Anand Classes provides the most accurate, step-by-step, and exam-focused NCERT Solutions for Matrices – Miscellaneous Exercise, Class 12 Chapter 3, helping students strengthen conceptual clarity and score higher in board examinations. These solutions are designed to simplify challenging matrix operations, properties, and numerical applications with clear explanations and solved examples. With well-structured study material and downloadable PDF notes, Anand Classes ensures a seamless learning experience for Class 12 Math students. Chapter 3 of the Class 12 NCERT Mathematics textbook, “Matrices,” explores various matrix operations and concepts. The Miscellaneous Exercise provides diverse problems on matrix addition, multiplication, determinants, and inverses. It challenges students to apply these concepts and reinforces their understanding of matrix theory through detailed problem-solving. Click the print button to download study material and notes.
Access NCERT Solutions for Matrices – Miscellaneous Exercise, Class 12 Chapter 3
NCERT Question 1: If $A, B$ are symmetric matrices of the same order, then prove that $(AB-BA)$ is a Skew symmetric matrix
Solution
Since $A$ and $B$ are symmetric matrices,
$$A’ = A\qquad\text{and}\qquad B’ = B$$
Consider the transpose of $(AB-BA)$.
Using $(X-Y)’=X’-Y’$ and $(XY)’=Y’X’$
$$
(AB-BA)’ = (AB)’- (BA)’ $$
$$ (AB-BA)’ = B’A’ – A’B’ $$
since $A’=A, B’=B$
$$(AB-BA)’= B A – A B $$
$$(AB-BA)’= -(AB-BA)$$
We have shown
$$ (AB-BA)’ = -(AB-BA) $$
which is precisely the condition for a matrix to be skew-symmetric.
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NCERT Question 2 : Show that the matrix $B’AB$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.
Solution :
Let’s take $A$ as symmetric matrix.
Since $A$ is symmetric,
$$A’ = A$$
Using $(AB)’ = B’A’$,
$$ (B’AB)’ = (AB)'(B’)’ $$
Again, $(B’)’ = B$, so
$$ (B’AB)’ = B’A’B $$
Substituting $A’ = A$,
$$ (B’AB)’ = B’AB $$
So it is a symmetric matrix.
Let’s take $A$ as skew matrix.
For a skew-symmetric matrix,
$$A’ = -A$$
Using $(AB)’ = B’A’$,
$$ (B’AB)’ = (AB)’ (B’)’ $$
$$ (B’AB)’= B’A’B $$
Substituting $A’ = -A$,
$$ (B’AB)’ = B'(-A)B $$
$$ (B’AB)’= -B’AB $$
Thus, it is a skew-symmetric matrix.
Hence, we can conclude that $B’AB$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.
NCERT Question.3 : Find the values of $x,y,z$ if the matrix
$$
A=\begin{bmatrix}0 & 2y & z\\ x & y & -z\\ x & -y & z\end{bmatrix}
$$
satisfies the equation $A’A = I$
Solution
Given
$$
A=\begin{bmatrix}0 & 2y & z\\ x & y & -z\\ x & -y & z\end{bmatrix}
$$
First compute the transpose $A’=A^{T}$. Interchange rows and columns:
$$
A’=\begin{bmatrix}
0 & x & x\\
2y & y & -y\\
z & -z & z
\end{bmatrix}
$$
Now form the product $A’A$ and set it equal to the identity matrix $I_3$. Compute $A’A$ entrywise using
$$A’A = I_3$$
$$
A’A=
\begin{bmatrix}
0 & x & x\\
2y & y & -y\\
z & -z & z
\end{bmatrix}
\begin{bmatrix}
0 & 2y & z\\
x & y & -z\\
x & -y & z
\end{bmatrix}=\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}$$
Evaluate the product (showing the diagonal entries which must equal 1 and the off-diagonals which must equal 0:
((1,1)) entry: $ (0)\cdot 0 + x\cdot x + x\cdot x = 2x^{2}.$
This must equal (1), so
$2x^{2}=1 \quad\Rightarrow\quad x=\pm\frac{1}{\sqrt{2}}$
((1,2)) and ((1,3)) entries (off-diagonals) evaluate to 0 automatically because of cancellations; you may check one, e.g.
$(1,2): 0\cdot 2y + x\cdot y + x\cdot(-y)=xy-xy=0$
((2,2)) entry: $(2y)\cdot(2y) + y\cdot y + (-y)\cdot(-y) = 4y^{2}+y^{2}+y^{2}=6y^{2}$
This must equal (1), so
$6y^{2}=1 \quad\Rightarrow\quad y=\pm\frac{1}{\sqrt{6}}$
((3,3)) entry: $z\cdot z + (-z)\cdot(-z) + z\cdot z = z^{2}+z^{2}+z^{2}=3z^{2}$
This must equal (1), so
$3z^{2}=1 \quad\Rightarrow\quad z=\pm\frac{1}{\sqrt{3}}$
All off-diagonal entries are zero by similar cancellations (you can verify any if needed), so the only constraints are the diagonal ones above.
