Anand Classes provides the most accurate and exam-focused Determinants NCERT Solutions for Exercise 4.4 of Class 12 Maths Chapter 4, designed to help students strengthen conceptual understanding and score higher in board exams. These solutions are prepared by expert faculty using step-by-step methods that strictly follow the latest NCERT guidelines, ensuring clarity, precision, and ease of revision. This downloadable PDF is ideal for quick practice, last-minute preparation, and gaining confidence in determinant properties and applications. Click the print button to download study material and notes.

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Access NCERT Solutions for Exercise 4.4 of Class 12 Maths Chapter 4 Determinants

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NCERT Question.1 : Find the adjoint of the matrix
$$A=\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}.$$

Solution
Compute the cofactors $A_{ij}$ of $A$.

The minor and cofactor values are

$$M_{11}=4\quad\Rightarrow\quad A_{11}=(-1)^{1+1}M_{11}=4$$

$$M_{12}=3\quad\Rightarrow\quad A_{12}=(-1)^{1+2}M_{12}=-3$$

$$M_{21}=2\quad\Rightarrow\quad A_{21}=(-1)^{2+1}M_{21}=-2$$

$$M_{22}=1\quad\Rightarrow\quad A_{22}=(-1)^{2+2}M_{22}=1$$

The matrix of cofactors is

$$\begin{bmatrix}A_{11} & A_{12}\\A_{21} & A_{22}\end{bmatrix}
=\begin{bmatrix}4 & -3\\-2 & 1\end{bmatrix}$$

The adjoint (adjugate) is the transpose of the cofactor matrix, so

$$\operatorname{adj}A=\begin{bmatrix}4 & -2\\-3 & 1\end{bmatrix}.$$

Final Result

$$\boxed{\operatorname{adj}A=\begin{bmatrix}4 & -2\\-3 & 1\end{bmatrix}}$$

Published by Anand Classes, Authored by Neeraj Anand — concise NCERT Class 12 Maths solutions for adjoint and inverse matrices, crafted to help students prepare effectively for CBSE, JEE and CUET with clear step-by-step explanations.

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NCERT Question.2 : Find the adjoint of the matrix
$$A=\begin{bmatrix}1 & -1 & 2\\2 & 3 & 5\\-2 & 0 & 1\end{bmatrix}$$

Solution
Compute cofactors $A_{ij}=(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the minor of $a_{ij}$.

$$M_{11}=\begin{vmatrix}3 & 5\\0 & 1\end{vmatrix}=3\cdot 1-5\cdot 0=3\quad\Rightarrow\quad A_{11}=3$$

$$M_{12}=\begin{vmatrix}2 & 5\\-2 & 1\end{vmatrix}=2\cdot 1-5\cdot(-2)=2+10=12$$

$$A_{12}=(-1)^{1+2}(12)=-12$$

$$M_{13}=\begin{vmatrix}2 & 3\\-2 & 0\end{vmatrix}=2\cdot 0-3\cdot(-2)=0+6=6\quad\Rightarrow\quad A_{13}=6$$

$$M_{21}=\begin{vmatrix}-1 & 2\\0 & 1\end{vmatrix}=(-1)\cdot 1-2\cdot 0=-1\quad\Rightarrow\quad A_{21}=(-1)^{2+1}(-1)=1$$

$$M_{22}=\begin{vmatrix}1 & 2\\-2 & 1\end{vmatrix}=1\cdot 1-2\cdot(-2)=1+4=5\quad\Rightarrow\quad A_{22}=5$$

$$M_{23}=\begin{vmatrix}1 & -1\\-2 & 0\end{vmatrix}=1\cdot 0-(-1)\cdot(-2)=0-2=-2$$

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

$$M_{31}=\begin{vmatrix}-1 & 2\\3 & 5\end{vmatrix}=(-1)\cdot 5-2\cdot 3=-5-6=-11\quad\Rightarrow\quad A_{31}=-11$$

$$M_{32}=\begin{vmatrix}1 & 2\\2 & 5\end{vmatrix}=1\cdot 5-2\cdot 2=5-4=1\quad\Rightarrow\quad A_{32}=(-1)^{3+2}(1)=-1$$

$$M_{33}=\begin{vmatrix}1 & -1\\2 & 3\end{vmatrix}=1\cdot 3-(-1)\cdot 2=3+2=5\quad\Rightarrow\quad A_{33}=5$$

