This page on Determinants Miscellaneous Exercise of Class 12 Maths Chapter 4 would give you a complete revision of all the important concepts studied in this chapter. In this exercise, you will solve questions based on evaluation of determinants, properties of determinants, minors and cofactors, adjoint and inverse of matrices, area of triangle using determinants, and solving system of linear equations.
Through these questions, you will also revise important properties related to determinants and inverse matrices. Besides that, you shall also use concepts like expansion along rows and columns, determinant properties for simplification, adjoint method for finding inverse, and results such as |A^{-1}|=\dfrac{1}{|A|} . You will also apply determinants in solving matrix equations and proving various algebraic results.
This miscellaneous exercise also works as a quick chapter recap before examinations, since it combines a variety of conceptual and computational problems from the complete chapter. By practising these questions, you can strengthen your understanding of determinant properties, improve calculation accuracy and revise the important concepts of Chapter 4 in one place.
Key Concepts
- Determinant: A determinant is a numerical value associated with a square matrix.
- Determinant of Order 2: If A=\begin{bmatrix}a&b\\c&d\end{bmatrix} , then |A|=ad-bc .
- Determinant of Order 3: A determinant of order 3\times3 can be evaluated by expansion along any row or column.
- Expansion of Determinants: A determinant can be expanded along any row or column using minors and cofactors.
- Minor of an Element: The minor of an element is the determinant obtained after deleting its row and column.
- Cofactor of an Element: The cofactor of a_{ij} is A_{ij}=(-1)^{i+j}M_{ij} .
- Sign Pattern of Cofactors: The signs in cofactor expansion follow the pattern
\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}
- Area of Triangle: The area of triangle with vertices (x_1,y_1),(x_2,y_2),(x_3,y_3) is
\text{Area}=\dfrac{1}{2}\left|\begin{matrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{matrix}\right|
- Transpose Property: The determinant remains unchanged on taking transpose, i.e. |A'|=|A| .
- Product Property: For square matrices, |AB|=|A||B| .
- Singular Matrix: A matrix is singular if |A|=0
- Non-Singular Matrix: A matrix is non-singular if |A|\neq0 .
- Adjoint of a Matrix: The transpose of cofactor matrix is called the adjoint of the matrix.
- Inverse of a Matrix: If |A|\neq0 , then A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A) .
- Determinant of Inverse: |A^{-1}|=\dfrac{1}{|A|}=|A|^{-1}
- Inverse Verification: To verify inverse, we check whether AA^{-1}=A^{-1}A=I .
- Consistency of Equations: Determinants can be used to solve and study consistency of system of linear equations.
- Inverse Matrix Method: If \Delta\neq0 , then only the system of equations has a unique solution.
💡 Always try to simplify determinants using properties before expanding them fully.
- Interchanging Rows or Columns: If any two rows or columns are interchanged, then the sign of determinant changes.
- Identical Rows or Columns: If two rows or columns are identical, then the determinant is zero.
- Proportional Rows or Columns: If two rows or columns are proportional, then the determinant is zero.
- Common Factor Property: A common factor from a row or column can be taken outside the determinant.
- Row/Column Operations: Adding a multiple of one row or column to another does not change the value of determinant.
Now, let’s begin this exercise with Question 1
Question 1: Determinants Miscellaneous Exercise
1. Prove that the determinant
\left|\begin{matrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{matrix}\right|
is independent of \theta .
Solution
Let
\Delta=\left|\begin{matrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{matrix}\right|
Expanding along the first row,
We get, \Delta=x\left|\begin{matrix}-x&1\\1&x\end{matrix}\right|-\sin\theta\left|\begin{matrix}-\sin\theta&1\\\cos\theta&x\end{matrix}\right|+\cos\theta\left|\begin{matrix}-\sin\theta&-x\\\cos\theta&1\end{matrix}\right|
=x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)
=-x^3-x+x\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^2\theta
So, \Delta =-x^3-x+x(\sin^2\theta+\cos^2\theta)
Using \sin^2\theta+\cos^2\theta=1 ,
\Delta=-x^3-x+x
or \Delta=-x^3
Since the final value does not contain \theta , the determinant is independent of \theta .
You may already be following Maths Better for important concepts and exam-oriented questions, like integration tricks and solving linear equations using the Matrix Method etc. Similarly, this NCERT Solutions series for Class 12 Maths will help strengthen your concepts and improve problem-solving skills.
Now, let’s move on to the next question of Determinants Miscellaneous Exercise.
