Anand Classes provides well-explained and exam-oriented NCERT Solutions for Determinants Exercise 4.4 Class 12 Math Chapter-4 (Set-2) to help students master important concepts with ease. These solutions are prepared strictly according to the latest CBSE syllabus and NCERT guidelines, ensuring accuracy, step-by-step explanations, and clarity for board exam preparation. The notes are ideal for quick revision, concept strengthening, and problem-solving practice. Click the print button to download study material and notes.
Access NCERT Solutions for Determinants Exercise 4.4 Class 12 Math Chapter-4
NCERT Question.11 : Find the inverse of the matrix
$$A=\begin{bmatrix}
1 & 0 & 0\\
0 & \cos\alpha & \sin\alpha\\
0 & \sin\alpha & -\cos\alpha
\end{bmatrix}.$$
Solution
Compute the determinant (expand along the first row):
$$|A|=1\cdot\begin{vmatrix}\cos\alpha & \sin\alpha\\\sin\alpha & -\cos\alpha\end{vmatrix}$$
$$|A|=\cos\alpha(-\cos\alpha)-\sin\alpha\sin\alpha
=-(\cos^2\alpha+\sin^2\alpha)=-1.$$
Compute cofactors $A_{ij}=(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the minor of $a_{ij}$.
$M_{11}=\begin{vmatrix}\cos\alpha & \sin\alpha\\\sin\alpha & -\cos\alpha\end{vmatrix}=-(\cos^2\alpha+\sin^2\alpha)=-1$
$ A_{11}=(-1)^{2}(-1)=-1.$
$M_{12}=M_{13}=0\quad\Rightarrow\quad A_{12}=0,A_{13}=0.$
$M_{21}=0\quad\Rightarrow\quad A_{21}=0.$
$M_{22}=\begin{vmatrix}1 & 0\\0 & -\cos\alpha\end{vmatrix}=-\cos\alpha\quad\Rightarrow\quad A_{22}=(-1)^{4}(-\cos\alpha)=-\cos\alpha.$
$M_{23}=\begin{vmatrix}1 & 0\\0 & \sin\alpha\end{vmatrix}=\sin\alpha\quad\Rightarrow\quad A_{23}=(-1)^{2+3}(\sin\alpha)=-\sin\alpha.$
$M_{31}=0\quad\Rightarrow\quad A_{31}=0.$
$M_{32}=\begin{vmatrix}1 & 0\\0 & \sin\alpha\end{vmatrix}=\sin\alpha\quad\Rightarrow\quad A_{32}=(-1)^{3+2}(\sin\alpha)=-\sin\alpha.$
$M_{33}=\begin{vmatrix}1 & 0\\0 & \cos\alpha\end{vmatrix}=\cos\alpha\quad\Rightarrow\quad A_{33}=(-1)^{6}(\cos\alpha)=\cos\alpha.$
Thus the cofactor matrix is
$$\begin{bmatrix}
-1 & 0 & 0\\
0 & -\cos\alpha & -\sin\alpha\\
0 & -\sin\alpha & \cos\alpha
\end{bmatrix}$$
and the adjoint (transpose of cofactor matrix) is
$$\operatorname{adj}A=
\begin{bmatrix}
-1 & 0 & 0\\
0 & -\cos\alpha & -\sin\alpha\\
0 & -\sin\alpha & \cos\alpha
\end{bmatrix}.$$
Inverse using $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A$ with $|A|=-1$:
$$A^{-1}=\dfrac{1}{-1}\begin{bmatrix}
-1 & 0 & 0\\
0 & -\cos\alpha & -\sin\alpha\\
0 & -\sin\alpha & \cos\alpha
\end{bmatrix}
=\begin{bmatrix}
1 & 0 & 0\\
0 & \cos\alpha & \sin\alpha\\
0 & \sin\alpha & -\cos\alpha
\end{bmatrix}=A.$$
Final Result
$$\boxed{A^{-1}=A=\begin{bmatrix}
1 & 0 & 0\\
0 & \cos\alpha & \sin\alpha\\
0 & \sin\alpha & -\cos\alpha
\end{bmatrix}}$$
Published by Anand Classes, Written by Neeraj Anand — a clear NCERT Class 12 solution showing this matrix is its own inverse, ideal for CBSE, JEE and CUET revision with step-by-step adjoint and determinant computations.
NCERT Question.12 : Let
$$A=\begin{bmatrix}3 & 7\\2 & 5\end{bmatrix}\quad \text{and} \quad B=\begin{bmatrix}6 & 8\\7 & 9\end{bmatrix}.$$
Verify that $(AB)^{-1}=B^{-1}A^{-1}.$
Solution
Step 1: Compute $A^{-1}$
Determinant of $A$:
$$|A|=3\cdot 5-7\cdot 2=15-14=1.$$
Since $|A|\ne 0$, inverse exists.
