Matrices: Chapter 3 Links
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This post provides clear and step-by-step NCERT Solutions for Class 12 Maths Chapter 3 Matrices 3.2. These solutions will cover matrix operations, properties of matrices and matrix multiplication in an easy-to-understand manner to help you practice and solve questions confidently.
This exercise is based on different operations on matrices and their important algebraic properties. Let’s first understand the core concepts used in this exercise of Matrices.
Key Concepts
1. Addition and Subtraction of Matrices
Two matrices can be added or subtracted only when they are of the same order. The corresponding elements of the matrices are added or subtracted.
If A=[a_{ij}] \text{ and } B=[b_{ij}]
then A+B=[a_{ij}+b_{ij}]
and A-B=[a_{ij}-b_{ij}]
2. Scalar Multiplication
When every element of a matrix is multiplied by a constant, the resulting matrix is called the scalar multiple of the given matrix.
If k is a scalar and A=[a_{ij}] , then
kA=[ka_{ij}]
3. Multiplication of Matrices
Two matrices can be multiplied only when the number of columns of the first matrix is equal to the number of rows of the second matrix.
If A is of order m\times n and B is of order n\times p , then the product AB is defined and its order is m\times p .
4. Properties of Matrix Addition
- Commutative Property: A+B=B+A
- Associative Property: A+(B+C)=(A+B)+C
- Additive Identity: A+O=A , where O is the zero matrix.
- Additive Inverse: A+(-A)=O
5. Properties of Scalar Multiplication
- k(A+B)=kA+kB
- (k+l)A=kA+lA
- k(lA)=(kl)A
6. Properties of Matrix Multiplication
- Associative Property: A(BC)=(AB)C
- Distributive Property: A(B+C)=AB+AC
- Multiplicative Identity: AI=IA=A , where I is the identity matrix.
- Matrix multiplication is generally not commutative, that is, AB\ne BA .
7. Product of Two Non-Zero Matrices
It is possible for the product of two non-zero matrices to be a zero matrix.
That is, even if A\ne O \text{ and } B\ne O , we may still have
AB=O
Now that we have understood the important concepts and properties used in this exercise matrices 3.2, let’s start solving the questions one by one.
Question 1: Matrices 3.2
1. Let A=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix},\ B=\begin{bmatrix}1 & 3 \\ -2 & 5\end{bmatrix},\ C=\begin{bmatrix}-2 & 5 \\ 3 & 4\end{bmatrix}
Find each of the following:
- (i) A+B
- (ii) A-B
- (iii) 3A-C
- (iv) AB
- (v) BA
Solution
(i) Finding A+B
We add the corresponding elements of the matrices.
A+B=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}+\begin{bmatrix}1 & 3 \\ -2 & 5\end{bmatrix}
=\begin{bmatrix}2+1 & 4+3 \\ 3+(-2) & 2+5\end{bmatrix}
So, we get A+B=\begin{bmatrix}3 & 7 \\ 1 & 7\end{bmatrix}
(ii) Finding A-B
We subtract the corresponding elements of the matrices.
A-B=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}-\begin{bmatrix}1 & 3 \\ -2 & 5\end{bmatrix}
=\begin{bmatrix}2-1 & 4-3 \\ 3-(-2) & 2-5\end{bmatrix}
So, we get A-B=\begin{bmatrix}1 & 1 \\ 5 & -3\end{bmatrix}
(iii) Finding 3A-C
First, multiply each element of matrix A by 3.
3A=3\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}=\begin{bmatrix}6 & 12 \\ 9 & 6\end{bmatrix}
Now subtract matrix C .
3A-C=\begin{bmatrix}6 & 12 \\ 9 & 6\end{bmatrix}-\begin{bmatrix}-2 & 5 \\ 3 & 4\end{bmatrix}
=\begin{bmatrix}6-(-2) & 12-5 \\ 9-3 & 6-4\end{bmatrix}
Therefore, 3A-C=\begin{bmatrix}8 & 7 \\ 6 & 2\end{bmatrix}
(iv) Finding AB
We multiply the rows of matrix A by the columns of matrix B .
AB=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}\begin{bmatrix}1 & 3 \\ -2 & 5\end{bmatrix}
=\begin{bmatrix}(2)(1)+(4)(-2) & (2)(3)+(4)(5) \\ (3)(1)+(2)(-2) & (3)(3)+(2)(5)\end{bmatrix}
i.e. AB=\begin{bmatrix}2-8 & 6+20 \\ 3-4 & 9+10\end{bmatrix}
=\begin{bmatrix}-6 & 26 \\ -1 & 19\end{bmatrix}
(v) Finding BA
We multiply the rows of matrix B by the columns of matrix A .