Question 2: Miscellaneous Exercise Determinants
2. Evaluate
\left|\begin{matrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{matrix}\right|
Solution
Expanding along the second row,
We get, \Delta=-(-\sin\beta)\left|\begin{matrix}\cos\alpha\sin\beta&-\sin\alpha\\\sin\alpha\sin\beta&\cos\alpha\end{matrix}\right|+\cos\beta\left|\begin{matrix}\cos\alpha\cos\beta&-\sin\alpha\\\sin\alpha\cos\beta&\cos\alpha\end{matrix}\right|
=\sin\beta\left(\cos^2\alpha\sin\beta+\sin^2\alpha\sin\beta\right)+\cos\beta\left(\cos^2\alpha\cos\beta+\sin^2\alpha\cos\beta\right)
=\sin\beta\cdot\sin\beta(\cos^2\alpha+\sin^2\alpha)+\cos\beta\cdot\cos\beta(\cos^2\alpha+\sin^2\alpha)
Using \cos^2\alpha+\sin^2\alpha=1 ,
We get, \Delta =\sin^2\beta+\cos^2\beta
Again using \sin^2\beta+\cos^2\beta=1 ,
Therefore, \Delta=1
Question 3: Miscellaneous Exercise Chapter 4
3. If A^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix} and B=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix} , find (AB)^{-1} .
Solution
We know that
(AB)^{-1}=B^{-1}A^{-1}
First, let us find B^{-1} .
|B|=\left|\begin{matrix}1&2&-2\\-1&3&0\\0&-2&1\end{matrix}\right|
Expanding along the first row,
We get, |B|=1\left|\begin{matrix}3&0\\-2&1\end{matrix}\right|-2\left|\begin{matrix}-1&0\\0&1\end{matrix}\right|+(-2)\left|\begin{matrix}-1&3\\0&-2\end{matrix}\right|
=1(3)-2(-1)+(-2)(2)
=3+2-4
i.e. |B| =1
Now, cofactors of B are
A_{11}=3,\quad A_{12}=1,\quad A_{13}=2
A_{21}=2,\quad A_{22}=1,\quad A_{23}=2
and A_{31}=6,\quad A_{32}=2,\quad A_{33}=5
Therefore, cofactor matrix of B is
\begin{bmatrix}3&1&2\\2&1&2\\6&2&5\end{bmatrix}
Hence,
\operatorname{adj}(B)=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
Therefore,
B^{-1}=\dfrac{1}{|B|}\operatorname{adj}(B)
=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
Now,
(AB)^{-1}=B^{-1}A^{-1}
=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
Multiplying the matrices, we get
(AB)^{-1}=\begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}
Question 4: Determinants Miscellaneous
4. Let A=\begin{bmatrix}1&2&1\\2&3&1\\1&1&5\end{bmatrix} . Verify that
(i) [\operatorname{adj}A]^{-1}=\operatorname{adj}(A^{-1})
(ii) (A^{-1})^{-1}=A
Solution
First, let us find A^{-1} .
|A|=\left|\begin{matrix}1&2&1\\2&3&1\\1&1&5\end{matrix}\right|
Expanding along the first row,
We get, |A|=1\left|\begin{matrix}3&1\\1&5\end{matrix}\right|-2\left|\begin{matrix}2&1\\1&5\end{matrix}\right|+1\left|\begin{matrix}2&3\\1&1\end{matrix}\right|
=1(15-1)-2(10-1)+(2-3)
=14-18-1
i.e. |A|=-5
Now, cofactors of A are
A_{11}=14,\quad A_{12}=-9,\quad A_{13}=-1
A_{21}=-9,\quad A_{22}=4,\quad A_{23}=1
and A_{31}=-1,\quad A_{32}=1,\quad A_{33}=-1
Therefore,
\operatorname{adj}(A)=\begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}
Hence,
A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A)
=-\dfrac{1}{5}\begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}