Final Answer
$$
\boxed{x=\pm\frac{1}{\sqrt{2}},\qquad y=\pm\frac{1}{\sqrt{6}},\qquad z=\pm\frac{1}{\sqrt{3}}}$$
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NCERT Question.4 : For what values of $x$ if
$$
\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1 & 0 & 2\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}=O $$
Solution
First multiply the row vector and the matrix:
$$
\begin{bmatrix}1 & 2 & 1\end{bmatrix}
\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1 & 0 & 2\end{bmatrix}=$$
$$=\begin{bmatrix}
1\cdot1+2\cdot2+1\cdot1 & 1\cdot2+2\cdot0+1\cdot0 & 1\cdot0+2\cdot1+1\cdot2
\end{bmatrix}
=\begin{bmatrix}6 & 2 & 4\end{bmatrix}.
$$
Now multiply this row vector by the column vector:
$$
\begin{bmatrix}6 & 2 & 4\end{bmatrix}\begin{bmatrix}0\\2\\x\end{bmatrix}
=6\cdot0+2\cdot2+4\cdot x=4+4x.
$$
Set the result equal to zero and solve for $x$ :
$$
4+4x=0\quad\Rightarrow\quad x+1=0\quad\Rightarrow\quad x=-1.
$$
$$
\boxed{x=-1}
$$
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NCERT Question.8 : If
$$A=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}$$
show that $$A^{2}-5A+7I=0$$
Solution
Compute $A^{2}=A\cdot A$:
$$
A^{2}=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}$$
$$
A^{2}=\begin{bmatrix}3\cdot3+1\cdot(-1) & 3\cdot1+1\cdot2\\
(-1)\cdot3+2\cdot(-1) & (-1)\cdot1+2\cdot2\end{bmatrix}$$
$$A^{2}=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}.$$
Compute $5A$ and $7I$ :
$$
5A=5\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}=\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}$$
$$7I=7\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}$$
Now form $A^{2}-5A+7I$:
$$
A^{2}-5A+7I=
\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}+\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}$$
$$
A^{2}-5A+7I=\begin{bmatrix}8-15+7 & 5-5+0\\-5+5+0 & 3-10+7\end{bmatrix}$$
$$
A^{2}-5A+7I=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$$
Therefore
$$
\boxed{A^{2}-5A+7I=0}.
$$
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NCERT Question.9 : Find $x$, if
$$
\begin{bmatrix}x & -5 & -1\end{bmatrix}
\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0 & 3\end{bmatrix}
\begin{bmatrix}x\\4\\1\end{bmatrix}
=0$$
Solution
Multiply the row matrix ($1\times3$) and the $3\times3$ matrix to get a $1\times3$ row matrix. Compute componentwise:
$$
\begin{bmatrix}x & -5 & -1\end{bmatrix}
\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0 & 3\end{bmatrix}$$
$$\begin{bmatrix}
x\cdot 1 + (-5)\cdot 0 + (-1)\cdot 2 &
x\cdot 0 + (-5)\cdot 2 + (-1)\cdot 0 &
x\cdot 2 + (-5)\cdot 1 + (-1)\cdot 3
\end{bmatrix}
$$
So the product is
$$
\begin{bmatrix}x-2 & -10 & 2x-8\end{bmatrix}.
$$
Now multiply this row matrix by the column matrix
$$\begin{bmatrix}x\\4\\1\end{bmatrix}$$
We get
$$
\begin{bmatrix}x-2 & -10 & 2x-8\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}
$$
This gives a scalar:
$$
(x-2)x + (-10)\cdot 4 + (2x-8)\cdot 1 = 0.
$$
Simplify the scalar equation:
$$
x(x-2) -40 + 2x -8 = 0
$$
$$
x^2 -2x -40 +2x -8 = 0
$$
$$
x^2 -48 = 0.
$$
Solve for $x$:
$$
x^2 = 48 \quad\Rightarrow\quad x = \pm \sqrt{48} = \pm 4\sqrt{3}.
$$
Final Result
$$
\boxed{x=\pm 4\sqrt{3}}
$$
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NCERT Question.10 : A manufacturer produces three products $x,y,z$ which he sells in two markets. Annual sales are indicated below :
| Market | Product x | Product y | Product z |
|---|---|---|---|
| I | 10,000 | 2,000 | 18,000 |
| II | 6,000 | 20,000 | 8,000 |
(a) If unit sale prices of $x,y,z$ are ₹2.50, ₹1.50 and ₹1.00 respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the unit costs are ₹2.00, ₹1.00 and ₹0.50 respectively, find the gross profit for each market.