BA=\begin{bmatrix}1 & 3 \\ -2 & 5\end{bmatrix}\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}
=\begin{bmatrix}(1)(2)+(3)(3) & (1)(4)+(3)(2) \\ (-2)(2)+(5)(3) & (-2)(4)+(5)(2)\end{bmatrix}
So, BA=\begin{bmatrix}2+9 & 4+6 \\ -4+15 & -8+10\end{bmatrix}
=\begin{bmatrix}11 & 10 \\ 11 & 2\end{bmatrix}
Question 2: Addition of Matrices
2. Compute the following:
- (i) \begin{bmatrix}a & b \\ -b & a\end{bmatrix}+\begin{bmatrix}a & b \\ b & a\end{bmatrix}
- (ii) \begin{bmatrix}a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2\end{bmatrix}+\begin{bmatrix}2ab & 2bc \\ -2ac & -2ab\end{bmatrix}
- (iii) \begin{bmatrix}-1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{bmatrix}+\begin{bmatrix}12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4\end{bmatrix}
- (iv) \begin{bmatrix}\cos^2x & \sin^2x \\ \sin^2x & \cos^2x\end{bmatrix}+\begin{bmatrix}\sin^2x & \cos^2x \\ \cos^2x & \sin^2x\end{bmatrix}
Solution
(i) Both matrices are of order 2\times2 , therefore addition is possible.
\begin{bmatrix}a & b \\ -b & a\end{bmatrix}+\begin{bmatrix}a & b \\ b & a\end{bmatrix}
=\begin{bmatrix}a+a & b+b \\ -b+b & a+a\end{bmatrix}
So, Sum =\begin{bmatrix}2a & 2b \\ 0 & 2a\end{bmatrix}
(ii) As the order of first matrix is equal to the order of second matrix, therefore addition is possible.
\begin{bmatrix}a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2\end{bmatrix}+\begin{bmatrix}2ab & 2bc \\ -2ac & -2ab\end{bmatrix}
=\begin{bmatrix}a^2+b^2+2ab & b^2+c^2+2bc \\ a^2+c^2-2ac & a^2+b^2-2ab\end{bmatrix}
We get, Sum =\begin{bmatrix}(a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2\end{bmatrix}
(iii) Both matrices are of order 3\times3 , therefore addition is possible.
\begin{bmatrix}-1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{bmatrix}+\begin{bmatrix}12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4\end{bmatrix}
=\begin{bmatrix}-1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4\end{bmatrix}
Therefore, Sum =\begin{bmatrix}11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9\end{bmatrix}
(iv) Order of both the matrices is 2\times2 , therefore addition is possible.
\begin{bmatrix}\cos^2x & \sin^2x \\ \sin^2x & \cos^2x\end{bmatrix}+\begin{bmatrix}\sin^2x & \cos^2x \\ \cos^2x & \sin^2x\end{bmatrix}
=\begin{bmatrix}\cos^2x+\sin^2x & \sin^2x+\cos^2x \\ \sin^2x+\cos^2x & \cos^2x+\sin^2x\end{bmatrix}
Using the identity:
\sin^2x+\cos^2x=1
=\begin{bmatrix}1 & 1 \\ 1 & 1\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 Matrices 3.2.
Question 3: Multiplication of Matrices
3. Compute the indicated products.
- (i) \left[\begin{array}{cc}a & b \\ -b & a\end{array}\right]\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]
- (ii) \left[\begin{array}{c}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{ccc}2 & 3 & 4\end{array}\right]
- (iii) \left[\begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]
- (iv) \left[\begin{array}{ccc}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]
- (v) \left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]
- (vi) \left[\begin{array}{ccc}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]
Solution
(i) The number of columns of the first matrix is equal to the number of rows of the second matrix, therefore multiplication is possible.
\left[\begin{array}{cc}a & b \\ -b & a\end{array}\right]\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]
=\left[\begin{array}{cc}a(a)+b(b) & a(-b)+b(a) \\ -b(a)+a(b) & (-b)(-b)+a(a)\end{array}\right]
On simplifying, we get the required product
=\left[\begin{array}{cc}a^2+b^2 & -ab+ab \\ -ab+ab & b^2+a^2\end{array}\right]
=\left[\begin{array}{cc}a^2+b^2 & 0 \\ 0 & a^2+b^2\end{array}\right]
(ii) The first matrix is of order 3\times1 and second is 1\times3 , so multiplication is possible.
\left[\begin{array}{c}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{ccc}2 & 3 & 4\end{array}\right]
=\left[\begin{array}{ccc}(1)(2) & (1)(3) & (1)(4) \\ (2)(2) & (2)(3) & (2)(4) \\ (3)(2) & (3)(3) & (3)(4)\end{array}\right]
So, required product is
=\left[\begin{array}{ccc}2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12\end{array}\right]
(iii) The first matrix is 2\times2 and second is 2\times3 , so multiplication is possible.
\left[\begin{array}{cc}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]
=\left[\begin{array}{ccc}1\!\cdot\!1+(-2)\!\cdot\!2&1\!\cdot\!2+(-2)\!\cdot\!3&1\!\cdot\!3+(-2)\!\cdot\!1\\2\!\cdot\!1+3\!\cdot\!2&2\!\cdot\!2+3\!\cdot\!3&2\!\cdot\!3+3\!\cdot\!1\end{array}\right]
So, the required product is
=\left[\begin{array}{ccc}1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3\end{array}\right]
=\left[\begin{array}{ccc}-3 & -4 & 1 \\ 8 & 13 & 9\end{array}\right]
(iv) Both matrices are 3\times3 , so multiplication is possible.