For a $2\times 2$ matrix $A=\begin{bmatrix}a & b\\c & d\end{bmatrix}$,
$$A^{-1}=\frac{1}{|A|}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}.$$
So,
$$A^{-1}=\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix}.$$
Step 2: Compute $B^{-1}$
Determinant of $B$:
$$|B|=6\cdot 9-8\cdot 7=54-56=-2.$$
So,
$$B^{-1}=\frac{1}{-2}\begin{bmatrix}9 & -8\\-7 & 6\end{bmatrix}
=\begin{bmatrix}-\frac{9}{2} & 4\\\frac{7}{2} & -3\end{bmatrix}.$$
Step 3: Compute $B^{-1}A^{-1}$
Multiply $B^{-1}A^{-1}$:
$$B^{-1}A^{-1}=
\begin{bmatrix}-\frac{9}{2} & 4\\\frac{7}{2} & -3\end{bmatrix}
\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix}.$$
Compute product entries:
- First row, first column:
$$-\frac{9}{2}\cdot 5+4(-2)=-\frac{45}{2}-8=-\frac{45}{2}-\frac{16}{2}=-\frac{61}{2},$$ - First row, second column:
$$-\frac{9}{2}(-7)+4\cdot 3=\frac{63}{2}+12=\frac{63}{2}+\frac{24}{2}=\frac{87}{2},$$ - Second row, first column:
$$\frac{7}{2}\cdot 5+(-3)(-2)=\frac{35}{2}+6=\frac{35}{2}+\frac{12}{2}=\frac{47}{2},$$ - Second row, second column:
$$\frac{7}{2}(-7)+(-3)\cdot 3=-\frac{49}{2}-9=-\frac{49}{2}-\frac{18}{2}=-\frac{67}{2}.$$
So,
$$B^{-1}A^{-1}=
\begin{bmatrix}
-\frac{61}{2} & \frac{87}{2}\\
\frac{47}{2} & -\frac{67}{2}
\end{bmatrix}.$$
Step 4: Compute $(AB)^{-1}$
First find $AB$:
$$AB=
\begin{bmatrix}3 & 7\\2 & 5\end{bmatrix}\begin{bmatrix}6 & 8\\7 & 9\end{bmatrix}$$
$$AB=\begin{bmatrix}18+49 & 24+63\\12+35 & 16+45\end{bmatrix}
=\begin{bmatrix}67 & 87\\47 & 61\end{bmatrix}.$$
Determinant of $AB$:
$$|AB|=67\cdot 61-87\cdot 47=-2.$$
So,
$$(AB)^{-1}=\frac{1}{-2}
\begin{bmatrix}61 & -87\\-47 & 67\end{bmatrix}$$
$$(AB)^{-1}=\begin{bmatrix}
-\frac{61}{2} & \frac{87}{2}\\
\frac{47}{2} & -\frac{67}{2}
\end{bmatrix}.$$
We see that
$$(AB)^{-1}=B^{-1}A^{-1}.$$
Final Result
$$\boxed{(AB)^{-1}=B^{-1}A^{-1}}$$
Published by Anand Classes, Director by Neeraj Anand — detailed NCERT Class 12 solution verifying the matrix identity $(AB)^{-1}=B^{-1}A^{-1}$ using determinants and adjoint, ideal for CBSE, JEE and CUET preparation.
NCERT Question.13 : Let
$$A = \begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}.$$
Show that $A^2 – 5A + 7I = O$ and hence find $A^{-1}$.
Solution
Step 1: Compute $A^2$
$$
A^2 = A \cdot A =
\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}
\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}.
$$
Multiply:
$$ A^2 =
\begin{bmatrix}
3\cdot 3 + 1\cdot (-1) & 3\cdot 1 + 1\cdot 2\\
-1\cdot 3 + 2\cdot (-1) & -1\cdot 1 + 2\cdot 2
\end{bmatrix}$$
$$ A^2 =\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}$$
Step 2: Compute $A^2 – 5A + 7I$
$$
5A = 5 \begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} = \begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix},\quad
7I = 7 \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} = \begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}.
$$
Now,
$$
A^2 – 5A + 7I =\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}+\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}$$
Compute element-wise:
$$
A^2 – 5A + 7I =\begin{bmatrix}8-15+7 & 5-5+0\\-5+5+0 & 3-10+7\end{bmatrix} =
\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix} = O.
