Determinants: Chapter 4 Links
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In this exercise of Determinants 4.4, we will learn about adjoint and inverse of matrices, along with some very important properties of determinants.
You will study concepts like singular and non-singular matrices, inverse of a matrix, adjoint of a matrix, and useful results involving determinants.
We will also use these concepts to solve NCERT questions step by step in a simple and clear way.
Key Concepts
-
- Adjoint of a Matrix: The adjoint of a square matrix A=[a_{ij}] is defined as the transpose of the cofactor matrix.
\operatorname{adj}(A)=\left[A_{ij}\right]^T
-
- Singular Matrix: A square matrix A is said to be singular if
|A|=0
-
- Non-Singular Matrix: A square matrix A is said to be non-singular if
|A|\neq0
-
- Important Result:
A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I
-
- Product of Determinants:
|AB|=|A||B|
-
- Inverse of a Matrix: A square matrix A is invertible if and only if A is non-singular.
A^{-1}=\dfrac{\operatorname{adj}A}{|A|}
-
- Reversal Law:
(AB)^{-1}=B^{-1}A^{-1}
-
- Useful Result: For a matrix A of order 3,
|\operatorname{adj}A|=|A|^2
In general, if A is a square matrix of order n, then
|\operatorname{adj}(A)|=|A|^{n-1}
-
- Determinant of inverse matrix: If A is invertible, then the determinant of its inverse is the reciprocal of the determinant of the matrix itself.
|A^{-1}|=\dfrac{1}{|A|}
\det(A^{-1})=\dfrac{1}{\det(A)}
- Shortcut to Find Adjoint of a Matrix of Order 2: Interchange the main diagonal elements and change the signs of off-diagonal elements.
Before You Begin
- Remember that inverse exists only for non-singular matrices.
- Be careful while finding cofactors because sign mistakes are very common.
- Always take transpose of the cofactor matrix while finding adjoint.
- Use the shortcut carefully for matrices of order 2 only.
Now that we have understood the important concepts and properties used in this exercise Determinants 4.4, let’s start solving the questions one by one.
Question 1: Determinants 4.4
1. Find the adjoint of A=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}
Solution
For a matrix A=\begin{bmatrix}a & b \\ c & d\end{bmatrix} ,
\operatorname{adj}(A)=\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}
Here,
A=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}
Interchange the main diagonal elements and change the signs of the off-diagonal elements.
\operatorname{adj}(A)=\begin{bmatrix}4 & -2 \\ -3 & 1\end{bmatrix}
Question 2: Adjoint of a Matrix
2. Find the adjoint of A=\begin{bmatrix}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{bmatrix}
Solution
We first find the cofactors of all the elements of the matrix.
A=\begin{bmatrix}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{bmatrix}
Cofactors of first row:
A_{11}=\begin{vmatrix}3 & 5 \\ 0 & 1\end{vmatrix}=3
A_{12}=-\begin{vmatrix}2 & 5 \\ -2 & 1\end{vmatrix}=-(2+10)=-12
and A_{13}=\begin{vmatrix}2 & 3 \\ -2 & 0\end{vmatrix}=0+6=6
Cofactors of second row:
A_{21}=-\begin{vmatrix}-1 & 2 \\ 0 & 1\end{vmatrix}=1
A_{22}=\begin{vmatrix}1 & 2 \\ -2 & 1\end{vmatrix}=1+4=5
and A_{23}=-\begin{vmatrix}1 & -1 \\ -2 & 0\end{vmatrix}=-(0-2)=2
Cofactors of third row:
A_{31}=\begin{vmatrix}-1 & 2 \\ 3 & 5\end{vmatrix}=-5-6=-11
A_{32}=-\begin{vmatrix}1 & 2 \\ 2 & 5\end{vmatrix}=-(5-4)=-1
and A_{33}=\begin{vmatrix}1 & -1 \\ 2 & 3\end{vmatrix}=3+2=5
Thus, the cofactor matrix is
\begin{bmatrix}3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{bmatrix}
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 4.4.
Question 3: Determinants Ex 4.4
3. Verify A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I for A=\begin{bmatrix}2 & 3 \\ -4 & -6\end{bmatrix}
Solution
Given,
A=\begin{bmatrix}2 & 3 \\ -4 & -6\end{bmatrix}
First, find the adjoint of A.