Form the cofactor matrix and transpose it to get the adjoint:

Cofactor matrix:
$$\begin{bmatrix}A_{11} & A_{12} & A_{13}\\A_{21} & A_{22} & A_{23}\\A_{31} & A_{32} & A_{33}\end{bmatrix}
=\begin{bmatrix}3 & -12 & 6\\1 & 5 & 2\\-11 & -1 & 5\end{bmatrix}$$

Therefore the adjoint (adjugate) is the transpose:

$$\operatorname{adj}A=\begin{bmatrix}3 & 1 & -11\\-12 & 5 & -1\\6 & 2 & 5\end{bmatrix}.$$

Final Result

$$\boxed{\operatorname{adj}A=\begin{bmatrix}3 & 1 & -11\\-12 & 5 & -1\\6 & 2 & 5\end{bmatrix}}$$

Published by Anand Classes, Written by Neeraj Anand — dependable NCERT Class 12 Maths adjoint solutions with clear cofactor computation to boost your CBSE, JEE, and CUET preparation.

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NCERT Question.3 : Verify $$A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I.$$ for
$$A=\begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix}$$

Solution
Determinant of given matrix is

$$A=\begin{vmatrix}2 & 3\\-4 & -6\end{vmatrix}$$

First compute the determinant $|A|$:

$$|A|=2(-6)-3(-4)=-12+12=0$$

Next compute cofactors $A_{ij}=(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the minor of $a_{ij}$.

$$M_{11}=-6\quad\Rightarrow\quad A_{11}=(-1)^{1+1}(-6)=-6$$

$$M_{12}=-4\quad\Rightarrow\quad A_{12}=(-1)^{1+2}(-4)=4$$

$$M_{21}=3\quad\Rightarrow\quad A_{21}=(-1)^{2+1}(3)=-3$$

$$M_{22}=2\quad\Rightarrow\quad A_{22}=(-1)^{2+2}(2)=2$$

Cofactor matrix and adjoint (transpose of cofactor matrix):

$$\begin{bmatrix}A_{11} & A_{12}\\A_{21} & A_{22}\end{bmatrix}
=\begin{bmatrix}-6 & 4\\-3 & 2\end{bmatrix}$$

$$\operatorname{adj}A=\begin{bmatrix}-6 & -3\\4 & 2\end{bmatrix}$$

Now compute $A(\operatorname{adj}A)$:

$$A(\operatorname{adj}A)=
\begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix}
\begin{bmatrix}-6 & -3\\4 & 2\end{bmatrix}$$

Multiply matrices entrywise:

$$A(\operatorname{adj}A)=
\begin{bmatrix}
2(-6)+3(4) & 2(-3)+3(2)\\
-4(-6)+(-6)(4) & -4(-3)+(-6)(2)
\end{bmatrix}$$

$$A(\operatorname{adj}A)=
\begin{bmatrix}
-12+12 & -6+6\\
24-24 & 12-12
\end{bmatrix}$$

$$A(\operatorname{adj}A)=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$$

Since $|A|=0$ we have $|A|I=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$, so $A(\operatorname{adj}A)=|A|I$.

Similarly compute $(\operatorname{adj}A)A$ (or note symmetry for $2\times2$ case):

$$ (\operatorname{adj}A)A=
\begin{bmatrix}-6 & -3\\4 & 2\end{bmatrix}
\begin{bmatrix}2 & 3\\-4 & -6\end{bmatrix}$$

$$ (\operatorname{adj}A)A=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}
=|A|I $$

Final Result

$$\boxed{A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}}$$

Published by Anand Classes, Authored by Neeraj Anand — clear NCERT verification showing matrix and adjoint multiplication yields $|A|I$, ideal for strengthening your understanding of adjoint properties for CBSE, JEE and CUET preparation.