$$
Thus,
$$\boxed{A^2 – 5A + 7I = O}.$$
Step 3: Find $A^{-1}$
From $A^2 – 5A + 7I = O$, we have:
$$A^2 – 5A = -7I \quad \Rightarrow \quad A(A-5I) = -7I.$$
Multiply both sides by $A^{-1}$:
$$A^{-1} A (A-5I) = -7 A^{-1} \quad \Rightarrow \quad (A-5I) = -7 A^{-1}.$$
So,
$$A^{-1} = -\frac{1}{7}(A-5I) = \frac{1}{7}(5I – A).$$
Compute $5I – A$:
$$
5I – A = \begin{bmatrix}5 & 0\\0 & 5\end{bmatrix} – \begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} = \begin{bmatrix}2 & -1\\1 & 3\end{bmatrix}.
$$
Hence,
$$
A^{-1} = \frac{1}{7} \begin{bmatrix}2 & -1\\1 & 3\end{bmatrix}.
$$
Final Result
$$\boxed{A^{-1} = \frac{1}{7} \begin{bmatrix}2 & -1\\1 & 3\end{bmatrix}}$$
Published by Anand Classes, Written by Neeraj Anand — step-by-step NCERT Class 12 solution illustrating how to compute the inverse of a matrix using the characteristic equation method, perfect for CBSE, JEE and CUET exam preparation.
NCERT Question.14 : For the matrix
$$A = \begin{bmatrix}3 & 2\\1 & 1\end{bmatrix},$$
find the numbers $a$ and $b$ such that
$$A^2 + aA + bI = O.$$
Solution
Step 1: Compute $A^2$
$$
A^2 = A \cdot A =
\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix}
\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix}.
$$
Multiply:
$$
A^2 =
\begin{bmatrix}
3\cdot 3 + 2\cdot 1 & 3\cdot 2 + 2\cdot 1\\
1\cdot 3 + 1\cdot 1 & 1\cdot 2 + 1\cdot 1
\end{bmatrix}$$
$$
A^2 =\begin{bmatrix}11 & 8\\4 & 3\end{bmatrix}.
$$
Step 2: Assume $A^2 + aA + bI = O$
We can write:
$$
A^2 + aA + bI = O \quad \Rightarrow \quad A^2 + aA = -bI.
$$
Step 3: Find $A^{-1}$
Inverse of $A$:
$$
A^{-1} = \frac{1}{|A|}\begin{bmatrix}d & -b\\-c & a\end{bmatrix},
$$
where $|A| = 3\cdot 1 – 2\cdot 1 = 1$, so
$$
A^{-1} = \begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}.
$$
Step 4: Express $A^{-1}$ in terms of $a$ and $b$
From Step 2, multiply both sides by $-1/b$:
$$
A^{-1} = -\frac{1}{b}(A + aI)
= -\frac{1}{b}
\begin{bmatrix}3+a & 2\\1 & 1+a\end{bmatrix}.
$$
Compare with $A^{-1} = \begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}$:
- Top-left entry: $-(3+a)/b = 1 \Rightarrow -3 – a = b$.
- Top-right entry: $-2/b = -2 \Rightarrow b = 1$.
Then $-3 – a = 1 \Rightarrow a = -4$.
Final Result
$$\boxed{a = -4, b = 1}$$
Published by Anand Classes, Written by Neeraj Anand — detailed NCERT Class 12 solution showing how to find coefficients $a$ and $b$ in $A^2 + aA + bI = O$ using matrix operations, ideal for CBSE, JEE, and CUET exam preparation.
NCERT Question.15 : Let
$$
A=\begin{bmatrix}
1 & 1 & 2\\
1 & 2 & -1\\
1 & -3 & 3
\end{bmatrix}
$$
Show that $A^3-6A^2+5A+11I=O$
and hence find $A^{-1}$.
Solution
Step 1: Find $A^2$
$$
A^2=A\cdot A=
\begin{bmatrix}
1 & 1 & 2\\
1 & 2 & -1\\
1 & -3 & 3
\end{bmatrix}\begin{bmatrix}
1 & 1 & 2\\
1 & 2 & -1\\
1 & -3 & 3
\end{bmatrix}.
$$
After multiplication,
$$
A^2=
\begin{bmatrix}
4 & 2 & 1\\
-3 & 8 & -14\\
7 & -3 & 14
\end{bmatrix}.
$$
Step 2: Find $A^3$
$$
A^3=A^2\cdot A.
$$
$$
A^3=
\begin{bmatrix}
4 & 2 & 1\\
-3 & 8 & -14\\
7 & -3 & 14
\end{bmatrix}\begin{bmatrix}
1 & 1 & 2\\
1 & 2 & -1\\
1 & -3 & 3
\end{bmatrix}.