\operatorname{adj}(A)=\begin{bmatrix}-6 & -3 \\ 4 & 2\end{bmatrix}
Now, find A(\operatorname{adj}A)
Now, A(\operatorname{adj}A)=\begin{bmatrix}2 & 3 \\ -4 & -6\end{bmatrix}\begin{bmatrix}-6 & -3 \\ 4 & 2\end{bmatrix}
=\begin{bmatrix}(2)(-6)+(3)(4) & (2)(-3)+(3)(2) \\ (-4)(-6)+(-6)(4) & (-4)(-3)+(-6)(2)\end{bmatrix}
=\begin{bmatrix}-12+12 & -6+6 \\ 24-24 & 12-12\end{bmatrix}
So, A(\operatorname{adj}A)=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
Now, find (\operatorname{adj}A)A
We have, (\operatorname{adj}A)A=\begin{bmatrix}-6 & -3 \\ 4 & 2\end{bmatrix}\begin{bmatrix}2 & 3 \\ -4 & -6\end{bmatrix}
=\begin{bmatrix}(-6)(2)+(-3)(-4) & (-6)(3)+(-3)(-6) \\ (4)(2)+(2)(-4) & (4)(3)+(2)(-6)\end{bmatrix}
=\begin{bmatrix}-12+12 & -18+18 \\ 8-8 & 12-12\end{bmatrix}
So, (\operatorname{adj}A)A =\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
Now, find |A|
|A|=\begin{vmatrix}2 & 3 \\ -4 & -6\end{vmatrix}=(2)(-6)-(3)(-4)
=-12+12=0
Therefore,
|A|I=0\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
Hence verified that,
A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I
Question 4: Determinants 4.4
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
Given,
A=\begin{bmatrix}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{bmatrix}
First, find the cofactors of the matrix.
Cofactors of first row:
A_{11}=\begin{vmatrix}0 & -2 \\ 0 & 3\end{vmatrix}=0
A_{12}=-\begin{vmatrix}3 & -2 \\ 1 & 3\end{vmatrix}=-(9+2)=-11
and A_{13}=\begin{vmatrix}3 & 0 \\ 1 & 0\end{vmatrix}=0
Cofactors of second row:
A_{21}=-\begin{vmatrix}-1 & 2 \\ 0 & 3\end{vmatrix}=3
A_{22}=\begin{vmatrix}1 & 2 \\ 1 & 3\end{vmatrix}=3-2=1
and A_{23}=-\begin{vmatrix}1 & -1 \\ 1 & 0\end{vmatrix}=-(0+1)=-1
Cofactors of third row:
A_{31}=\begin{vmatrix}-1 & 2 \\ 0 & -2\end{vmatrix}=2
A_{32}=-\begin{vmatrix}1 & 2 \\ 3 & -2\end{vmatrix}=-(-2-6)=8
and A_{33}=\begin{vmatrix}1 & -1 \\ 3 & 0\end{vmatrix}=0+3=3
Thus, the cofactor matrix is
\begin{bmatrix}0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3\end{bmatrix}
Therefore,
\operatorname{adj}(A)=\begin{bmatrix}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{bmatrix}
Now, find A(\operatorname{adj}A)
We have, 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}
=\begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix}
=11I
Now, find (\operatorname{adj}A)A
We have, (\operatorname{adj}A)A=\begin{bmatrix}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{bmatrix}\begin{bmatrix}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{bmatrix}
=\begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix}
=11I
Now, find |A|
|A|=\begin{vmatrix}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{vmatrix}
Expanding along the first row,
We get, |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}
=1(0)-(-1)(9+2)+2(0)
=11
Therefore,
|A|I=11\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix}
Hence verified that,
A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I
Question 5: Determinants 4.4
5. Find the inverse of A=\begin{bmatrix}2 & -2 \\ 4 & 3\end{bmatrix}
Solution
Given,
A=\begin{bmatrix}2 & -2 \\ 4 & 3\end{bmatrix}
First, find the determinant of A.
|A|=\begin{vmatrix}2 & -2 \\ 4 & 3\end{vmatrix}=(2)(3)-(-2)(4)
=6+8=14
Since |A|\neq0 , the matrix is non-singular and inverse exists.