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NCERT Question.4 : Verify $$A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I.$$ for
$$A=\begin{bmatrix}1 & -1 & 2\\3 & 0 & -2\\1 & 0 & 3\end{bmatrix}$$

Solution
Compute the determinant $|A|$ by expansion (here using first row):

$$|A|=1\begin{vmatrix}0 & -2\\0 & 3\end{vmatrix}-(-1)\begin{vmatrix}3 & -2\\1 & 3\end{vmatrix}+2\begin{vmatrix}3 & 0\\1 & 0\end{vmatrix}$$

Evaluate minors:

$$\begin{vmatrix}0 & -2\\0 & 3\end{vmatrix}=0\cdot 3-(-2)\cdot 0=0$$

$$\begin{vmatrix}3 & -2\\1 & 3\end{vmatrix}=3\cdot 3-(-2)\cdot 1=9+2=11$$

$$\begin{vmatrix}3 & 0\\1 & 0\end{vmatrix}=3\cdot 0-0\cdot 1=0$$

So

$$|A|=1\cdot 0+1\cdot 11+2\cdot 0=11$$

Compute cofactors $A_{ij}=(-1)^{i+j}M_{ij}$ and form the cofactor matrix.

$M_{11}=\begin{vmatrix}0 & -2\\0 & 3\end{vmatrix}=0\quad\Rightarrow\quad A_{11}=0$

$M_{12}=\begin{vmatrix}3 & -2\\1 & 3\end{vmatrix}=11\quad\Rightarrow\quad A_{12}=(-1)^{1+2}(11)=-11$

$M_{13}=\begin{vmatrix}3 & 0\\1 & 0\end{vmatrix}=0\quad\Rightarrow\quad A_{13}=0$

$M_{21}=\begin{vmatrix}-1 & 2\\0 & 3\end{vmatrix}=(-1)\cdot 3-2\cdot 0=-3\quad\Rightarrow\quad A_{21}=(-1)^{2+1}(-3)=3$

$M_{22}=\begin{vmatrix}1 & 2\\1 & 3\end{vmatrix}=1\cdot 3-2\cdot 1=1\quad\Rightarrow\quad A_{22}=1$

$M_{23}=\begin{vmatrix}1 & -1\\1 & 0\end{vmatrix}=1\cdot 0-(-1)\cdot 1=1\quad\Rightarrow\quad A_{23}=(-1)^{2+3}(1)=-1$

$M_{31}=\begin{vmatrix}-1 & 2\\0 & -2\end{vmatrix}=(-1)(-2)-2\cdot 0=2\quad\Rightarrow\quad A_{31}=2$

$M_{32}=\begin{vmatrix}1 & 2\\3 & -2\end{vmatrix}=1(-2)-2\cdot 3=-8\quad\Rightarrow\quad A_{32}=(-1)^{3+2}(-8)=8$

$M_{33}=\begin{vmatrix}1 & -1\\3 & 0\end{vmatrix}=1\cdot 0-(-1)\cdot 3=3\quad\Rightarrow\quad A_{33}=3$

Cofactor matrix:

$$\begin{bmatrix}
A_{11} & A_{12} & A_{13}\\
A_{21} & A_{22} & A_{23}\\
A_{31} & A_{32} & A_{33}
\end{bmatrix}=\begin{bmatrix}
0 & -11 & 0\\
3 & 1 & -1\\
2 & 8 & 3
\end{bmatrix}$$

Adjoint (adjugate) is transpose of cofactor matrix:

$$\operatorname{adj}A=\begin{bmatrix}
0 & 3 & 2\\
-11 & 1 & 8\\
0 & -1 & 3
\end{bmatrix}$$

Compute $A(\operatorname{adj}A)$

$$A(\operatorname{adj}A)=
\begin{bmatrix}1 & -1 & 2\\3 & 0 & -2\\1 & 0 & 3\end{bmatrix}
\begin{bmatrix}0 & 3 & 2\\-11 & 1 & 8\\0 & -1 & 3\end{bmatrix}$$

Multiply to get each entry (showing key sums):