(i) Verification of [\operatorname{adj}A]^{-1}=\operatorname{adj}(A^{-1})
We have
\operatorname{adj}(A)=\begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}
Now,
|\operatorname{adj}(A)|=\left|\begin{matrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{matrix}\right|
Expanding along the first row,
We get, |\operatorname{adj}(A)| =14\left|\begin{matrix}4&1\\1&-1\end{matrix}\right|-(-9)\left|\begin{matrix}-9&1\\-1&-1\end{matrix}\right|+(-1)\left|\begin{matrix}-9&4\\-1&1\end{matrix}\right|
=14(-5)+9(10)+(-1)(-5)
=-70+90+5
So, |\operatorname{adj}(A)|=25
Now, cofactors of \operatorname{adj}(A) are
A_{11}=-5,\quad A_{12}=-10,\quad A_{13}=-5
A_{21}=-10,\quad A_{22}=-15,\quad A_{23}=-5
and A_{31}=-5,\quad A_{32}=-5,\quad A_{33}=-25
Therefore,
\operatorname{adj}(\operatorname{adj}A)=\begin{bmatrix}-5&-10&-5\\-10&-15&-5\\-5&-5&-25\end{bmatrix}
Hence,
[\operatorname{adj}A]^{-1}=\dfrac{1}{25}\begin{bmatrix}-5&-10&-5\\-10&-15&-5\\-5&-5&-25\end{bmatrix}
=-\dfrac{1}{5}\begin{bmatrix}1&2&1\\2&3&1\\1&1&5\end{bmatrix}
Now,
A^{-1}=-\dfrac{1}{5}\begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}
=\begin{bmatrix}-14/5&9/5&1/5\\9/5&-4/5&-1/5\\1/5&-1/5&1/5\end{bmatrix}
Now, cofactors of A^{-1} are
A_{11}=-1/5,\quad A_{12}=-2/5,\quad A_{13}=-1/5
A_{21}=-2/5,\quad A_{22}=-3/5,\quad A_{23}=-1/5
and A_{31}=-1/5,\quad A_{32}=-1/5,\quad A_{33}=-1
Therefore,
\operatorname{adj}(A^{-1})=\begin{bmatrix}-1/5&-2/5&-1/5\\-2/5&-3/5&-1/5\\-1/5&-1/5&-1\end{bmatrix}
=-\dfrac{1}{5}\begin{bmatrix}1&2&1\\2&3&1\\1&1&5\end{bmatrix}
Thus,
[\operatorname{adj}A]^{-1}=\operatorname{adj}(A^{-1})
(ii) Verification of (A^{-1})^{-1}=A
We have
A^{-1}=\begin{bmatrix}-14/5&9/5&1/5\\9/5&-4/5&-1/5\\1/5&-1/5&1/5\end{bmatrix}
Also,
|A^{-1}|=\dfrac{1}{|A|}=-\dfrac{1}{5}
Now, (A^{-1})^{-1}=\dfrac{1}{|A^{-1}|}\operatorname{adj}(A^{-1})
=(-5)\left[-\dfrac{1}{5}\begin{bmatrix}1&2&1\\2&3&1\\1&1&5\end{bmatrix}\right]
=\begin{bmatrix}1&2&1\\2&3&1\\1&1&5\end{bmatrix}
i.e. (A^{-1})^{-1}=A
Hence verified.
💡Another way to actually prove it!
(i) Verification of [\operatorname{adj}A]^{-1}=\operatorname{adj}(A^{-1})
We know that
\operatorname{adj}(A)=|A|A^{-1}
=(-5)A^{-1}
Taking inverse on both sides,
[\operatorname{adj}(A)]^{-1}=\dfrac{-1}{5}(A^{-1})^{-1}
Since (A^{-1})^{-1}=A ,
[\operatorname{adj}(A)]^{-1}=\dfrac{-1}{5}A
Also, \operatorname{adj}(A^{-1})=|A^{-1}|(A^{-1})^{-1}
=\dfrac{1}{|A|}A
=\dfrac{-1}{5}A
Thus,
[\operatorname{adj}A]^{-1}=\operatorname{adj}(A^{-1})
(ii) Verification of (A^{-1})^{-1}=A
Since AA^{-1}=I , taking inverse on both sides and using reversal law, we get
(A^{-1})^{-1}A^{-1}=I
Post-multiplying both sides by A , we get
(A^{-1})^{-1}A^{-1}A=IA
Therefore, by using
A^{-1}A=I, AI=IA=A
We get, (A^{-1})^{-1}=A
Try solving the next question on your own first by expanding along any row or column, and then compare your approach with the solution given below.