Now, find the adjoint of A.
\operatorname{adj}(A)=\begin{bmatrix}3 & 2 \\ -4 & 2\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{14}\begin{bmatrix}3 & 2 \\ -4 & 2\end{bmatrix}
These types of questions are considered scoring if the basic concepts and operations are clear.
Question 6: Matrices Ex 3.2
6. Find the inverse of A=\begin{bmatrix}-1 & 5 \\ -3 & 2\end{bmatrix}
Solution
Given,
A=\begin{bmatrix}-1 & 5 \\ -3 & 2\end{bmatrix}
First, find the determinant of A.
i.e. |A|=\begin{vmatrix}-1 & 5 \\ -3 & 2\end{vmatrix}=
=(-1)(2)-(5)(-3)
=-2+15=13
Since |A|\neq0 , the inverse exists.
Now, find the adjoint of A.
For a matrix of order 2\times2 , we interchange the diagonal elements and change the signs of off-diagonal elements.
Here, the adjoint can also be shown by finding minors and cofactors.
\operatorname{adj}(A)=\begin{bmatrix}2 & -5 \\ 3 & -1\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{13}\begin{bmatrix}2 & -5 \\ 3 & -1\end{bmatrix}
Question 7: Ex 4.4 Determinants
7. Find the inverse of A=\begin{bmatrix}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{bmatrix}
Solution
Given,
A=\begin{bmatrix}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{bmatrix}
First, find the determinant of A.
Expanding along the first column,
|A|=1\begin{vmatrix}2 & 4 \\ 0 & 5\end{vmatrix}-0\begin{vmatrix}2 & 3 \\ 0 & 5\end{vmatrix}+0\begin{vmatrix}2 & 3 \\ 2 & 4\end{vmatrix}
💡Moreover, since A is an upper triangular matrix, its determinant is simply the product of its diagonal elements. This is a useful shortcut (though not directly discussed in NCERT).
So, |A|=(1)(2)(5)=10
Since |A|\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 \\ 0 & 5\end{vmatrix}=10
A_{12}=-\begin{vmatrix}0 & 4 \\ 0 & 5\end{vmatrix}=0
and A_{13}=\begin{vmatrix}0 & 2 \\ 0 & 0\end{vmatrix}=0
Cofactors of second row:
A_{21}=-\begin{vmatrix}2 & 3 \\ 0 & 5\end{vmatrix}=-10
A_{22}=\begin{vmatrix}1 & 3 \\ 0 & 5\end{vmatrix}=5
and A_{23}=-\begin{vmatrix}1 & 2 \\ 0 & 0\end{vmatrix}=0
Cofactors of third row:
A_{31}=\begin{vmatrix}2 & 3 \\ 2 & 4\end{vmatrix}=8-6=2
A_{32}=-\begin{vmatrix}1 & 3 \\ 0 & 4\end{vmatrix}=-4
and A_{33}=\begin{vmatrix}1 & 2 \\ 0 & 2\end{vmatrix}=2
Thus, the cofactor matrix is
\begin{bmatrix}10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{10}\begin{bmatrix}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{bmatrix}
💡You can leave your answer in this form, or simplify further by multiplying the scalar with each element of the matrix.
=\begin{bmatrix}1 & -1 & \frac{1}{5} \\ 0 & \frac{1}{2} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{5}\end{bmatrix}
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 4.4
8. Find the inverse of A=\begin{bmatrix}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{bmatrix}
Solution
Given,
A=\begin{bmatrix}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{bmatrix}
First, find the determinant of A.
Expanding along the first row,
|A|=1\begin{vmatrix}3 & 0 \\ 2 & -1\end{vmatrix}-0\begin{vmatrix}3 & 0 \\ 5 & -1\end{vmatrix}+0\begin{vmatrix}3 & 3 \\ 5 & 2\end{vmatrix}
💡Moreover, since A is an lower triangular matrix, its determinant is simply the product of its diagonal elements. This is a useful shortcut (though not directly discussed in NCERT).
So, |A|=(1)(3)(-1)=-3
Since |A|\neq0 , the matrix is non-singular and inverse exists.