• First row:
  entry (1,1): $1\cdot 0+(-1)(-11)+2\cdot 0=0+11+0=11$
  entry (1,2): $1\cdot 3+(-1)(1)+2(-1)=3-1-2=0$
  entry (1,3): $1\cdot 2+(-1)(8)+2\cdot 3=2-8+6=0$
• Second row:
  entry (2,1): $3\cdot 0+0(-11)+(-2)\cdot 0=0$
  entry (2,2): $3\cdot 3+0\cdot 1+(-2)(-1)=9+0+2=11$
  entry (2,3): $3\cdot 2+0\cdot 8+(-2)\cdot 3=6+0-6=0$
• Third row:
  entry (3,1): $1\cdot 0+0(-11)+3\cdot 0=0$
  entry (3,2): $1\cdot 3+0\cdot 1+3(-1)=3+0-3=0$
  entry (3,3): $1\cdot 2+0\cdot 8+3\cdot 3=2+0+9=11$

Thus

$$A(\operatorname{adj}A)=\begin{bmatrix}11 & 0 & 0\\0 & 11 & 0\\0 & 0 & 11\end{bmatrix}=11I=|A|I.$$

Similarly compute $(\operatorname{adj}A)A$ (or note matrix algebra symmetry) to obtain the same result:

$$(\operatorname{adj}A)A=\begin{bmatrix}11 & 0 & 0\\0 & 11 & 0\\0 & 0 & 11\end{bmatrix}=|A|I.$$

Final Result

$$\boxed{A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I=11I}$$

Published by Anand Classes, Main Head Faculty by Neeraj Anand — clear NCERT Class 12 Determinants and adjoint verifications, crafted to help students excel in CBSE, JEE and CUET preparation with precise step-by-step computations.

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NCERT Question.5 : Find the inverse of the matrix
$$A=\begin{bmatrix}2 & -2\\4 & 3\end{bmatrix}$$

Solution
Compute the determinant $|A|$:

$$|A|=2\cdot 3-(-2)\cdot 4=6+8=14$$

Form the adjoint (adjugate) matrix for a $2\times 2$ matrix $A=\begin{bmatrix}a & b\\c & d\end{bmatrix}$ as $\operatorname{adj}A=\begin{bmatrix}d & -b\\-c & a\end{bmatrix}$. Thus

$$\operatorname{adj}A=\begin{bmatrix}3 & 2\\-4 & 2\end{bmatrix}$$

The inverse exists since $|A|\neq 0$ and is given by $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A$. Therefore

$$A^{-1}=\dfrac{1}{14}\begin{bmatrix}3 & 2\\-4 & 2\end{bmatrix}
=\begin{bmatrix}\dfrac{3}{14} & \dfrac{1}{7}\\-\dfrac{2}{7} & \dfrac{1}{7}\end{bmatrix}$$

Final Result

$$\boxed{A^{-1}=\begin{bmatrix}\dfrac{3}{14} & \dfrac{1}{7}\\-\dfrac{2}{7} & \dfrac{1}{7}\end{bmatrix}}$$

Published by Anand Classes, Director by Neeraj Anand — comprehensive NCERT Class 12 Maths solutions with precise inverse and adjoint computations to boost your CBSE, JEE and CUET preparation.

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NCERT Question.6 : Find the inverse of the matrix
$$A=\begin{bmatrix}-1 & 5\\-3 & 2\end{bmatrix}.$$

Solution
Compute the determinant $|A|$:

$$|A|=(-1)\cdot 2-5\cdot(-3)=-2+15=13$$

Since $|A|\neq 0$ the inverse exists.

For a $2\times 2$ matrix $A=\begin{bmatrix}a & b\\c & d\end{bmatrix}$ we have

$$\operatorname{adj}A=\begin{bmatrix}d & -b\\-c & a\end{bmatrix}$$

so for the given matrix

$$\operatorname{adj}A=\begin{bmatrix}2 & -5\\3 & -1\end{bmatrix}$$

Therefore

$$A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A
=\dfrac{1}{13}\begin{bmatrix}2 & -5\\3 & -1\end{bmatrix}$$

Final Result

$$\boxed{A^{-1}=\dfrac{1}{13}\begin{bmatrix}2 & -5\\3 & -1\end{bmatrix}}$$

Published by Anand Classes, Authored by Neeraj Anand — concise NCERT Class 12 Maths inverse matrix solution, crafted to help you practise adjoint and inverse computations for CBSE, JEE and CUET preparation.