Solution
Let the sales matrix $S$ (markets × products) and price / cost column matrices be:
$$
S=\begin{bmatrix}10000 & 2000 & 18000\\6000 & 20000 & 8000\end{bmatrix}
$$
(a) Unit sale prices:
$$
P=\begin{bmatrix}2.5\\1.5\\1\end{bmatrix}
$$
Total revenue vector $R = SP$:
$$ R=
\begin{bmatrix}10000 & 2000 & 18000\\6000 & 20000 & 8000\end{bmatrix}\begin{bmatrix}2.5\\1.5\\1\end{bmatrix}$$
$$ R=\begin{bmatrix}
10000\cdot2.5 + 2000\cdot1.5 + 18000\cdot1\\
6000\cdot2.5 + 20000\cdot1.5 + 8000\cdot1
\end{bmatrix}$$
$$ R=\begin{bmatrix}46000\\53000\end{bmatrix}$$
So revenue in Market I = ₹46,000 and in Market II = ₹53,000.
(b) Unit costs:
$$
C=\begin{bmatrix}2\\1\\0.5\end{bmatrix}
$$
Total cost vector $T = SC$:
$$
T=
\begin{bmatrix}10000 & 2000 & 18000\\6000 & 20000 & 8000\end{bmatrix}\begin{bmatrix}2\\1\\0.5\end{bmatrix}$$
$$T=\begin{bmatrix}10000\cdot2 + 2000\cdot1 + 18000\cdot0.5\\6000\cdot2 + 20000\cdot1 + 8000\cdot0.5
\end{bmatrix}$$
$$
T=\begin{bmatrix}31000\\36000\end{bmatrix}
$$
Gross profit $G = R – T$ :
$$
G=\begin{bmatrix}46000\\53000\end{bmatrix}-\begin{bmatrix}31000\\36000\end{bmatrix}=\begin{bmatrix}15000\\17000\end{bmatrix}$$
Total gross profit = ₹15,000 (Market I) and ₹17,000 (Market II).
Combined profit = ₹32,000.
Final Result
$$
\boxed{R=\begin{bmatrix}46000\\53000\end{bmatrix},\qquad
T=\begin{bmatrix}31000\\36000\end{bmatrix},\qquad
G=\begin{bmatrix}15000\\17000\end{bmatrix}}
$$
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NCERT Question 11 : Find the matrix $X$ so that
$$
X\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9 \\ 2 & 4 & 6\end{bmatrix}.
$$
Solution
Let
$$X=\begin{bmatrix}p & q \\ r & s\end{bmatrix}$$
Then,
$$X\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix}$$
$$\begin{bmatrix}p & q \\ r & s\end{bmatrix}\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix}$$
$$\begin{bmatrix}
p+4q & 2p+5q & 3p+6q \\
r+4s & 2r+5s & 3r+6s
\end{bmatrix}.
$$
This must equal
$$
\begin{bmatrix}-7 & -8 & -9 \\ 2 & 4 & 6\end{bmatrix}.
$$
So we equate corresponding entries:
$$\begin{bmatrix}
p+4q & 2p+5q & 3p+6q \\
r+4s & 2r+5s & 3r+6s
\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9 \\ 2 & 4 & 6\end{bmatrix}
$$
First row:
$$
p+4q=-7,\qquad 2p+5q=-8,\qquad 3p+6q=-9.
$$
Subtracting the first two equations:
$$
(2p+5q)-(p+4q)=-8-(-7)
$$
$$
p+q=-1.
$$
Subtracting the next pair:
$$
(3p+6q)-(2p+5q)=-9-(-8)
$$
$$
p+q=-1.
$$
Now use $p+4q=-7$ and $p+q=-1$:
$$
(p+4q)-(p+q)=-7-(-1)
$$
$$
3q=-6 ;\Rightarrow; q=-2.
$$
Then:
$$
p+(-2)=-1 \Rightarrow p=1.
$$
Second row:
$$
r+4s=2,\qquad 2r+5s=4,\qquad 3r+6s=6.
$$
Subtracting the first two:
$$
(2r+5s)-(r+4s)=4-2
$$
$$
r+s=2.
$$
Subtracting the next pair:
$$
(3r+6s)-(2r+5s)=6-4
$$
$$
r+s=2.
$$
Now solve $r+4s=2$ and $r+s=2$:
$$
(r+4s)-(r+s)=2-2
$$
$$
3s=0\Rightarrow s=0.
$$
Then:
$$
r+s=2 \Rightarrow r=2.
$$
Thus,
$$
X=\begin{bmatrix}
1 & -2 \\
2 & 0
\end{bmatrix}
$$
Final Result
$$
\boxed{X=\begin{bmatrix}1 & -2 \\ 2 & 0\end{bmatrix}}
$$
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NCERT Question.13 : If
$$A=\begin{bmatrix}\alpha & \beta\\\gamma & -\alpha\end{bmatrix}$$
is such that $A^{2}=I$, then which of the following is true?