Now, find the cofactors of A.
Cofactors of first row:
A_{11}=\begin{vmatrix}3 & 0 \\ 2 & -1\end{vmatrix}=-3
A_{12}=-\begin{vmatrix}3 & 0 \\ 5 & -1\end{vmatrix}=3
and A_{13}=\begin{vmatrix}3 & 3 \\ 5 & 2\end{vmatrix}=6-15=-9
Cofactors of second row:
A_{21}=-\begin{vmatrix}0 & 0 \\ 2 & -1\end{vmatrix}=0
A_{22}=\begin{vmatrix}1 & 0 \\ 5 & -1\end{vmatrix}=-1
and A_{23}=-\begin{vmatrix}1 & 0 \\ 5 & 2\end{vmatrix}=-2
Cofactors of third row:
A_{31}=\begin{vmatrix}0 & 0 \\ 3 & 0\end{vmatrix}=0
A_{32}=-\begin{vmatrix}1 & 0 \\ 3 & 0\end{vmatrix}=0
and A_{33}=\begin{vmatrix}1 & 0 \\ 3 & 3\end{vmatrix}=3
Thus, the cofactor matrix is
\begin{bmatrix}-3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{-3}\begin{bmatrix}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{bmatrix}
💡You can leave your answer in this form, or simplify further by multiplying the scalar with each element of the matrix.
=\begin{bmatrix}1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1\end{bmatrix}
Question 9: Inverse of a Matrix
9. Find the inverse of A=\begin{bmatrix}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{bmatrix}
Solution
Given,
A=\begin{bmatrix}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{bmatrix}
First, find the determinant of A.
|A|=\begin{vmatrix}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{vmatrix}
Expanding along the first row,
We get, |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}
=2(-1)-1(4)+3(8-7)
=-2-4+3=-3
Since |A|\neq0 , the matrix is non-singular and inverse exists.
Now, find the cofactors of A.
Cofactors of first row:
A_{11}=\begin{vmatrix}-1 & 0 \\ 2 & 1\end{vmatrix}=-1
A_{12}=-\begin{vmatrix}4 & 0 \\ -7 & 1\end{vmatrix}=-4
and A_{13}=\begin{vmatrix}4 & -1 \\ -7 & 2\end{vmatrix}=8-7=1
Cofactors of second row:
A_{21}=-\begin{vmatrix}1 & 3 \\ 2 & 1\end{vmatrix}=-(1-6)=5
A_{22}=\begin{vmatrix}2 & 3 \\ -7 & 1\end{vmatrix}=2+21=23
and A_{23}=-\begin{vmatrix}2 & 1 \\ -7 & 2\end{vmatrix}=-(4+7)=-11
Cofactors of third row:
A_{31}=\begin{vmatrix}1 & 3 \\ -1 & 0\end{vmatrix}=3
A_{32}=-\begin{vmatrix}2 & 3 \\ 4 & 0\end{vmatrix}=-(-12)=12
and A_{33}=\begin{vmatrix}2 & 1 \\ 4 & -1\end{vmatrix}=-2-4=-6
Thus, the cofactor matrix is
\begin{bmatrix}-1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
A^{-1}=\dfrac{1}{-3}\begin{bmatrix}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6\end{bmatrix}
💡You can leave your answer in this form, or simplify further by multiplying the scalar with each element of the matrix.
=\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}
Question 10: Determinants 4.4 Exercise
10. Find the inverse of A=\begin{bmatrix}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{bmatrix}
Solution
Given,
A=\begin{bmatrix}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{bmatrix}
First, find the determinant of A.
|A|=\begin{vmatrix}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{vmatrix}
Expanding along the first row,
We get, |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}
=1(8-6)+1(0+9)+2(0-6)
=2+9-12=-1
Since |A|\neq0 , the matrix is non-singular and inverse exists.
Now, find the cofactors of A.