\left[\begin{array}{ccc}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]
=\left[\begin{array}{ccc}2\!\cdot\!1+3\!\cdot\!0+4\!\cdot\!3&2\!\cdot\!(-3)+3\!\cdot\!2+4\!\cdot\!0&2\!\cdot\!5+3\!\cdot\!4+4\!\cdot\!5\\3\!\cdot\!1+4\!\cdot\!0+5\!\cdot\!3&3\!\cdot\!(-3)+4\!\cdot\!2+5\!\cdot\!0&3\!\cdot\!5+4\!\cdot\!4+5\!\cdot\!5\\4\!\cdot\!1+5\!\cdot\!0+6\!\cdot\!3&4\!\cdot\!(-3)+5\!\cdot\!2+6\!\cdot\!0&4\!\cdot\!5+5\!\cdot\!4+6\!\cdot\!5\end{array}\right]
=\left[\begin{array}{ccc}14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70\end{array}\right]
(v) First matrix is 3\times2 , second is 2\times3 , so multiplication is possible.
\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]
=\left[\begin{array}{ccc}2\!\cdot\!1+1\!\cdot\!(-1)&2\!\cdot\!0+1\!\cdot\!2&2\!\cdot\!1+1\!\cdot\!1\\3\!\cdot\!1+2\!\cdot\!(-1)&3\!\cdot\!0+2\!\cdot\!2&3\!\cdot\!1+2\!\cdot\!1\\(-1)\!\cdot\!1+1\!\cdot\!(-1)&(-1)\!\cdot\!0+1\!\cdot\!2&(-1)\!\cdot\!1+1\!\cdot\!1\end{array}\right]
Or, the product is
=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0\end{array}\right]
(vi) First matrix is 2\times3 , second is 3\times2 , so multiplication is possible.
\left[\begin{array}{ccc}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]
=\left[\begin{array}{cc}3\cdot2+(-1)\cdot1+3\cdot3 & 3\cdot(-3)+(-1)\cdot0+3\cdot1 \\ (-1)\cdot2+0\cdot1+2\cdot3 & (-1)\cdot(-3)+0\cdot0+2\cdot1\end{array}\right]
i.e., required product is
=\left[\begin{array}{cc}6-1+9 & -9+0+3 \\ -2+0+6 & 3+0+2\end{array}\right]
=\left[\begin{array}{cc}14 & -6 \\ 4 & 5\end{array}\right]
Question 4: Matrices 3.2
4. If A=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right],\ B=\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right] \text{ and } C=\left[\begin{array}{ccc}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]
then compute (A+B) and (B-C) . Also, verify that
A+(B-C)=(A+B)-C
Solution
Finding A+B
Both matrices are of the same order, therefore addition is possible.
A+B=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right]+\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right]
=\left[\begin{array}{ccc}1+3 & 2+(-1) & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3\end{array}\right]
So, A+B=\left[\begin{array}{ccc}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4\end{array}\right]
Finding B-C
Similarly,
B-C=\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right]-\left[\begin{array}{ccc}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]
=\left[\begin{array}{ccc}3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3\end{array}\right]
=\left[\begin{array}{ccc}-1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0\end{array}\right]
Verifying A+(B-C)=(A+B)-C
First, find A+(B-C) .
A+(B-C)=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right]+\left[\begin{array}{ccc}-1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0\end{array}\right]
=\left[\begin{array}{ccc}1-1 & 2-2 & -3+0 \\ 5+4 & 0-1 & 2+3 \\ 1+1 & -1+2 & 1+0\end{array}\right]
So, L.H.S. =\left[\begin{array}{ccc}0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1\end{array}\right]
Now, find (A+B)-C .
(A+B)-C=\left[\begin{array}{ccc}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4\end{array}\right]-\left[\begin{array}{ccc}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]
=\left[\begin{array}{ccc}4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1-(-2) & 4-3\end{array}\right]
So, R.H.S. =\left[\begin{array}{ccc}0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1\end{array}\right]
Since both sides are equal, therefore
A+(B-C)=(A+B)-C is verified.
Question 5: Operations on Matrices
5. If A=\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right] \text{ and } B=\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]
then compute 3A-5B .
Solution
Finding 3A
Multiply each element of matrix A by 3.
3A=3\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]
=\left[\begin{array}{ccc}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]
Finding 5B
Multiply each element of matrix B by 5.
5B=5\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]
=\left[\begin{array}{ccc}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]
Now finding 3A-5B
Both matrices are of the same order, therefore subtraction is possible.
3A-5B=\left[\begin{array}{ccc}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]-\left[\begin{array}{ccc}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]
=\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]
Hence, 3A-5B is a zero matrix.
These types of questions are considered scoring if the basic concepts and operations are clear.
Question 6: Matrices Ex 3.2
6. Simplify \cos\theta\left[\begin{array}{cc}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{array}\right]+\sin\theta\left[\begin{array}{cc}\sin\theta & -\cos\theta \\ \cos\theta & \sin\theta\end{array}\right]
Solution
First, multiply each matrix by the corresponding scalar.
\cos\theta\left[\begin{array}{cc}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{array}\right]
=\left[\begin{array}{cc}\cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta\end{array}\right]
and
\sin\theta\left[\begin{array}{cc}\sin\theta & -\cos\theta \\ \cos\theta & \sin\theta\end{array}\right]
=\left[\begin{array}{cc}\sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta\end{array}\right]
Now add the two matrices.