(A) $1+\alpha^{2}+\beta\gamma=0$
(B) $1-\alpha^{2}+\beta\gamma=0$
(C) $1-\alpha^{2}-\beta\gamma=0$
(D) $1+\alpha^{2}-\beta\gamma=0$
Solution
Compute $A^{2}=AA$:
$$
A^{2}=
\begin{bmatrix}\alpha & \beta\\\gamma & -\alpha\end{bmatrix}
\begin{bmatrix}\alpha & \beta\\\gamma & -\alpha\end{bmatrix}$$
$$A^{2}=\begin{bmatrix}
\alpha^{2}+\beta\gamma & \alpha\beta – \alpha\beta\\
\alpha\gamma – \alpha\gamma & \beta\gamma+\alpha^{2}
\end{bmatrix}$$
$$A^{2}=\begin{bmatrix}
\alpha^{2}+\beta\gamma & 0\\
0 & \alpha^{2}+\beta\gamma
\end{bmatrix}.
$$
Given $A^{2}=I$, where $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$, we must have
$$\alpha^{2}+\beta\gamma=1.$$
Rewriting this yields
$$1-\alpha^{2}-\beta\gamma=0.$$
So the correct option is (C).
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NCERT Question.14 :
If the matrix $A$ is both symmetric and skew symmetric, then
(A) $A$ is a diagonal matrix
(B) $A$ is a zero matrix
(C) $A$ is a square matrix
(D) None of these
Solution
If $A$ is symmetric then
$$A’ = A$$
If $A$ is skew-symmetric then
$$A’ = -A$$
Combining these two relations gives
$$A = A’ = -A.$$
Hence
$$A + A = 0 \quad\Rightarrow\quad 2A = 0 \quad\Rightarrow\quad A = 0.$$
Therefore (A) must be the zero matrix.
Answer: $$\boxed{\text{(B) }A\text{ is a zero matrix}}$$
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NCERT Question 15 : If $A$ is a square matrix such that $A^{2}=A$, then
$$(I+A)^{3}-7A$$
is equal to
(A) $A \quad$ (B) $I-A \quad$ (C) $I \quad$ (D) $3A$
Solution
Expand $(I+A)^{3}$ using the binomial-like expansion (valid for matrices that commute with $I$, which every matrix does):
$$
(I+A)^{3}=I^{3}+3I^{2}A+3IA^{2}+A^{3}=I+3A+3A^{2}+A^{3}.
$$
Now form the required expression:
$$
(I+A)^{3}-7A = I+3A+3A^{2}+A^{3}-7A.
$$
Use the given condition $A^{2}=A$. Then $A^{3}=A^{2}A=AA=A$. Substitute these:
$$
I+3A+3A^{2}+A^{3}-7A = I+3A+3A+ A -7A $$
$$(I+A)^{3}-7A= I+(3A+3A+A-7A)$$
Combine the $A$-terms:
$$
3A+3A+A-7A=0,
$$
so the whole expression equals
$$(I+A)^{3}-7A= I$$
$$
\boxed{I}.
$$
Answer: (C) $I$.
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✅ FAQ Section
Q1. What is included in the Matrices Miscellaneous Exercise NCERT Solutions for Class 12?
The miscellaneous exercise covers advanced problems on matrix operations, properties, transposes, inverses, and applications. Solutions include step-by-step explanations.
Q2. Are these NCERT Matrices solutions helpful for CBSE Board Exams?
Yes. Anand Classes provides accurate and exam-oriented solutions strictly following the latest NCERT & CBSE guidelines, making them ideal for board exam preparation.
Q3. Can I download the Class 12 Matrices Miscellaneous Exercise PDF for free?
Yes, you can download the complete solutions and notes for free in PDF format from Anand Classes.
Q4. Do these solutions include step-by-step explanations?
Absolutely. Every question is solved logically with clear steps to help students understand matrix concepts easily.
Q5. Are these solutions useful for competitive exams as well?
Yes. Matrices is a key topic in JEE, NDA, CUET, and other competitive exams, and these solutions help strengthen conceptual understanding.