Cofactors of first row:
A_{11}=\begin{vmatrix}2 & -3 \\ -2 & 4\end{vmatrix}=8-6=2
A_{12}=-\begin{vmatrix}0 & -3 \\ 3 & 4\end{vmatrix}=-(0+9)=-9
and A_{13}=\begin{vmatrix}0 & 2 \\ 3 & -2\end{vmatrix}=0-6=-6
Cofactors of second row:
A_{21}=-\begin{vmatrix}-1 & 2 \\ -2 & 4\end{vmatrix}=-(-4+4)=0
A_{22}=\begin{vmatrix}1 & 2 \\ 3 & 4\end{vmatrix}=4-6=-2
and A_{23}=-\begin{vmatrix}1 & -1 \\ 3 & -2\end{vmatrix}=-(-2+3)=-1
Cofactors of third row:
A_{31}=\begin{vmatrix}-1 & 2 \\ 2 & -3\end{vmatrix}=3-4=-1
A_{32}=-\begin{vmatrix}1 & 2 \\ 0 & -3\end{vmatrix}=-(-3)=3
and A_{33}=\begin{vmatrix}1 & -1 \\ 0 & 2\end{vmatrix}=2
Thus, the cofactor matrix is
\begin{bmatrix}2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
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}
The questions in Determinants 4.4 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 4.4
11. Find the inverse of A=\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha\end{bmatrix}
Solution
Given,
A=\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha\end{bmatrix}
First, find the determinant of A.
Expanding along the first row,
We get, |A|=1\begin{vmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha\end{vmatrix}
=(1)\left[-\cos^2\alpha-\sin^2\alpha\right]
=-(\cos^2\alpha+\sin^2\alpha)
Using \sin^2\alpha+\cos^2\alpha=1 ,
|A|=-1
Since |A|\neq0 , the matrix is non-singular and inverse exists.
Now, find the cofactors of A.
Cofactors of first row:
A_{11}=\begin{vmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha\end{vmatrix}=-(\cos^2\alpha+\sin^2\alpha)=-1
A_{12}=-\begin{vmatrix}0 & \sin\alpha \\ 0 & -\cos\alpha\end{vmatrix}=0
and A_{13}=\begin{vmatrix}0 & \cos\alpha \\ 0 & \sin\alpha\end{vmatrix}=0
Cofactors of second row:
A_{21}=-\begin{vmatrix}0 & 0 \\ \sin\alpha & -\cos\alpha\end{vmatrix}=0
A_{22}=\begin{vmatrix}1 & 0 \\ 0 & -\cos\alpha\end{vmatrix}=-\cos\alpha
and A_{23}=-\begin{vmatrix}1 & 0 \\ 0 & \sin\alpha\end{vmatrix}=-\sin\alpha
Cofactors of third row:
A_{31}=\begin{vmatrix}0 & 0 \\ \cos\alpha & \sin\alpha\end{vmatrix}=0
A_{32}=-\begin{vmatrix}1 & 0 \\ 0 & \sin\alpha\end{vmatrix}=-\sin\alpha
and A_{33}=\begin{vmatrix}1 & 0 \\ 0 & \cos\alpha\end{vmatrix}=\cos\alpha
Thus, the cofactor matrix is
\begin{bmatrix}-1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha\end{bmatrix}
Taking transpose, we get
\operatorname{adj}(A)=\begin{bmatrix}-1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha\end{bmatrix}
Using the formula
A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|}
we get
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}
Here, we have, A^{-1}=A
💡 Interesting Observation: Since A^{-1}=A , the matrix is called a self-inverse (involutory) matrix. Such matrices satisfy A^2=I .
Question 12: Determinants 4.4
12. Let A=\begin{bmatrix}3 & 7 \\ 2 & 5\end{bmatrix} and B=\begin{bmatrix}6 & 8 \\ 7 & 9\end{bmatrix} . Verify that (AB)^{-1}=B^{-1}A^{-1}
Solution
First, find AB .
Now, AB=\begin{bmatrix}3 & 7 \\ 2 & 5\end{bmatrix}\begin{bmatrix}6 & 8 \\ 7 & 9\end{bmatrix}
=\begin{bmatrix}(3)(6)+(7)(7) & (3)(8)+(7)(9) \\ (2)(6)+(5)(7) & (2)(8)+(5)(9)\end{bmatrix}
=\begin{bmatrix}18+49 & 24+63 \\ 12+35 & 16+45\end{bmatrix}
So, AB=\begin{bmatrix}67 & 87 \\ 47 & 61\end{bmatrix}
Now, find (AB)^{-1} .