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NCERT Question.7 : Find the adjoint and inverse of the matrix
$$A=\begin{bmatrix}1 & 2 & 3\\0 & 2 & 4\\0 & 0 & 5\end{bmatrix}.$$

Solution
Compute the determinant and cofactors to get the adjoint.

Determinant (product of diagonal for an upper-triangular matrix):

$$|A|=1\cdot 2\cdot 5=10.$$

Compute minors and cofactors $A_{ij}=(-1)^{i+j}M_{ij}$:

$M_{11}=\begin{vmatrix}2 & 4\\0 & 5\end{vmatrix}=2\cdot 5-4\cdot 0=10\quad\Rightarrow\quad A_{11}=10.$

$M_{12}=\begin{vmatrix}0 & 4\\0 & 5\end{vmatrix}=0\quad\Rightarrow\quad A_{12}=(-1)^{1+2}0=0.$

$M_{13}=\begin{vmatrix}0 & 2\\0 & 0\end{vmatrix}=0\quad\Rightarrow\quad A_{13}=0.$

$M_{21}=\begin{vmatrix}2 & 3\\0 & 5\end{vmatrix}=2\cdot 5-3\cdot 0=10\quad\Rightarrow\quad A_{21}=(-1)^{2+1}10=-10.$

$M_{22}=\begin{vmatrix}1 & 3\\0 & 5\end{vmatrix}=1\cdot 5-3\cdot 0=5\quad\Rightarrow\quad A_{22}=5.$

$M_{23}=\begin{vmatrix}1 & 2\\0 & 0\end{vmatrix}=0\quad\Rightarrow\quad A_{23}=(-1)^{2+3}0=0.$

$M_{31}=\begin{vmatrix}2 & 3\\2 & 4\end{vmatrix}=2\cdot 4-3\cdot 2=8-6=2\quad\Rightarrow\quad A_{31}=2.$

$M_{32}=\begin{vmatrix}1 & 3\\0 & 4\end{vmatrix}=1\cdot 4-3\cdot 0=4\quad\Rightarrow\quad A_{32}=(-1)^{3+2}4=-4.$

$M_{33}=\begin{vmatrix}1 & 2\\0 & 2\end{vmatrix}=1\cdot 2-2\cdot 0=2\quad\Rightarrow\quad A_{33}=2.$

Cofactor matrix:
$$\begin{bmatrix}
A_{11} & A_{12} & A_{13}\\
A_{21} & A_{22} & A_{23}\\
A_{31} & A_{32} & A_{33}
\end{bmatrix}=\begin{bmatrix}
10 & 0 & 0\\
-10 & 5 & 0\\
2 & -4 & 2
\end{bmatrix}.$$

Adjoint (transpose of cofactor matrix):

$$\operatorname{adj}A=
\begin{bmatrix}
10 & -10 & 2\\
0 & 5 & -4\\
0 & 0 & 2
\end{bmatrix}.$$

Since $|A|=10\neq 0$ the inverse exists and

$$A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A
=\dfrac{1}{10}\begin{bmatrix}
10 & -10 & 2\\
0 & 5 & -4\\
0 & 0 & 2
\end{bmatrix}$$

Simplify entries:

$$\boxed{\operatorname{adj}A=\begin{bmatrix}
10 & -10 & 2\\
0 & 5 & -4\\
0 & 0 & 2
\end{bmatrix}\quad\text{and}\quad
A^{-1}=\begin{bmatrix}
1 & -1 & \tfrac{1}{5}\\
0 & \tfrac{1}{2} & -\tfrac{2}{5}\\
0 & 0 & \tfrac{1}{5}
\end{bmatrix}}$$

Published by Anand Classes, Authored by Neeraj Anand — concise NCERT Class 12 Determinants solution with adjoint and inverse computed step-by-step to help you master triangular matrix inversion for CBSE, JEE and CUET preparation.