$$
On multiplication,
$$
A^3=
\begin{bmatrix}
8 & 7 & 1\\
-23 & 27 & -69\\
32 & -13 & 58
\end{bmatrix}.
$$
Step 3: Verify $A^3-6A^2+5A+11I=O$
$$A^3-6A^2+5A+11I=\begin{bmatrix}
8 & 7 & 1\\
-23 & 27 & -69\\
32 & -13 & 58
\end{bmatrix}
-6\begin{bmatrix}
4 & 2 & 1\\
-3 & 8 & -14\\
7 & -3 & 14
\end{bmatrix}+\\[1em]+5\begin{bmatrix}
1 & 1 & 2\\
1 & 2 & -1\\
1 & -3 & 3
\end{bmatrix}+11\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}.
$$
Simplifying term by term,
$$
A^3-6A^2+5A+11I=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
=O.
$$
Hence,
$$
\boxed{A^3-6A^2+5A+11I=O}.
$$
Step 4: Find $A^{-1}$
From
$$
A^3-6A^2+5A+11I=O,
$$
multiply both sides by $A^{-1}$,
$$
A^2-6A+5I+11IA^{-1}=OA^{-1}
$$
$$
A^2-6A+5I+11A^{-1}=O
$$
$$
A^2-6A+5I=O-11A^{-1}
$$
$$
A^2-6A+5I=-11A^{-1}.
$$
Therefore,
$$
A^{-1}=-\frac{1}{11}(A^2-6A+5I).
$$
Now,
$$
A^2-6A+5I=
\begin{bmatrix}
3 & -4 & -5\\
-9 & 1 & 4\\
-5 & 3 & 1
\end{bmatrix}.
$$
Hence,
$$
\boxed{
A^{-1}=-\frac{1}{11}
\begin{bmatrix}
3 & -4 & -5\\
-9 & 1 & 4\\
-5 & 3 & 1
\end{bmatrix}
}
$$
Published by Anand Classes, Authored by Neeraj Anand — this complete NCERT Class 12 solution demonstrates the use of matrix polynomial identities to verify cubic equations and efficiently compute the inverse of a matrix, highly useful for CBSE board exams and competitive exams like JEE.
NCERT Question.16 : Let
$$
A=\begin{bmatrix}
2 & -1 & 1\\
-1 & 2 & -1\\
1 & -1 & 2
\end{bmatrix}.
$$
Verify that $A^3-6A^2+9A-4I=O$
and hence find $A^{-1}$.
Solution
Step 1: Find $A^2$
$$
A^2=A\cdot A=
\begin{bmatrix}
2 & -1 & 1\\
-1 & 2 & -1\\
1 & -1 & 2
\end{bmatrix}\begin{bmatrix}
2 & -1 & 1\\
-1 & 2 & -1\\
1 & -1 & 2
\end{bmatrix}.
$$
After multiplication,
$$
A^2=
\begin{bmatrix}
6 & -5 & 5\\
-5 & 6 & -5\\
5 & -5 & 6
\end{bmatrix}.
$$
Step 2: Find $A^3$
$$
A^3=A^2\cdot A.
$$
$$A^3=\begin{bmatrix}
6 & -5 & 5\\
-5 & 6 & -5\\
5 & -5 & 6
\end{bmatrix}\begin{bmatrix}
2 & -1 & 1\\
-1 & 2 & -1\\
1 & -1 & 2
\end{bmatrix}.
$$
On multiplication,
$$
A^3=\begin{bmatrix}
22 & -21 & 21\\
-21 & 22 & -21\\
21 & -21 & 22
\end{bmatrix}.
$$
Step 3: Verify $A^3-6A^2+9A-4I=O$
$$A^3-6A^2+9A-4I=\begin{bmatrix}
22 & -21 & 21\\
-21 & 22 & -21\\
21 & -21 & 22
\end{bmatrix}-6\begin{bmatrix}
6 & -5 & 5\\
-5 & 6 & -5\\
5 & -5 & 6
\end{bmatrix}+\\[1em]+9\begin{bmatrix}
2 & -1 & 1\\
-1 & 2 & -1\\
1 & -1 & 2
\end{bmatrix}-4\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}.
$$
Simplifying,
$$
A^3-6A^2+9A-4I=\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
=O
$$
Hence,
$$
\boxed{A^3-6A^2+9A-4I=O}.
$$
Step 4: Find $A^{-1}$
From
$$
A^3-6A^2+9A-4I=O
$$
multiplying both sides by $A^{-1}$,
$$
A^2-6A+9I=4A^{-1}.