We have, |AB|=\begin{vmatrix}67 & 87 \\ 47 & 61\end{vmatrix}=(67)(61)-(87)(47)
=4087-4089=-2
\operatorname{adj}(AB)=\begin{bmatrix}61 & -87 \\ -47 & 67\end{bmatrix}
Therefore,
(AB)^{-1}=\dfrac{1}{-2}\begin{bmatrix}61 & -87 \\ -47 & 67\end{bmatrix}
=\begin{bmatrix}-\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2}\end{bmatrix}
Now, find A^{-1} .
|A|=\begin{vmatrix}3 & 7 \\ 2 & 5\end{vmatrix}=15-14=1
A^{-1}=\begin{bmatrix}5 & -7 \\ -2 & 3\end{bmatrix}
Next, find B^{-1} .
|B|=\begin{vmatrix}6 & 8 \\ 7 & 9\end{vmatrix}=54-56=-2
So we get, B^{-1}=\dfrac{1}{-2}\begin{bmatrix}9 & -8 \\ -7 & 6\end{bmatrix}
=\begin{bmatrix}-\frac{9}{2} & 4 \\ \frac{7}{2} & -3\end{bmatrix}
Now, calculate B^{-1}A^{-1} .
We have, B^{-1}A^{-1}=\begin{bmatrix}-\frac{9}{2} & 4 \\ \frac{7}{2} & -3\end{bmatrix}\begin{bmatrix}5 & -7 \\ -2 & 3\end{bmatrix}
=\begin{bmatrix}-\frac{45}{2}-8 & \frac{63}{2}+12 \\ \frac{35}{2}+6 & -\frac{49}{2}-9\end{bmatrix}
=\begin{bmatrix}-\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2}\end{bmatrix}
Thus,
(AB)^{-1}=B^{-1}A^{-1}
Now, let’s move on to the next question of Determinants 4.4.
Question 13: Ex. 4.4 Determinants
13. If A=\begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix} , show that A^2-5A+7I=O . Hence find A^{-1}
Solution
Given,
A=\begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}
First, find A^2 .
Now, A^2=\begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}\begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}
=\begin{bmatrix}(3)(3)+(1)(-1) & (3)(1)+(1)(2) \\ (-1)(3)+(2)(-1) & (-1)(1)+(2)(2)\end{bmatrix}
=\begin{bmatrix}9-1 & 3+2 \\ -3-2 & -1+4\end{bmatrix}
So, A^2=\begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix}
Now,
5A=5\begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}=\begin{bmatrix}15 & 5 \\ -5 & 10\end{bmatrix}
Also,
7I=\begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}
Therefore,
We get, 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}
=\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
Hence proved that
A^2-5A+7I=O
Now,
A^2-5A=-7I
A(A-5I)=-7I
Pre-multiplying both sides by A^{-1} ,
A^{-1}A(A - 5I)=A^{-1}(-7I)
I(A - 5I)=(-7)A^{-1}I
or (AI - 5I^2)=(-7)A^{-1}I
or (A - 5I)=(-7)A^{-1}
since IA = A, A^{-1}I = A^{-1}, I^2 = I
Multiplying both sides by -\frac{1}{7} ,
-\frac{1}{7}(A-5I)= A^{-1}
Hence,
A^{-1}=-\frac{1}{7}(A-5I)
Putting the values, we get
=-\frac{1}{7}\left(\begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}-\begin{bmatrix}5 & 0 \\ 0 & 5\end{bmatrix}\right)
=-\frac{1}{7}\begin{bmatrix}-2 & 1 \\ -1 & -3\end{bmatrix}
i.e. A^{-1}=\frac{1}{7}\begin{bmatrix}2 & -1 \\ 1 & 3\end{bmatrix}
💡 Exam Tip: Questions of the form A^2+aA+bI=O , like the next one, are often solved using the Cayley-Hamilton Theorem. Even if you solve them by direct multiplication, the final relation can be used as a quick verification tool during exams to check whether your values of a and b are correct after all calculations.
🎥 You can also watch this YouTube video for a clearer understanding and faster approach to such questions.