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NCERT Question.8 : Find the adjoint and inverse of the matrix
$$A=\begin{bmatrix}1 & 0 & 0\\3 & 3 & 0\\5 & 2 & -1\end{bmatrix}.$$

Solution
Compute the determinant (expand along the first row since it has zeros):

$$|A|=1\begin{vmatrix}3 & 0\\2 & -1\end{vmatrix}-0+0
=1\bigl(3(-1)-0\cdot 2\bigr)= -3.$$

So $|A|=-3\neq 0$, hence $A^{-1}$ exists.

Compute minors and cofactors $A_{ij}=(-1)^{i+j}M_{ij}$.

Minors (deleting row $i$ and column $j$):

$$\begin{aligned}
M_{11}&=\begin{vmatrix}3 & 0\\2 & -1\end{vmatrix}=3(-1)-0\cdot 2=-3,\\
M_{12}&=\begin{vmatrix}3 & 0\\5 & -1\end{vmatrix}=3(-1)-0\cdot 5=-3,\\
M_{13}&=\begin{vmatrix}3 & 3\\5 & 2\end{vmatrix}=3\cdot 2-3\cdot 5=6-15=-9,\\
M_{21}&=\begin{vmatrix}0 & 0\\2 & -1\end{vmatrix}=0\cdot(-1)-0\cdot 2=0,\\
M_{22}&=\begin{vmatrix}1 & 0\\5 & -1\end{vmatrix}=1(-1)-0\cdot 5=-1,\\
M_{23}&=\begin{vmatrix}1 & 0\\5 & 2\end{vmatrix}=1\cdot 2-0\cdot 5=2,\\
M_{31}&=\begin{vmatrix}0 & 0\\3 & 0\end{vmatrix}=0\cdot 0-0\cdot 3=0,\\
M_{32}&=\begin{vmatrix}1 & 0\\3 & 0\end{vmatrix}=1\cdot 0-0\cdot 3=0,\\
M_{33}&=\begin{vmatrix}1 & 0\\3 & 3\end{vmatrix}=1\cdot 3-0\cdot 3=3.
\end{aligned}$$

Cofactors $A_{ij}=(-1)^{i+j}M_{ij}$:

$A_{11}=(-1)^{2}(-3)=-3,\quad A_{12}=(-1)^{3}(-3)=3, $
$A_{13}=(-1)^{4}(-9)=-9,\quad A_{21}=(-1)^{3}(0)=0,$
$A_{22}=(-1)^{4}(-1)=-1, \quad A_{23}=(-1)^{5}(2)=-2,$
$A_{31}=(-1)^{4}(0)=0,\quad A_{32}=(-1)^{5}(0)=0,$
$A_{33}=(-1)^{6}(3)=3.$

Form the cofactor matrix and take its transpose to get the adjoint (adjugate):

Cofactor matrix:
$$\begin{bmatrix}
-3 & 3 & -9\\
0 & -1 & -2\\
0 & 0 & 3
\end{bmatrix}$$

So the adjoint is its transpose:

$$\operatorname{adj}A=
\begin{bmatrix}
-3 & 0 & 0\\
3 & -1 & 0\\
-9 & -2 & 3
\end{bmatrix}.$$

Compute the inverse using $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A$:

$$A^{-1}=\dfrac{1}{-3}\begin{bmatrix}
-3 & 0 & 0\\
3 & -1 & 0\\
-9 & -2 & 3
\end{bmatrix}
=\begin{bmatrix}
1 & 0 & 0\\
-1 & \tfrac{1}{3} & 0\\
3 & \tfrac{2}{3} & -1
\end{bmatrix}.$$

Final Result

$$\boxed{\operatorname{adj}A=\begin{bmatrix}
-3 & 0 & 0\\
3 & -1 & 0\\
-9 & -2 & 3
\end{bmatrix}\quad\text{and}\quad
A^{-1}=\begin{bmatrix}
1 & 0 & 0\\
-1 & \tfrac{1}{3} & 0\\
3 & \tfrac{2}{3} & -1
\end{bmatrix}}$$

Published by Anand Classes, Main Head Faculty by Neeraj Anand — step-by-step NCERT Class 12 adjoint and inverse computation for triangular-structured matrices, designed to strengthen your CBSE, JEE and CUET preparation.