\left[\begin{array}{cc}\cos^2\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^2\theta\end{array}\right]+\left[\begin{array}{cc}\sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta\end{array}\right]
=\left[\begin{array}{cc}\cos^2\theta+\sin^2\theta & 0 \\ 0 & \cos^2\theta+\sin^2\theta\end{array}\right]
Using the identity:
\sin^2\theta+\cos^2\theta=1
=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]
Question 7: Exercise 3.2 Matrices
7. Find X and Y , if
- (i) X+Y=\left[\begin{array}{cc}7 & 0 \\ 2 & 5\end{array}\right] \text{ and } X-Y=\left[\begin{array}{cc}3 & 0 \\ 0 & 3\end{array}\right]
- (ii) 2X+3Y=\left[\begin{array}{cc}2 & 3 \\ 4 & 0\end{array}\right] \text{ and } 3X+2Y=\left[\begin{array}{cc}2 & -2 \\ -1 & 5\end{array}\right]
Solution
(i) Given,
X+Y=\left[\begin{array}{cc}7 & 0 \\ 2 & 5\end{array}\right] \quad ...(1)
and
X-Y=\left[\begin{array}{cc}3 & 0 \\ 0 & 3\end{array}\right] \quad ...(2)
Adding equations (1) and (2),
We get, (X+Y)+(X-Y)=\left[\begin{array}{cc}7 & 0 \\ 2 & 5\end{array}\right]+\left[\begin{array}{cc}3 & 0 \\ 0 & 3\end{array}\right]
2X=\left[\begin{array}{cc}10 & 0 \\ 2 & 8\end{array}\right]
X=\frac{1}{2}\left[\begin{array}{cc}10 & 0 \\ 2 & 8\end{array}\right]
Therefore, X=\left[\begin{array}{cc}5 & 0 \\ 1 & 4\end{array}\right]
Now, subtract equation (2) from equation (1).
We get, (X+Y)-(X-Y)=\left[\begin{array}{cc}7 & 0 \\ 2 & 5\end{array}\right]-\left[\begin{array}{cc}3 & 0 \\ 0 & 3\end{array}\right]
2Y=\left[\begin{array}{cc}4 & 0 \\ 2 & 2\end{array}\right]
Y=\frac{1}{2}\left[\begin{array}{cc}4 & 0 \\ 2 & 2\end{array}\right]
Therefore, Y=\left[\begin{array}{cc}2 & 0 \\ 1 & 1\end{array}\right]
(This is just like Solving Linear Equations in Two Variables)
(ii) Given that,
2X+3Y=\left[\begin{array}{cc}2 & 3 \\ 4 & 0\end{array}\right] \quad ...(1)
and
3X+2Y=\left[\begin{array}{cc}2 & -2 \\ -1 & 5\end{array}\right] \quad ...(2)
Multiply equation (1) by 3 and equation (2) by 2.
We get, 6X+9Y=\left[\begin{array}{cc}6 & 9 \\ 12 & 0\end{array}\right] \quad ...(3)
6X+4Y=\left[\begin{array}{cc}4 & -4 \\ -2 & 10\end{array}\right] \quad ...(4)
Subtract equation (4) from equation (3).
5Y=\left[\begin{array}{cc}2 & 13 \\ 14 & -10\end{array}\right]
Y=\frac{1}{5}\left[\begin{array}{cc}2 & 13 \\ 14 & -10\end{array}\right]
We get, Y=\left[\begin{array}{cc}\frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2\end{array}\right]
Substitute the value of Y in equation (1).
We get, 2X+3\left[\begin{array}{cc}\frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ 4 & 0\end{array}\right]
2X+\left[\begin{array}{cc}\frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ 4 & 0\end{array}\right]
2X=\left[\begin{array}{cc}2-\frac{6}{5} & 3-\frac{39}{5} \\ 4-\frac{42}{5} & 0-(-6)\end{array}\right]
i.e. 2X=\left[\begin{array}{cc}\frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{array}\right]
X=\frac{1}{2}\left[\begin{array}{cc}\frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{array}\right]
Therefore, X=\left[\begin{array}{cc}\frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{array}\right]
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: Matrices
8. Find X , if
Y=\left[\begin{array}{cc}3 & 2 \\ 1 & 4\end{array}\right] \text{ and } 2X+Y=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]
Solution
Given,
Y=\left[\begin{array}{cc}3 & 2 \\ 1 & 4\end{array}\right]
and
2X+Y=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]
Substituting the value of Y in the equation,
2X+\left[\begin{array}{cc}3 & 2 \\ 1 & 4\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]
Subtracting the matrices,
2X=\left[\begin{array}{cc}1-3 & 0-2 \\ -3-1 & 2-4\end{array}\right]
2X=\left[\begin{array}{cc}-2 & -2 \\ -4 & -2\end{array}\right]
Dividing each element by 2,
X=\frac{1}{2}\left[\begin{array}{cc}-2 & -2 \\ -4 & -2\end{array}\right]
We get, X=\left[\begin{array}{cc}-1 & -1 \\ -2 & -1\end{array}\right]
Question 9: Matrices 3.2
9. Find x and y , if
2\left[\begin{array}{cc}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{cc}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}5 & 6 \\ 1 & 8\end{array}\right]
Solution
First, multiply the matrix by 2.