Question 14: Determinants 4.4
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
Given,
A=\begin{bmatrix}3 & 2 \\ 1 & 1\end{bmatrix}
First, find A^2 .
Now, A^2=\begin{bmatrix}3 & 2 \\ 1 & 1\end{bmatrix}\begin{bmatrix}3 & 2 \\ 1 & 1\end{bmatrix}
=\begin{bmatrix}(3)(3)+(2)(1) & (3)(2)+(2)(1) \\ (1)(3)+(1)(1) & (1)(2)+(1)(1)\end{bmatrix}
=\begin{bmatrix}9+2 & 6+2 \\ 3+1 & 2+1\end{bmatrix}
So, A^2=\begin{bmatrix}11 & 8 \\ 4 & 3\end{bmatrix}
Now, let
A^2+aA+bI=O
Substituting the matrices,
\begin{bmatrix}11 & 8 \\ 4 & 3\end{bmatrix}+a\begin{bmatrix}3 & 2 \\ 1 & 1\end{bmatrix}+b\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
\begin{bmatrix}11+3a+b & 8+2a \\ 4+a & 3+a+b\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
Equating corresponding elements,
8+2a=0
2a=-8
i.e. a=-4
Also,
11+3a+b=0
Putting a = -4
11+3(-4)+b=0
11-12+b=0
i.e. b=1
Hence,
a=-4,\quad b=1
Question 15: Inverse without Cofactors
15. For the matrix A=\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{bmatrix} Show that A^3-6A^2+5A+11I=O . Hence, find A^{-1}
Solution
Given,
A=\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{bmatrix}
First, find A^2 .
A^2=\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{bmatrix}\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{bmatrix}
=\begin{bmatrix}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{bmatrix}
Now, find A^3 .
We know, A^3=A^2A
=\begin{bmatrix}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{bmatrix}\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{bmatrix}
=\begin{bmatrix}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{bmatrix}
Also,
6A^2=\begin{bmatrix}24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{bmatrix}
5A=\begin{bmatrix}5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15\end{bmatrix}
and 11I=\begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix}
Therefore,
A^3-6A^2+5A+11I
=\begin{bmatrix}8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58\end{bmatrix}-\begin{bmatrix}24 & 12 & 6 \\ -18 & 48 & -84 \\ 42 & -18 & 84\end{bmatrix}+\begin{bmatrix}5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15\end{bmatrix}+\begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix}
So, LHS =\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}=O
Hence proved that
A^3-6A^2+5A+11I=O
Now,
A^3-6A^2+5A=-11I
A(A^2-6A+5I)=-11I
Pre-multiplying both sides by A^{-1} ,
A^{-1}A(A^2-6A+5I)=A^{-1}(-11I)
We know, A^{-1}A = I, A^{-1}I = A^{-1}, AI = IA = I, I^n = I
Therefore, I(A^2-6A+5I)=-11A^{-1}I
gives, A^2-6A+5I=-11A^{-1}
or A^{-1}=-\frac{1}{11}(A^2-6A+5I)
Substituting the values, we get
A^{-1}=-\frac{1}{11}\left(\begin{bmatrix}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{bmatrix}-\begin{bmatrix}6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18\end{bmatrix}+\begin{bmatrix}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{bmatrix}\right)
=-\frac{1}{11}\begin{bmatrix}3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1\end{bmatrix}
So, A^{-1}=\frac{1}{11}\begin{bmatrix}-3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1\end{bmatrix}
Question 16: Determinants 4.4
16. If 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
Given,
A=\begin{bmatrix}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{bmatrix}
First, find A^2 .
A^2=\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}
=\begin{bmatrix}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{bmatrix}
Now, find A^3 .