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NCERT Question.9 : Find the adjoint and inverse of the matrix
$$A=\begin{bmatrix}2 & 1 & 3\\4 & -1 & 0\\-7 & 2 & 1\end{bmatrix}.$$

Solution
Compute the determinant by expanding along the first row:

$$|A|=2\begin{vmatrix}-1 & 0\\2 & 1\end{vmatrix}-1\begin{vmatrix}4 & 0\\-7 & 1\end{vmatrix}+3\begin{vmatrix}4 & -1\\-7 & 2\end{vmatrix}$$

Evaluate the $2\times2$ minors:

$\begin{vmatrix}-1 & 0\\2 & 1\end{vmatrix}=(-1)(1)-0\cdot 2=-1,$

$\begin{vmatrix}4 & 0\\-7 & 1\end{vmatrix}=4\cdot 1-0\cdot(-7)=4,$

$\begin{vmatrix}4 & -1\\-7 & 2\end{vmatrix}=4\cdot 2-(-1)(-7)=8-7=1$

Substitute:

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

So $A^{-1}$ exists.

Compute cofactors $A_{ij}=(-1)^{i+j}M_{ij}$ (where $M_{ij}$ is the corresponding minor).

The cofactor (adjugate before transpose) entries evaluate to:

$$\begin{aligned}
A_{11}&= -1,\quad &A_{12}&= -4,\quad &A_{13}&= 1,\\
A_{21}&= 5,\quad &A_{22}&= 23,\quad &A_{23}&= -11,\\
A_{31}&= 3,\quad &A_{32}&= 12,\quad &A_{33}&= -6.
\end{aligned}$$

Form the cofactor matrix and transpose it to get the adjoint (adjugate):

$$\operatorname{adj}A=
\begin{bmatrix}
-1 & 5 & 3\\
-4 & 23 & 12\\
1 & -11 & -6
\end{bmatrix}.$$

Inverse using $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A$:

$$A^{-1}=\dfrac{1}{-3}\begin{bmatrix}
-1 & 5 & 3\\
-4 & 23 & 12\\
1 & -11 & -6
\end{bmatrix}
=\begin{bmatrix}
\frac{1}{3} & -\frac{5}{3} & -1\\
\frac{4}{3} & -\frac{23}{3} & -4\\
-\frac{1}{3} & \frac{11}{3} & 2
\end{bmatrix}.$$

Final Result

$$\boxed{\operatorname{adj}A=\begin{bmatrix}
-1 & 5 & 3\\
-4 & 23 & 12\\
1 & -11 & -6
\end{bmatrix}\quad\text{and}\quad
A^{-1}=\begin{bmatrix}
\frac{1}{3} & -\frac{5}{3} & -1\\
\frac{4}{3} & -\frac{23}{3} & -4\\
-\frac{1}{3} & \frac{11}{3} & 2
\end{bmatrix}}$$

Published by Anand Classes, Authored by Neeraj Anand — concise and carefully computed NCERT Class 12 adjoint and inverse solution, ideal for strengthening your CBSE, JEE and CUET preparation with step-by-step cofactor computations.

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NCERT Question.10 : Find the adjoint and inverse of the matrix
$$A=\begin{bmatrix}1 & -1 & 2\\0 & 2 & -3\\3 & -2 & 4\end{bmatrix}.$$

Solution
Compute the determinant by expansion along the first row:

$$
|A|=1\begin{vmatrix}2 & -3\\-2 & 4\end{vmatrix}-(-1)\begin{vmatrix}0 & -3\\3 & 4\end{vmatrix}+2\begin{vmatrix}0 & 2\\3 & -2\end{vmatrix}$$

$$|A|=1\bigl(2\cdot 4-(-3)(-2)\bigr)+1\bigl(0\cdot 4-(-3)\cdot 3\bigr)+2\bigl(0\cdot(-2)-2\cdot 3\bigr)$$

$$|A|=1(8-6)+1(9)+2(0-6)=2+9-12=-1$$

So $|A|=-1\neq 0$, hence the inverse exists.