$$
Thus,
$$
A^{-1}=\frac{1}{4}(A^2-6A+9I).
$$
Now,
$$
A^2-6A+9I=
\begin{bmatrix}
3 & 1 & -1\\
1 & 3 & 1\\
-1 & 1 & 3
\end{bmatrix}.
$$
Therefore,
$$
\boxed{
A^{-1}=\frac{1}{4}
\begin{bmatrix}
3 & 1 & -1\\
1 & 3 & 1\\
-1 & 1 & 3
\end{bmatrix}
}
$$
Published by Anand Classes, Director by Neeraj Anand — this NCERT Class 12 solution clearly verifies a cubic matrix equation and derives the inverse using polynomial identities, making it extremely helpful for CBSE board preparation and competitive exams such as JEE and CUET.
NCERT Question.17 : Let $A$ be a non-singular matrix of order $3 \times 3$. Then $|\operatorname{adj} A|$ is equal to
(A) $|A|$
(B) $|A|^2$
(C) $|A|^3$
(D) $3|A|$
Solution
We know the standard identity:
$$
(\operatorname{adj} A)A = |A|I.
$$
Taking determinant on both sides,
$$
|(\operatorname{adj} A)A| = ||A|I|.
$$
Using the property $|AB| = |A||B|$,
$$
|\operatorname{adj} A||A| = |A|^3,
$$
since for a $3 \times 3$ identity matrix $I$,
$$
||A|I| = |A|^3.
$$
Dividing both sides by $|A|$ (as $A$ is non-singular, $|A|\neq 0$),
$$
|\operatorname{adj} A| = |A|^2.
$$
Final Result
$$
\boxed{|\operatorname{adj} A| = |A|^2}
$$
Hence, option (B) is correct.
In general for $A$ be a non-singular matrix of order $n \times n$.
Then $|\operatorname{adj} A|$ =$|A|^{n-1}$
Published by Anand Classes, Main Head Faculty by Neeraj Anand — this concise NCERT Class 12 solution explains determinant properties of adjoint matrices, essential for CBSE board exams and competitive exams like JEE and CUET.
NCERT Question.18 : If $A$ is an invertible matrix of order $2$, then $\det(A^{-1})$ is equal to
(A) $\det(A)$
(B) $\dfrac{1}{\det A}$
(C) $1$
(D) $0$
Solution
Since $A$ is invertible, $A^{-1}$ exists.
We know,
$$
A^{-1}=\frac{\operatorname{adj}A}{|A|}.
$$
Let
$$
A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}.
$$
Then,
$$
|A|=ad-bc
$$
and
$$
\operatorname{adj}A=\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}.
$$
Hence,
$$
A^{-1}=\frac{1}{|A|}
\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}.
$$
$$
A^{-1}=
\begin{bmatrix} \frac{d}{|A|} & \frac{-b}{|A|}\\ \frac{-c}{|A|} & \frac{a}{|A|} \end{bmatrix}.
$$
Now, taking determinant on both sides,
$$
|A^{-1}|=\frac{1}{|A|^2}
\begin{vmatrix}
d & -b\\
-c & a
\end{vmatrix}.
$$
$$
|A^{-1}|=\frac{1}{|A|^2}(ad-bc).
$$
Since $ad-bc=|A|$,
$$
|A^{-1}|=\frac{|A|}{|A|^2}=\frac{1}{|A|}.
$$
Thus,
$$
\boxed{\det(A^{-1})=\frac{1}{\det A}}.
$$
Final Result
Correct option is (B).
Published by Anand Classes, Written by Neeraj Anand — this NCERT Class 12 solution explains the determinant of an inverse matrix using adjoint properties, a key concept for CBSE board exams and competitive exams such as JEE and CUET.
FAQ Section
Q1. What is covered in Determinants Exercise 4.4 Class 12 Maths?
Exercise 4.4 focuses on applications of determinants, including solving problems related to areas of triangles and properties of determinants.
Q2. Are these NCERT Solutions based on the latest CBSE syllabus?
Yes, all solutions provided by Anand Classes are fully aligned with the latest CBSE syllabus and NCERT textbooks.
Q3. Is the PDF download for Determinants Exercise 4.4 Set-2 free?
Yes, students can download the complete NCERT Solutions PDF for free.
Q4. How are Anand Classes notes helpful for board exams?
The notes include step-by-step solutions, exam-oriented explanations, and clear methods that help students score better in CBSE board exams.
Q5. Can these solutions be used for revision before exams?
Absolutely. These solutions are ideal for last-minute revision and quick concept clarification before exams.