2\left[\begin{array}{cc}1 & 3 \\ 0 & x\end{array}\right]=\left[\begin{array}{cc}2 & 6 \\ 0 & 2x\end{array}\right]
Now add the matrices.
\left[\begin{array}{cc}2 & 6 \\ 0 & 2x\end{array}\right]+\left[\begin{array}{cc}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}5 & 6 \\ 1 & 8\end{array}\right]
\left[\begin{array}{cc}2+y & 6 \\ 1 & 2x+2\end{array}\right]=\left[\begin{array}{cc}5 & 6 \\ 1 & 8\end{array}\right]
By comparing the corresponding elements,
2+y=5
y=3
and
2x+2=8
2x=6
Therefore, x=3
Question 10: Matrices Ex 3.2
10. Solve the equation for x,\ y,\ z \text{ and } t , if
2\left[\begin{array}{cc}x & z \\ y & t\end{array}\right]+3\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]=3\left[\begin{array}{cc}3 & 5 \\ 4 & 6\end{array}\right]
Solution
First, multiply the matrices by the corresponding scalars.
We get, 2\left[\begin{array}{cc}x & z \\ y & t\end{array}\right]=\left[\begin{array}{cc}2x & 2z \\ 2y & 2t\end{array}\right]
3\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}3 & -3 \\ 0 & 6\end{array}\right]
and 3\left[\begin{array}{cc}3 & 5 \\ 4 & 6\end{array}\right]=\left[\begin{array}{cc}9 & 15 \\ 12 & 18\end{array}\right]
Substituting these values in the given equation,
\left[\begin{array}{cc}2x & 2z \\ 2y & 2t\end{array}\right]+\left[\begin{array}{cc}3 & -3 \\ 0 & 6\end{array}\right]=\left[\begin{array}{cc}9 & 15 \\ 12 & 18\end{array}\right]
\left[\begin{array}{cc}2x+3 & 2z-3 \\ 2y & 2t+6\end{array}\right]=\left[\begin{array}{cc}9 & 15 \\ 12 & 18\end{array}\right]
Comparing the corresponding elements, we get
- 2x+3=9 \Rightarrow x=3
- 2z-3=15 \Rightarrow z=9
- 2y=12 \Rightarrow y=6
- 2t+6=18 \Rightarrow t=6
The questions in Matrices 3.2 are mostly based on matrix operations, properties and multiplication 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: Operations on Matrices
11. If
x\left[\begin{array}{c}2 \\ 3\end{array}\right]+y\left[\begin{array}{c}-1 \\ 1\end{array}\right]=\left[\begin{array}{c}10 \\ 5\end{array}\right]
find the values of x and y .
Solution
Multiplying the matrices by the corresponding scalars,
\left[\begin{array}{c}2x \\ 3x\end{array}\right]+\left[\begin{array}{c}-y \\ y\end{array}\right]=\left[\begin{array}{c}10 \\ 5\end{array}\right]
\left[\begin{array}{c}2x-y \\ 3x+y\end{array}\right]=\left[\begin{array}{c}10 \\ 5\end{array}\right]
Comparing the corresponding elements,
2x-y=10 \quad ...(1)
3x+y=5 \quad ...(2)
Adding equations (1) and (2),
5x=15
x=3
Substituting x=3 in equation (1),
We get, 2(3)-y=10
6-y=10
-y=4
Therefore, y=-4
Question 12: Matrices 3.2
12. Given 3\left[\begin{array}{cc}x & y \\ z & w\end{array}\right]=\left[\begin{array}{cc}x & 6 \\ -1 & 2w\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right]
find the values of x,\ y,\ z \text{ and } w .
Solution
First, add the matrices on the right-hand side.
\left[\begin{array}{cc}x & 6 \\ -1 & 2w\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right]
=\left[\begin{array}{cc}x+4 & x+y+6 \\ z+w-1 & 2w+3\end{array}\right]
Also,
3\left[\begin{array}{cc}x & y \\ z & w\end{array}\right]=\left[\begin{array}{cc}3x & 3y \\ 3z & 3w\end{array}\right]
Therefore,
\left[\begin{array}{cc}3x & 3y \\ 3z & 3w\end{array}\right]=\left[\begin{array}{cc}x+4 & x+y+6 \\ z+w-1 & 2w+3\end{array}\right]
Comparing the corresponding elements,
- 3x=x+4 \Rightarrow 2x=4 \Rightarrow x=2
- 3y=x+y+6 \Rightarrow 2y=x+6
- Substituting x=2 ,
2y=8 \Rightarrow y=4
- 3w=2w+3 \Rightarrow w=3
- 3z=z+w-1 \Rightarrow 2z=w-1
- Substituting w=3 ,
2z=2 \Rightarrow z=1
Hence,
x=2,\ y=4,\ z=1,\ w=3
Now, let’s move on to the next question of Matrices 3.2.
Question 13: Ex. 3.2
13. If F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]
show that F(x)F(y)=F(x+y) .
Solution
We have,
F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]
and F(y)=\left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]
Both matrices are of order 3\times3 , therefore multiplication is possible.