Compute minors and cofactors $A_{ij}=(-1)^{i+j}M_{ij}$.

$$\begin{aligned}
M_{11}&=\begin{vmatrix}2 & -3\\-2 & 4\end{vmatrix}=8-6=2 \Rightarrow A_{11}=2,\\
M_{12}&=\begin{vmatrix}0 & -3\\3 & 4\end{vmatrix}=0-(-9)=9 \Rightarrow A_{12}=-9,\\
M_{13}&=\begin{vmatrix}0 & 2\\3 & -2\end{vmatrix}=0-6=-6 \Rightarrow A_{13}=-6,\\
M_{21}&=\begin{vmatrix}-1 & 2\\-2 & 4\end{vmatrix}=(-1)(4)-2(-2)=-4+4=0 \Rightarrow A_{21}=0,\\
M_{22}&=\begin{vmatrix}1 & 2\\3 & 4\end{vmatrix}=4-6=-2 \Rightarrow A_{22}=-2,\\
M_{23}&=\begin{vmatrix}1 & -1\\3 & -2\end{vmatrix}=-2+3=1 \Rightarrow A_{23}=-1,\\
M_{31}&=\begin{vmatrix}-1 & 2\\2 & -3\end{vmatrix}=3-4=-1 \Rightarrow A_{31}=-1,\\
M_{32}&=\begin{vmatrix}1 & 2\\0 & -3\end{vmatrix}=-3-0=-3 \Rightarrow A_{32}=3,\\
M_{33}&=\begin{vmatrix}1 & -1\\0 & 2\end{vmatrix}=2-0=2 \Rightarrow A_{33}=2.
\end{aligned}$$

(Each $A_{ij}$ above uses the sign $(-1)^{i+j}$ applied to the corresponding minor.)

Form the cofactor matrix and transpose to obtain the adjoint (adjugate):

Cofactor matrix
$$\begin{bmatrix}
2 & -9 & -6\\
0 & -2 & -1\\
-1 & 3 & 2
\end{bmatrix}$$

Therefore the adjoint is its transpose:

$$\operatorname{adj}A=
\begin{bmatrix}
2 & 0 & -1\\
-9 & -2 & 3\\
-6 & -1 & 2
\end{bmatrix}.$$

Inverse using $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A$ and $|A|=-1$:

$$A^{-1}=\dfrac{1}{-1}\begin{bmatrix}
2 & 0 & -1\\
-9 & -2 & 3\\
-6 & -1 & 2
\end{bmatrix}
=\begin{bmatrix}
-2 & 0 & 1\\
9 & 2 & -3\\
6 & 1 & -2
\end{bmatrix}.$$

Final Result

$$\boxed{\operatorname{adj}A=\begin{bmatrix}
2 & 0 & -1\\
-9 & -2 & 3\\
-6 & -1 & 2
\end{bmatrix}\quad\text{and}\quad
A^{-1}=\begin{bmatrix}
-2 & 0 & 1\\
9 & 2 & -3\\
6 & 1 & -2
\end{bmatrix}}$$

Published by Anand Classes, Director by Neeraj Anand — step-by-step NCERT Class 12 adjoint and inverse solution, prepared to strengthen your CBSE, JEE and CUET preparation with clear cofactor computations.

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✅ FAQ Section

Q1. What is covered in Determinants Exercise 4.4 of Class 12 Maths?

Exercise 4.4 focuses on the application of determinant properties, evaluation using row/column operations, and solving related higher-order problems. It strengthens conceptual clarity for board exams.

Q2. Are the NCERT Solutions for Exercise 4.4 enough for board exam preparation?

Yes. NCERT exercises—including 4.4—form the foundation of board exam questions. Solving these thoroughly is essential for scoring full marks.

Q3. Can I download the Determinants Exercise 4.4 Class 12 PDF for free?

Yes. Anand Classes provides 100% free downloadable PDFs of all NCERT Solutions, including Exercise 4.4 of Chapter 4.

Q4. Who prepares the NCERT Solutions at Anand Classes?

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Q5. Is this PDF based on the latest CBSE NCERT syllabus?

Absolutely. All solutions strictly follow the latest NCERT Class 12 Maths syllabus and updated exam pattern.