\end{bmatrix},\quad
X=\begin{bmatrix}x\\ y\end{bmatrix},\quad
B=\begin{bmatrix}2\\ 3\end{bmatrix}.
$$
Thus,
$$
\begin{bmatrix}1 & 2\\ 2 & 3\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}2\\ 3\end{bmatrix}.
$$
Now,
$$
|A|=
\begin{vmatrix}
1 & 2\\
2 & 3
\end{vmatrix}=3-4=-1\neq0.
$$
Since $|A|\neq0$, the inverse of matrix $A$ exists.
Therefore, the given system of equations has a unique solution.
Final Result
The system of equations is consistent.
Published by Anand Classes, Authored by Neeraj Anand — this solution demonstrates the matrix method for checking consistency of linear equations, an important topic for NCERT Class 12 Mathematics, CBSE board exams, and competitive exams such as JEE and CUET.
NCERT Question 2 : Examine the consistency of the system of equations
$2x – y = 5$
$x + y = 4$
Solution
Matrix form of the given equations is $AX = B$.
where,
$$
A =
\begin{bmatrix}
2 & -1 \\
1 & 1
\end{bmatrix},
\quad
B =
\begin{bmatrix}
5 \\
4
\end{bmatrix}
\quad \text{and} \quad
X =
\begin{bmatrix}
x \\
y
\end{bmatrix}
$$
Therefore
$$\begin{bmatrix}
2 & -1 \\
1 & 1
\end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix}=\begin{bmatrix}
5 \\4\end{bmatrix}$$
Now,
$$
|A| =
\begin{vmatrix}
2 & -1 \\
1 & 1
\end{vmatrix}
= 2 – (-1) = 3 \ne 0
$$
$\because$ inverse of matrix exists, unique solution.
$\therefore$ System of equation is consistent.
This solution on consistency of linear equations using matrices is published by Anand Classes, carefully authored by Neeraj Anand, Main Head Faculty, and is ideal for Class 12 CBSE and competitive exam preparation with clear matrix-based explanations.
NCERT Question 3 : Examine the consistency of the system of equations
$x + 3y = 5$
$2x + 6y = 8$
Solution
Matrix form of the given equations is $AX = B$.
where,
$$
A =
\begin{bmatrix}
1 & 3 \\
2 & 6
\end{bmatrix},
\quad
B =
\begin{bmatrix}
5 \\
8
\end{bmatrix}
\quad \text{and} \quad
X =
\begin{bmatrix}
x \\
y
\end{bmatrix}
$$
Therefore
$$
\begin{bmatrix}
1 & 3 \\
2 & 6
\end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix}=\begin{bmatrix}
5 \\8\end{bmatrix}$$
Now,
$$
|A| =
\begin{vmatrix}
1 & 3 \\
2 & 6
\end{vmatrix}
= 6 – 6 = 0
$$
And,
$$
\text{adj } A =
\begin{bmatrix}
6 & -3 \\
-2 & 1
\end{bmatrix}
$$
Therefore
$$(\text{adj } A)B =
\begin{bmatrix}
6 & -3 \\
-2 & 1
\end{bmatrix}\begin{bmatrix}
5 \\
8
\end{bmatrix}$$
$$(\text{adj } A)B =\begin{bmatrix}30 – 24 \\-10 + 8
\end{bmatrix}$$
$$(\text{adj } A)B =\begin{bmatrix}
6 \\
-2
\end{bmatrix}
\ne 0
$$
$\because$ have no common solution.
$\therefore$ System of equation is inconsistent.
This detailed matrix-based consistency analysis is published by Anand Classes and clearly written by Neeraj Anand, Director and senior mathematics faculty, offering reliable Class 12 NCERT-aligned content for CBSE board exams and competitive entrance preparation.
NCERT Question 4 : Examine the consistency of the system of equations
$x + y + z = 1$
$2x + 3y + 2z = 2$
$ax + ay + 2az = 4$
Solution
Matrix form of the given equations is $AX = B$.
where,
$$
A =
\begin{bmatrix}
1 & 1 & 1 \\
2 & 3 & 2 \\
a & a & 2a
\end{bmatrix},
\quad
B =
\begin{bmatrix}
1 \\
2 \\
4
\end{bmatrix}
\quad \text{and} \quad
X =
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
$$
Therefore
$$
\begin{bmatrix}
1 & 1 & 1 \\
2 & 3 & 2 \\
a & a & 2a
\end{bmatrix}\begin{bmatrix}x \\y \\z\end{bmatrix}=\begin{bmatrix}1 \\2 \\4\end{bmatrix}$$
Now,
$$
|A| =
\begin{vmatrix}
1 & 1 & 1 \\
2 & 3 & 2 \\
a & a & 2a
\end{vmatrix}
$$
Expanding along the first row,
$$
|A|
= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)
$$
$$
|A| = 4a – 2a – a = a
$$
Since,
$$
|A| = a \ne 0
$$
$\because$ inverse of matrix exists, unique solution.
$\therefore$ System of equation is consistent (for $a \ne 0$).
This solution on consistency of linear equations using determinants is published by Anand Classes and thoughtfully authored by Neeraj Anand, Main Head Faculty, providing dependable NCERT-based guidance for CBSE Class 12 mathematics and competitive exam preparation.