Now,
F(x)F(y)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]
=\left[\begin{array}{ccc}\cos x\cos y-\sin x\sin y & -\cos x\sin y-\sin x\cos y & 0 \\ \sin x\cos y+\cos x\sin y & -\sin x\sin y+\cos x\cos y & 0 \\ 0 & 0 & 1\end{array}\right]
Using the trigonometric identities:
- \cos(x+y)=\cos x\cos y-\sin x\sin y
- \sin(x+y)=\sin x\cos y+\cos x\sin y
we get,
F(x)F(y)=\left[\begin{array}{ccc}\cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1\end{array}\right]
But,
F(x+y)=\left[\begin{array}{ccc}\cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1\end{array}\right]
Hence,
F(x)F(y)=F(x+y)
Remember that matrix multiplication is not commutative in general, that is, AB\ne BA . Look at the following example:
Question 14: Matrices 3.2
14. Show that
- (i) \left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]\ne\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]
- (ii) \left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]\ne\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]
Solution
(i) Let A=\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right] \text{ and } B=\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]
Both matrices are of order 2\times2 , therefore multiplication is possible.
Now,
AB=\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]
=\left[\begin{array}{cc}(5)(2)+(-1)(3) & (5)(1)+(-1)(4) \\ (6)(2)+(7)(3) & (6)(1)+(7)(4)\end{array}\right]
So, L.H.S. =\left[\begin{array}{cc}7 & 1 \\ 33 & 34\end{array}\right]
Also,
BA=\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]
=\left[\begin{array}{cc}(2)(5)+(1)(6) & (2)(-1)+(1)(7) \\ (3)(5)+(4)(6) & (3)(-1)+(4)(7)\end{array}\right]
So, R.H.S. =\left[\begin{array}{cc}16 & 5 \\ 39 & 25\end{array}\right]
Since
AB\ne BA
therefore matrix multiplication is not commutative.
(ii) Let A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right] \text{ and } B=\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]
Both matrices are square matrices of order 3, therefore multiplication is possible.
Now,
AB=\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]
=\left[\begin{array}{ccc}5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1\end{array}\right]
Also,
BA=\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]
=\left[\begin{array}{ccc}-1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6\end{array}\right]
Since
AB\ne BA
hence proved.
Question 15: Matrix Equations
15. Find A^2-5A+6I , if
A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]
Solution
First, find A^2 .
A^2=A\cdot A
=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]
So, A^2=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]
Now,
5A=5\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]=\left[\begin{array}{ccc}10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0\end{array}\right]
Also,
6I=\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]
Therefore,
A^2-5A+6I
=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-\left[\begin{array}{ccc}10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0\end{array}\right]+\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]
After simplification, we get, A^2-5A+6I
=\left[\begin{array}{ccc}1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4\end{array}\right]
The next three questions are very important from the examination point of view. So, pay close attention to the method and calculations.
Question 16: Matrices 3.2
16. If A=\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]
prove that A^3-6A^2+7A+2I=0 .
Solution
First, find A^2 .
Consider, A^2=A\cdot A
=\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]
=\left[\begin{array}{ccc}1\!\cdot\!1+0\!\cdot\!0+2\!\cdot\!2&1\!\cdot\!0+0\!\cdot\!2+2\!\cdot\!0&1\!\cdot\!2+0\!\cdot\!1+2\!\cdot\!3\\0\!\cdot\!1+2\!\cdot\!0+1\!\cdot\!2&0\!\cdot\!0+2\!\cdot\!2+1\!\cdot\!0&0\!\cdot\!2+2\!\cdot\!1+1\!\cdot\!3\\2\!\cdot\!1+0\!\cdot\!0+3\!\cdot\!2&2\!\cdot\!0+0\!\cdot\!2+3\!\cdot\!0&2\!\cdot\!2+0\!\cdot\!1+3\!\cdot\!3\end{array}\right]
So, A^2 =\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]
Now, find A^3 .
Consider, A^3=A^2\cdot A
=\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]
=\left[\begin{array}{ccc}5\!\cdot\!1+0\!\cdot\!0+8\!\cdot\!2&5\!\cdot\!0+0\!\cdot\!2+8\!\cdot\!0&5\!\cdot\!2+0\!\cdot\!1+8\!\cdot\!3\\2\!\cdot\!1+4\!\cdot\!0+5\!\cdot\!2&2\!\cdot\!0+4\!\cdot\!2+5\!\cdot\!0&2\!\cdot\!2+4\!\cdot\!1+5\!\cdot\!3\\8\!\cdot\!1+0\!\cdot\!0+13\!\cdot\!2&8\!\cdot\!0+0\!\cdot\!2+13\!\cdot\!0&8\!\cdot\!2+0\!\cdot\!1+13\!\cdot\!3\end{array}\right]
Therefore, A^3 =\left[\begin{array}{ccc}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]
Also,
6A^2=6\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]=\left[\begin{array}{ccc}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]
7A=7\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]=\left[\begin{array}{ccc}7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21\end{array}\right]
And, 2I=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]
Therefore,
A^3-6A^2+7A+2I
=\left[\begin{array}{ccc}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]-\left[\begin{array}{ccc}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]+\left[\begin{array}{ccc}7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21\end{array}\right]+\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]
=\left[\begin{array}{ccc}21-30+7+2 & 0-0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0-0+0+0 & 55-78+21+2\end{array}\right]
Therefore, A^3-6A^2+7A+2I .