Question 5: Determinants Miscellaneous
5. Evaluate
\left|\begin{matrix}x&y&x+y\\y&x+y&x\\x+y&x&y\end{matrix}\right|
Solution
Let
\Delta=\left|\begin{matrix}x&y&x+y\\y&x+y&x\\x+y&x&y\end{matrix}\right|
Expanding along the first row,
We get, \Delta=x\left|\begin{matrix}x+y&x\\x&y\end{matrix}\right|-y\left|\begin{matrix}y&x\\x+y&y\end{matrix}\right|+(x+y)\left|\begin{matrix}y&x+y\\x+y&x\end{matrix}\right|
=x\{y(x+y)-x^2\}-y\{y^2-x(x+y)\}+(x+y)\{xy-(x+y)^2\}
=x(xy+y^2-x^2)-y(y^2-x^2-xy)+(x+y)(-x^2-xy-y^2)
Or \Delta =x^2y+xy^2-x^3-y^3+x^2y+xy^2-(x+y)(x^2+xy+y^2)
Using
(x+y)(x^2+xy+y^2)=x^3+2x^2y+2xy^2+y^3
we get
\Delta=x^2y+xy^2-x^3-y^3+x^2y+xy^2-x^3-2x^2y-2xy^2-y^3
i.e. \Delta=-2x^3-2y^3
\therefore \Delta=-2(x^3+y^3)
Question 6: Determinants Miscellaneous
6. Evaluate
\left|\begin{matrix}1&x&y\\1&x+y&y\\1&x&x+y\end{matrix}\right|
Solution
Let
\Delta=\left|\begin{matrix}1&x&y\\1&x+y&y\\1&x&x+y\end{matrix}\right|
Expanding along the first row,
We get, \Delta=1\left|\begin{matrix}x+y&y\\x&x+y\end{matrix}\right|-x\left|\begin{matrix}1&y\\1&x+y\end{matrix}\right|+y\left|\begin{matrix}1&x+y\\1&x\end{matrix}\right|
=(x+y)^2-xy-x(x+y-y)+y(x-(x+y))
i.e. \Delta =x^2+2xy+y^2-xy-x^2-y^2
=xy
\therefore \Delta=xy
Solving system of linear equations’ type of questions are generally direct, but presentation and correct steps play an important role in scoring full marks. You can check that in the following question.
Question 7: Miscellaneous Ex Determinants
7. Solve the system of equations
\dfrac{2}{x}+\dfrac{3}{y}+\dfrac{10}{z}=4
\dfrac{4}{x}-\dfrac{6}{y}+\dfrac{5}{z}=1
and \dfrac{6}{x}+\dfrac{9}{y}-\dfrac{20}{z}=2
Solution
Let
\dfrac{1}{x}=u,\quad \dfrac{1}{y}=v,\quad \dfrac{1}{z}=w
Then the given equations become
2u+3v+10w=4
4u-6v+5w=1
and 6u+9v-20w=2
Writing in matrix form,
\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}\begin{bmatrix}u\\v\\w\end{bmatrix}=\begin{bmatrix}4\\1\\2\end{bmatrix}
Let
A=\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix},\quad X=\begin{bmatrix}u\\v\\w\end{bmatrix},\quad B=\begin{bmatrix}4\\1\\2\end{bmatrix}
Then
AX=B
Therefore,
X=A^{-1}B
Now,
|A|=\left|\begin{matrix}2&3&10\\4&-6&5\\6&9&-20\end{matrix}\right|
Expanding along the first row,
We get, |A|=2\left|\begin{matrix}-6&5\\9&-20\end{matrix}\right|-3\left|\begin{matrix}4&5\\6&-20\end{matrix}\right|+10\left|\begin{matrix}4&-6\\6&9\end{matrix}\right|
=2(120-45)-3(-80-30)+10(36+36)
=150+330+720
So, |A|=1200
Now, cofactors of A are
A_{11}=75,\quad A_{12}=110,\quad A_{13}=72
A_{21}=150,\quad A_{22}=-100,\quad A_{23}=0
and A_{31}=75,\quad A_{32}=30,\quad A_{33}=-24
Therefore,
\operatorname{adj}(A)=\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
Hence,
A^{-1}=\dfrac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}
Now, X=A^{-1}B
=\dfrac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\begin{bmatrix}4\\1\\2\end{bmatrix}
=\dfrac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix}
So, X=\begin{bmatrix}1/2\\1/3\\1/5\end{bmatrix}
Thus,
u=\dfrac{1}{2},\quad v=\dfrac{1}{3},\quad w=\dfrac{1}{5}
Since
u=\dfrac{1}{x},\quad v=\dfrac{1}{y},\quad w=\dfrac{1}{z}
Therefore,
x=2,\quad y=3,\quad z=5
NCERT Class 12 Maths has a total of 61 exercises across Part 1 and Part 2 including the miscellaneous ones. I’ll cover each exercise one by one with clear explanations and exam-oriented solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter.
Here’s the next question.