NCERT Question 5 : Examine the consistency of the system of equations
$3x – y – 2z = 2$
$2y – z = -1$
$3x – 5y = 3$
Solution
The given system of linear equations is:
$$
3x – y – 2z = 2 \\[1em]
0x + 2y – z = -1\\[1em]
3x – 5y + 0z = 3
$$
This system can be written in the matrix form $AX = B$, where:
$$
A =
\begin{bmatrix}
3 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{bmatrix}, \quad X =
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix} , \quad
B =\begin{bmatrix}
2 \\
-1 \\
3
\end{bmatrix}
$$
1. Calculate the determinant of $A$
$$
|A| =
3
\begin{vmatrix}
2 & -1 \\
-5 & 0
\end{vmatrix}-(-1)\begin{vmatrix}
0 & -1 \\
3 & 0
\end{vmatrix}+(-2)
\begin{vmatrix}
0 & 2 \\
3 & -5
\end{vmatrix}
$$
$$
|A|
= 3(2 \times 0 – (-1)\times (-5))+1(0 \times 0 – (-1)\times 3)-2(0 \times (-5) – 2 \times 3)
$$
$$
|A|
= 3(0 – 5) + 1(0 + 3) – 2(0 – 6)
$$
$$
|A| = 3(-5) + 3 – 2(-6)
$$
$$
|A| = -15 + 3 + 12 = 0
$$
Since $|A| = 0$, the matrix $A$ is singular, meaning its inverse $A^{-1}$ does not exist. Therefore, the system either has no solution or infinitely many solutions.
2. Check for consistency
To determine consistency when $|A| = 0$, calculate $(\operatorname{adj} A)B$.
The cofactors of $A$ are:
$$
C_{11} = -5, C_{12} = -3, C_{13} = -6
$$
$$
C_{21} = 5, C_{22} = 3, C_{23} = 6
$$
$$
C_{31} = 10, C_{32} = 6, C_{33} = 12
$$
The adjoint of $A$ is:
$$
\operatorname{adj}(A) =
\begin{bmatrix}
-5 & 5 & 10 \\
-3 & 3 & 6 \\
-6 & 6 & 12
\end{bmatrix}
$$
Now,
$$
(\operatorname{adj} A)B =
\begin{bmatrix}
-5 & 5 & 10 \\
-3 & 3 & 6 \\
-6 & 6 & 12
\end{bmatrix}\begin{bmatrix}
2 \\
-1 \\
3
\end{bmatrix}
$$
$$
(\operatorname{adj} A)B =
\begin{bmatrix}
(-5 \times 2) + (5 \times -1) + (10 \times 3) \\
(-3 \times 2) + (3 \times -1) + (6 \times 3) \\
(-6 \times 2) + (6 \times -1) + (12 \times 3)
\end{bmatrix}
$$
$$
(\operatorname{adj} A)B =
\begin{bmatrix}
-10 – 5 + 30 \\
-6 – 3 + 18 \\
-12 – 6 + 36
\end{bmatrix}=\begin{bmatrix}
15 \\
9 \\
18
\end{bmatrix}
$$
Since $(\operatorname{adj} A)B \ne O$, the system of equations is inconsistent and has no solution.
This detailed matrix-method solution on consistency of linear equations is published by Anand Classes and written by Neeraj Anand, Director and Main Head Faculty, crafted for strong NCERT conceptual clarity and exam-oriented learning for Class 12 mathematics.
$\therefore$ System of equation is inconsistent.
This NCERT-based solution on consistency of linear equations using matrices is published by Anand Classes and written by Neeraj Anand, Director, offering reliable conceptual clarity for Class 12 Mathematics and competitive exam preparation.
NCERT Question 6 : Examine the consistency of the system of equations
$5x – y + 4z = 5$
$2x + 3y + 5z = 2$
$5x – 2y + 6z = -1$
Solution
Matrix form of the given equations is $AX = B$.
where,
$$
A =
\begin{bmatrix}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{bmatrix},
\quad
B =
\begin{bmatrix}
5 \\
2 \\
-1
\end{bmatrix}
\quad \text{and} \quad
X =
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
$$
Therefore
$$
\begin{bmatrix}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}=\begin{bmatrix}
5 \\
2 \\
-1
\end{bmatrix}
$$
Now,
$$
|A| =
\begin{vmatrix}
5 & -1 & 4 \
2 & 3 & 5 \
5 & -2 & 6
\end{vmatrix}
$$
$$
|A|
= 5(8 + 10) – (-1)(12 – 25) + 4(-4 – 15)
$$
$$
|A|= 140 – 13 – 76
$$
$$
|A|= 140 – 89 = 51 \ne 0
$$
$\because$ inverse of matrix exists, unique solution.
$$
\therefore \text{System of equation is consistent.}
$$
This solution on consistency of simultaneous linear equations using matrix method is published by Anand Classes and authored by Neeraj Anand, Main Head Faculty, designed to strengthen Class 12 NCERT fundamentals and support higher-level entrance exam preparation.
FAQ Section
Q1. What is covered in Determinants Exercise 4.5 Class 12 Math?
Exercise 4.5 focuses on advanced applications of determinants, including solving problems using properties of determinants and evaluating determinant-based expressions.
Q2. Are these NCERT Solutions suitable for CBSE board exams?
Yes, the solutions are fully aligned with the latest CBSE syllabus and NCERT textbook, making them ideal for board exam preparation.
Q3. Can I download Determinants Exercise 4.5 solutions in PDF for free?
Yes, Anand Classes provides free downloadable PDF notes and solutions for easy access and revision.
Q4. Are step-by-step solutions provided?
Yes, each question is solved in a clear, step-by-step manner to help students understand concepts and improve accuracy.
Q5. Is this material useful for last-minute revision?
Absolutely. The concise explanations and exam-focused approach make these notes perfect for quick revision before exams.