=\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]
= O
Hence proved.
Question 17: Matrices 3.2
17. If A=\left[\begin{array}{cc}3 & -2 \\ 4 & -2\end{array}\right] \text{ and } I=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]
find k so that A^2=kA-2I .
Solution
First, find A^2 .
We know, A^2=A\cdot A
=\left[\begin{array}{cc}3 & -2 \\ 4 & -2\end{array}\right]\left[\begin{array}{cc}3 & -2 \\ 4 & -2\end{array}\right]
=\left[\begin{array}{cc}(3)(3)+(-2)(4) & (3)(-2)+(-2)(-2) \\ (4)(3)+(-2)(4) & (4)(-2)+(-2)(-2)\end{array}\right]
Thus, A^2 =\left[\begin{array}{cc}1 & -2 \\ 4 & -4\end{array}\right]
Now,
A^2=kA-2I
Therefore,
\left[\begin{array}{cc}1 & -2 \\ 4 & -4\end{array}\right]=k\left[\begin{array}{cc}3 & -2 \\ 4 & -2\end{array}\right]-2\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]
=\left[\begin{array}{cc}3k & -2k \\ 4k & -2k\end{array}\right]-\left[\begin{array}{cc}2 & 0 \\ 0 & 2\end{array}\right]
=\left[\begin{array}{cc}3k-2 & -2k \\ 4k & -2k-2\end{array}\right]
Comparing the corresponding elements,
We get, 3k-2=1
3k=3
i.e. k=1
Question 18: Ex 3.2 Matrices
18. If A=\left[\begin{array}{cc}0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0\end{array}\right]
and I is the identity matrix of order 2, show that
I+A=(I-A)\left[\begin{array}{cc}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{array}\right]
Solution
Let
\tan\frac{\alpha}{2}=t
Then,
A=\left[\begin{array}{cc}0 & -t \\ t & 0\end{array}\right]
Also, the identity matrix of order 2 is
I=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]
Therefore,
I+A=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}0 & -t \\ t & 0\end{array}\right]
=\left[\begin{array}{cc}1 & -t \\ t & 1\end{array}\right]
Now,
I-A=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -t \\ t & 0\end{array}\right]
=\left[\begin{array}{cc}1 & t \\ -t & 1\end{array}\right]
Using the half-angle formulae,
- \cos\alpha=\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}=\frac{1-t^2}{1+t^2}
- \sin\alpha=\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}=\frac{2t}{1+t^2}
Therefore,
\left[\begin{array}{cc}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{array}\right]=\left[\begin{array}{cc}\frac{1-t^2}{1+t^2} & -\frac{2t}{1+t^2} \\ \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2}\end{array}\right]
Now, consider the RHS.
(I-A)\left[\begin{array}{cc}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{array}\right]
By using the above formulae and substitution, we get
RHS =\left[\begin{array}{cc}1 & t \\ -t & 1\end{array}\right]\left[\begin{array}{cc}\frac{1-t^2}{1+t^2} & -\frac{2t}{1+t^2} \\ \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2}\end{array}\right]
=\frac{1}{1+t^2}\left[\begin{array}{cc}1 & t \\ -t & 1\end{array}\right]\left[\begin{array}{cc}1-t^2 & -2t \\ 2t & 1-t^2\end{array}\right]
Multiplying the matrices, we get
RHS =\frac{1}{1+t^2}\left[\begin{array}{cc}(1)(1-t^2)+t(2t) & (1)(-2t)+t(1-t^2) \\ (-t)(1-t^2)+(1)(2t) & (-t)(-2t)+(1)(1-t^2)\end{array}\right]
=\frac{1}{1+t^2}\left[\begin{array}{cc}1+t^2 & -2t+t-t^3 \\ -t+t^3+2t & 2t^2+1-t^2\end{array}\right]
=\frac{1}{1+t^2}\left[\begin{array}{cc}1+t^2 & -t(1+t^2) \\ t(1+t^2) & 1+t^2\end{array}\right]
Thus, RHS =\left[\begin{array}{cc}1 & -t \\ t & 1\end{array}\right]
But,
I+A=\left[\begin{array}{cc}1 & -t \\ t & 1\end{array}\right]
Hence,
I+A=(I-A)\left[\begin{array}{cc}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{array}\right]
Try solving the next question on your own first, and then compare your approach with the solution given below.
Question 19: Matrices Word Problem
19. A trust fund has ₹30,000 that must be invested in two different types of bonds.
The first bond pays 5% interest per year and the second bond pays 7% interest per year.
Using matrix multiplication, determine how to divide ₹30,000 among the two types of bonds if the trust fund must obtain an annual total interest of
- (a) ₹1800
- (b) ₹2000
Solution
Let ₹ x be invested in the 5% bond and ₹ y be invested in the 7% bond.
Then,
x+y=30000 \quad ...(1)
The interest obtained from the investments is
\frac{5x}{100}+\frac{7y}{100}=I
where I is the required annual interest.