Question 8: Determinants Misc. – MCQ
8. If x,y,z are nonzero real numbers, then the inverse of matrix
A=\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix} is
- (A) \begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}
- (B) xyz\begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}
- (C) \dfrac{1}{xyz}\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}
- (D) \dfrac{1}{xyz}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
Solution
We know that
A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A)
Now,
|A|=\left|\begin{matrix}x&0&0\\0&y&0\\0&0&z\end{matrix}\right|
Expanding along the first row,
|A|=x\left|\begin{matrix}y&0\\0&z\end{matrix}\right|
=xyz
Now, cofactors of A are
A_{11}=yz,\quad A_{22}=xz,\quad A_{33}=xy
All other cofactors are zero.
Therefore,
\operatorname{adj}(A)=\begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix}
Hence,
A^{-1}=\dfrac{1}{xyz}\begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix}
=\begin{bmatrix}\dfrac{1}{x}&0&0\\0&\dfrac{1}{y}&0\\0&0&\dfrac{1}{z}\end{bmatrix}
✅️ Thus, the correct answer is (A).
💡 Shortcut: To find the inverse of a diagonal matrix, simply replace each nonzero diagonal element by its reciprocal. The positions of the diagonal elements remain unchanged.
If
A=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}
where a,b,c\neq0 , then
A^{-1}=\begin{bmatrix}\dfrac1a&0&0\\0&\dfrac1b&0\\0&0&\dfrac1c\end{bmatrix}
Question 9: Determinants Miscellaneous – MCQ
9. Let
A=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix} , where 0\le\theta\le2\pi . Then
- (A) \operatorname{Det}(A)=0
- (B) \operatorname{Det}(A)\in(2,\infty)
- (C) \operatorname{Det}(A)\in(2,4)
- (D) \operatorname{Det}(A)\in[2,4]
Solution
Let
\Delta=\left|\begin{matrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{matrix}\right|
Expanding along the first row,
We get, \Delta=1\left|\begin{matrix}1&\sin\theta\\-\sin\theta&1\end{matrix}\right|-\sin\theta\left|\begin{matrix}-\sin\theta&\sin\theta\\-1&1\end{matrix}\right|+1\left|\begin{matrix}-\sin\theta&1\\-1&-\sin\theta\end{matrix}\right|
=(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+(\sin^2\theta+1)
=1+\sin^2\theta+0+1+\sin^2\theta
So, |A|=2+2\sin^2\theta
Since
0\le\sin^2\theta\le1
Therefore,
0\le2sin^2\theta\le2
or 2\le2+2\sin^2\theta\le4
\therefore \operatorname{Det}(A)\in[2,4]
✅️ Thus, the correct answer is (D).
Common Mistakes to Avoid
- Sign mistakes while expanding determinants: Be careful with the signs +,-,+ while expanding along a row or column.
- Incorrect minors and cofactors: Students often forget that cofactors include both the minor and the corresponding sign.
- Errors in matrix multiplication: Always multiply rows of the first matrix with columns of the second matrix carefully.
- Forgetting reversal law: While finding inverse of a product, remember that (AB)^{-1}=B^{-1}A^{-1} .
- Mistakes in transpose properties: The correct property is (AB)'=B'A' , not A'B' .
- Incorrect simplification of algebraic expressions: Simplify expressions carefully after determinant expansion to avoid cancellation mistakes.
- Forgetting determinant condition for inverse: A matrix is invertible only if |A|\neq0 .
- Confusing adjoint and inverse: Remember that A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A) .
- Not checking the final answer: After finding inverse or solving equations, verify the result whenever possible.
- Arithmetic calculation errors: Small multiplication or subtraction mistakes can completely change the determinant value, so calculate carefully.
Continue Learning
After completing this miscellaneous exercise, you should now be comfortable with the important concepts of determinants and inverse of matrices. This exercise combines algebraic simplification, determinant expansion, properties of inverse matrices and solving systems of equations using matrices.
To strengthen your preparation further, make sure that you revise:
- Expansion of determinants along different rows and columns
- Calculation of minors, cofactors and adjoint matrices
- Properties of determinants and inverse matrices
- Verification of matrix identities
- Application of inverse matrices in solving linear equations
- Trigonometric and algebraic simplifications inside determinants
Explore More
Regular practice of determinant calculations improves both speed and accuracy, which becomes very useful in board examinations as well as competitive exams.
You can now move on to solving additional determinant-based problems and application questions to build even stronger conceptual clarity.
All the best and keep learning 👍
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