A^3=A^2A
Putting the values, we get
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}
=\begin{bmatrix}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{bmatrix}
Also,
6A^2=\begin{bmatrix}36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36\end{bmatrix}
9A=\begin{bmatrix}18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18\end{bmatrix}
and 4I=\begin{bmatrix}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{bmatrix}
Therefore,
A^3-6A^2+9A-4I
=\begin{bmatrix}22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22\end{bmatrix}-\begin{bmatrix}36 & -30 & 30 \\ -30 & 36 & -30 \\ 30 & -30 & 36\end{bmatrix}+\begin{bmatrix}18 & -9 & 9 \\ -9 & 18 & -9 \\ 9 & -9 & 18\end{bmatrix}-\begin{bmatrix}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{bmatrix}
or LHS =\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}=O
Hence verified that
A^3-6A^2+9A-4I=O
Now,
A^3-6A^2+9A=4I
A(A^2-6A+9I)=4I
Pre-multiplying both sides by A^{-1} ,
We get, A^{-1}A(A^2-6A+9I)=A^{-1}(4I)
using IA = A, A^{-1}I = A^{-1}, I^2 = I
I(A^2-6A+9I)=4A^{-1}
A^2-6A+9I=4A^{-1}
or A^{-1}=\frac{1}{4}(A^2-6A+9I)
Substituting the values,
A^{-1}=\frac{1}{4}\left(\begin{bmatrix}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{bmatrix}-\begin{bmatrix}12 & -6 & 6 \\ -6 & 12 & -6 \\ 6 & -6 & 12\end{bmatrix}+\begin{bmatrix}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{bmatrix}\right)
=\frac{1}{4}\begin{bmatrix}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{bmatrix}
💡You can leave your answer in this form, or simplify further by multiplying the scalar with each element of the matrix.
A^{-1}=\begin{bmatrix}\frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{1}{4} & \frac{3}{4}\end{bmatrix}
The following two MCQs have already been discussed in the separate Determinants MCQs post. You can also check out other important MCQs from the chapter after completing this exercise for additional practice and revision.
Question 17: Determinants 4.4 – MCQ
17. Let A be a nonsingular square matrix of order 3\times3 . Then |\operatorname{adj}A| is equal to
- (A) |A|
- (B) |A|^2
- (C) |A|^3
- (D) 3|A|
Solution
For a square matrix of order n ,
|\operatorname{adj}A|=|A|^{n-1}
Here, the order of matrix A is 3\times3 .
So, n=3
Therefore,
|\operatorname{adj}A|=|A|^{3-1}
=|A|^2
✅️ Hence, the correct option is (B)
|A|^2
Question 18: 4.4 Determinants – MCQ
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
We know that for any invertible matrix A ,
\det(A^{-1})=\dfrac{1}{\det(A)}
Since A is invertible, \det(A)\neq0 .
Therefore,
\det(A^{-1})=\dfrac{1}{\det(A)}
✅️ Hence, the correct option is (B)
\dfrac{1}{\det(A)}
Common Mistakes to Avoid
-
- Confusing adjoint with inverse: Remember that \operatorname{adj}(A) and A^{-1} are different. Inverse is given by A^{-1}=\dfrac{\operatorname{adj}(A)}{|A|} .
- Finding inverse without checking determinant: A matrix has an inverse only if |A|\neq0 .
- Incorrect signs while finding cofactors: Always use the sign pattern \begin{bmatrix}+ & - & + \\ - & + & - \\ + & - & +\end{bmatrix} .
- Forgetting to transpose the cofactor matrix: Adjoint is obtained only after taking transpose of the cofactor matrix.
- Mistakes in determinant expansion: Be careful while expanding determinants along rows or columns, especially with negative signs.
- Wrong order in reversal law: (AB)^{-1}=B^{-1}A^{-1} , not A^{-1}B^{-1} .
- Errors in matrix multiplication: Matrix multiplication is not simply multiplication of corresponding elements. Multiply rows by columns carefully.
- Using shortcut for adjoint incorrectly: The shortcut of interchanging diagonal elements and changing signs of off-diagonal elements works only for matrices of order 2\times2 .
- Confusing singular and non-singular matrices: Singular matrices have |A|=0 , while non-singular matrices have |A|\neq0 .
- Skipping simplification: Always simplify the final inverse matrix properly, especially when fractions are involved.
Continue Learning
- Learn how to find minors and cofactors of matrices step by step.
- Practice finding adjoint and inverse of 2\times2 and 3\times3 matrices.
- Understand the difference between singular and non-singular matrices.
- Practice the reversal law (AB)^{-1}=B^{-1}A^{-1} using examples.
- Explore how inverse matrices are used to solve systems of linear equations.
If you practice these questions regularly and understand the logic behind each step, matrix problems will become much easier and more scoring.
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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