Anand Classes provides detailed and exam-oriented NCERT Solutions for Determinants Exercise 4.5 of Class 12 Mathematics Chapter-4, prepared strictly as per the latest CBSE syllabus and marking scheme. These step-by-step solutions help students master important determinant concepts, improve problem-solving accuracy, and build confidence for board examinations and competitive exams. Click the print button to download study material and notes.
Access NCERT Solutions for Determinants Exercise 4.5 of Class 12 Mathematics Chapter-4
NCERT Question.7 : Solve system of linear equations using Matrix Method from given system of linear equations:
$5x + 2y = 4$
$7x + 3y = 5$
Solution
The given system of linear equations is:
$$
5x + 2y = 4
$$
$$
7x + 3y = 5
$$
Matrix form of the given equations is $AX = B$
where,
$$
A =
\begin{bmatrix}
5 & 2 \\
7 & 3
\end{bmatrix},
\quad
X =
\begin{bmatrix}
x \\
y
\end{bmatrix},
\quad
B =
\begin{bmatrix}
4 \\
5
\end{bmatrix}
$$
Determinant of $A$
$$
|A| =
\begin{vmatrix}
5 & 2 \\
7 & 3
\end{vmatrix}
= (5 \times 3) – (2 \times 7)
= 15 – 14
= 1 \ne 0
$$
Since $|A| \ne 0$, the inverse of $A$ exists and the system has a unique solution.
Finding $$X = A^{-1}B$$
$$
A^{-1} = \frac{1}{|A|} \operatorname{adj}(A)
$$
Adjoint of $A$ is:
$$
\operatorname{adj}(A) =
\begin{bmatrix}
3 & -2 \\
-7 & 5
\end{bmatrix}
$$
Therefore,
$$
X = A^{-1}B
= \frac{1}{1}
\begin{bmatrix}
3 & -2 \\
-7 & 5
\end{bmatrix}
\begin{bmatrix}
4 \\
5
\end{bmatrix}
$$
$$
X =\begin{bmatrix}
3 \times 4 – 2 \times 5 \\
-7 \times 4 + 5 \times 5
\end{bmatrix}$$
$$
X =\begin{bmatrix}
12 – 10 \\
-28 + 25
\end{bmatrix}$$
$$
\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
2 \\
-3
\end{bmatrix}
$$
$$x = 2, y = -3$$
Final Result
$$
\boxed{x = 2, y = -3}
$$
This complete NCERT-style matrix method solution for simultaneous linear equations is published by Anand Classes and authored by Neeraj Anand, Director and Main Head Faculty, designed to support Class 12 students preparing for CBSE board exams, competitive exams, and concept-based mastery of matrices.
NCERT Question.8 : Solve the system of linear equations using Matrix Method
$
2x – y = -2
$
$
3x + 4y = 3
$
Solution :
The given system of linear equations is:
$$
2x – y = -2
$$
$$
3x + 4y = 3
$$
Matrix form of the given equations is $AX = B$
where,
$$
A =
\begin{bmatrix}
2 & -1 \\
3 & 4
\end{bmatrix}
\quad
X =
\begin{bmatrix}
x \\
y
\end{bmatrix}
\quad
B =
\begin{bmatrix}
-2 \\
3
\end{bmatrix}
$$
Thus,
$$
AX = B
$$
Determinant of $A$
$$
|A| =
\begin{vmatrix}
2 & -1 \\
3 & 4
\end{vmatrix}
= (2 \times 4) – (-1 \times 3)
= 8 + 3
= 11 \ne 0
$$
Since $|A| \ne 0$, the inverse of matrix $A$ exists and the system has a unique solution.
Finding $X = A^{-1}B$
$$
A^{-1} = \frac{1}{|A|}\operatorname{adj}(A)
$$
Adjoint of $A$ is:
$$
\operatorname{adj}(A) =
\begin{bmatrix}
4 & 1 \\
-3 & 2
\end{bmatrix}
$$
Therefore,
$$
X = A^{-1}B
= \frac{1}{11}
\begin{bmatrix}
4 & 1 \\
-3 & 2
\end{bmatrix}
\begin{bmatrix}
-2 \\
3
\end{bmatrix}
$$
$$
X = \frac{1}{11}
\begin{bmatrix}
(4)(-2) + (1)(3) \\
(-3)(-2) + (2)(3)
\end{bmatrix}$$
$$X =\frac{1}{11}
\begin{bmatrix}
-8 + 3 \\
6 + 6
\end{bmatrix}$$
$$\begin{bmatrix}
x \\
y
\end{bmatrix}
=\begin{bmatrix}
-\frac{5}{11} \\
\frac{12}{11}
\end{bmatrix}
$$
Final Result
$$
\boxed{x = -\frac{5}{11}, y = \frac{12}{11}}
$$
This NCERT-based solution using the matrix inverse method is published by Anand Classes and written by Neeraj Anand, Director and Main Head Faculty, offering reliable Class 12 Mathematics support for CBSE board preparation, competitive exams, and strong conceptual understanding of linear equations using matrices.
NCERT Question.9 : Solve the system of linear equations using Matrix Method
$
4x – 3y = 3
$
$
3x – 5y = 7
$
Solution
The given system of linear equations is:
$$
4x – 3y = 3
$$
$$
3x – 5y = 7
$$
Matrix form of the given equations is $AX = B$
where,
$$
A =
\begin{bmatrix}
4 & -3 \\
3 & -5
\end{bmatrix}
\quad
X =
\begin{bmatrix}
x \
y
\end{bmatrix}
\quad
B =
\begin{bmatrix}
3 \
7
\end{bmatrix}
$$
Thus,
$$
AX = B
$$
Determinant of $A$
$$
|A| =
\begin{vmatrix}
4 & -3 \
3 & -5
\end{vmatrix}
= (4 \times -5) – (-3 \times 3)
= -20 + 9
= -11 \ne 0
$$
Since $|A| \ne 0$, the inverse of matrix $A$ exists and the system has a unique solution.
Finding $X = A^{-1}B$
$$
A^{-1} = \frac{1}{|A|}\operatorname{adj}(A)
$$
Adjoint of $A$ is:
$$
\operatorname{adj}(A) =
\begin{bmatrix}
-5 & 3 \
-3 & 4
\end{bmatrix}
$$
Therefore,
$$
X = A^{-1}B
= \frac{1}{-11}
\begin{bmatrix}
-5 & 3 \
-3 & 4
\end{bmatrix}
\begin{bmatrix}
3 \
7
\end{bmatrix}
$$
$$
X = \frac{1}{-11}
\begin{bmatrix}
(-5)(3) + (3)(7) \
(-3)(3) + (4)(7)
\end{bmatrix}
\frac{1}{-11}
\begin{bmatrix}
-15 + 21 \
-9 + 28
\end{bmatrix}
\begin{bmatrix}
-\frac{6}{11} \
-\frac{19}{11}
\end{bmatrix}
$$
Final Result
$$
\boxed{x = -\frac{6}{11},; y = -\frac{19}{11}}
$$
This NCERT-style solution using the matrix inverse method is published by Anand Classes and authored by Neeraj Anand, Director and Main Head Faculty, providing trusted Class 12 Mathematics content ideal for CBSE board exams, school assessments, and competitive exam preparation with clear matrix-based techniques.