Writing in matrix form,
\left[\begin{array}{cc}1 & 1 \\ 5 & 7\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]=\left[\begin{array}{c}30000 \\ 100I\end{array}\right]
(a) When the annual interest is ₹1800
Here,
I=1800
Therefore,
5x+7y=180000 \quad ...(2)
From equation (1),
x=30000-y
Substituting in equation (2),
We get, 5(30000-y)+7y=180000
150000-5y+7y=180000
2y=30000
i.e. y=15000
Therefore,
x=30000-15000=15000
Hence, ₹15,000 should be invested in each bond.
(b) When the annual interest is ₹2000
Here,
I=2000
Therefore,
5x+7y=200000 \quad ...(3)
From equation (1),
x=30000-y
Substituting in equation (3),
We get, 5(30000-y)+7y=200000
150000-5y+7y=200000
2y=50000
i.e. y=25000
Therefore,
x=30000-25000=5000
Hence, ₹5,000 should be invested in the 5% bond and ₹25,000 in the 7% bond.
Question 20: Word Problem
20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books and 10 dozen economics books.
Their selling prices are ₹80, ₹60 and ₹40 each respectively.
Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution
Number of books:
- Chemistry books =10 \text{ dozen}=120
- Physics books =8 \text{ dozen}=96
- Economics books =10 \text{ dozen}=120
Writing in matrix form,
A=\left[\begin{array}{ccc}120 & 96 & 120\end{array}\right]
and
B=\left[\begin{array}{c}80 \\ 60 \\ 40\end{array}\right]
Therefore, total amount received =AB
Putting the values, we get
AB=\left[\begin{array}{ccc}120 & 96 & 120\end{array}\right]\left[\begin{array}{c}80 \\ 60 \\ 40\end{array}\right]
On multiplication,
We get, AB=120(80)+96(60)+120(40)
=9600+5760+4800
=20160
Hence, the total amount received by the bookshop is ₹20,160.
The following two MCQs have already been discussed in the separate Matrices MCQs post. You can also check out other important MCQs from the chapter after completing this exercise for additional practice and revision.
Question 21: Matrices 3.2 – MCQ
Assume X,\ Y,\ Z,\ W \text{ and } P are matrices of order 2\times n,\ 3\times k,\ 2\times p,\ n\times3 \text{ and } p\times k respectively.
Choose the correct answer in Questions 21 and 22.
21. The restriction on n,\ k \text{ and } p so that PY+WY will be defined are:
- (A) k=3,\ p=n
- (B) k \text{ is arbitrary},\ p=2
- (C) p \text{ is arbitrary},\ k=3
- (D) k=2,\ p=3
Solution
Given,
- P is of order p\times k
- Y is of order 3\times k
- W is of order n\times3
For PY to be defined, number of columns of P must equal number of rows of Y .
k=3
Then, order of PY=p\times k=p\times3 .
Also, for WY to be defined,
W(n\times3)\text{ and }Y(3\times k)
which is always possible, and order of WY=n\times k=n\times3 .
For addition, orders of PY \text{ and } WY must be same.
p\times3=n\times3
p=n
Hence, the correct option is (A).
Question 22: Matrices Ex 3.2 – MCQ
22. If n=p , then the order of the matrix 7X-5Z is:
- (A) p\times2
- (B) 2\times n
- (C) n\times3
- (D) p\times n
Solution
Given,
- X is of order 2\times n
- Z is of order 2\times p
Since n=p , the order of both matrices becomes 2\times n .
Therefore, subtraction is possible and the order of
7X-5Z
is 2\times n .
Hence, the correct option is (B).
Common Mistakes to Avoid
- Adding or subtracting matrices of different orders
Matrix addition and subtraction are possible only when both matrices have the same order. - Ignoring the condition for matrix multiplication
Before multiplying two matrices, always check whether the number of columns of the first matrix is equal to the number of rows of the second matrix. - Writing incorrect order of the product matrix
If A is of order m\times n and B is of order n\times p , then AB is of order m\times p . - Assuming matrix multiplication is commutative
In general, AB\ne BA . Always calculate both products separately if required. - Mistakes in row-column multiplication
Each element of the product matrix is obtained by multiplying corresponding elements of a row and a column and then adding them carefully. - Confusing zero matrix with identity matrix
A zero matrix has all elements zero, while an identity matrix has 1’s on the principal diagonal and 0’s elsewhere. - Forgetting properties of identity matrix
For a square matrix A , we have AI=IA=A . - Errors in scalar multiplication
While multiplying a matrix by a scalar, every element of the matrix must be multiplied by that scalar. - Incorrect simplification in matrix equations
While evaluating expressions like A^2-5A+6I , perform matrix multiplication first and then addition or subtraction carefully. - Assuming product of non-zero matrices cannot be zero
Two non-zero matrices can have a product equal to the zero matrix.
Continue Learning
Matrices 3.2 mainly focuses on matrix operations, properties of matrices and matrix multiplication. These concepts are not only important for board examinations but are also used in higher mathematics and competitive exams.
While solving matrix questions, always pay attention to the order of matrices, multiplication conditions and careful calculations. Most mistakes in this chapter happen due to sign errors or incorrect row-column multiplication.
If you practice these questions regularly and understand the logic behind each step, matrix problems will become much easier and more scoring.
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