Class 12 Mathematics Chapter 3 Matrices

Key concept: Matrix Multiplication Inverse

b) $$(PQ)^{-1} = Q^{-1} P^{-1}$$
[Solution Description] The inverse of the product of two invertible matrices PQ is given by $$(PQ)^{-1} = Q^{-1} P^{-1}$$. This property states that the order of multiplication reverses when taking inverses.

Key concept: Matrix Order

c), They must be square matrices of the same order.
[Solution Description] For matrices A and B to be invertible, they both need to be square matrices of the same order. This is required because only square matrices can have an inverse, and operations involving matrices require them to be of the same dimensions.

Key concept: Uniqueness of Inverse

b) Unique
[Solution Description] The inverse of a square matrix, if it exists, is unique. There cannot be two different matrices that both multiply with the original matrix to give the identity matrix.

Key concept: Commutative Property

a) $$A + B = B + A$$
[Solution Description] The commutative property of matrix addition states that $$A + B = B + A$$. Thus, the order in which two matrices are added does not affect the result.

Key concept: Skew-Symmetric Matrix Property

a) It is skew-symmetric.
[Solution Description] According to the properties of matrices, for any square matrix A, the difference $$A – A’$$ is skew-symmetric. This is because $$(A – A’)’ = A’ – A = -(A – A’)$$.

Key concept: Matrix Example

b) Skew-symmetric
[Solution Description]
For the given matrix $$B = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$, its transpose $$B’$$ is $$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$. Since $$B’ = -B$$, the matrix is skew-symmetric.

Key concept: Square Matrix Recognition

c) [4 5; 6 7]
[Solution Description] A square matrix has the same number of rows and columns. Checking the options, option c) [4 5; 6 7] is a square matrix of order 2×2.

Key concept: Scalar Multiplication

a) $$\begin{bmatrix} 4 & 12 \\ 8 & 16 \end{bmatrix}$$
[Solution Description]
According to the property of scalar multiplication with transposes:
$$
(kC)’ = kC’
$$

Calculate $$ kC $$:
$$
kC = 2 \times \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 12 & 16 \end{bmatrix}
$$

Now find the transpose:
$$
(kC)’ = \begin{bmatrix} 4 & 12 \\ 8 & 16 \end{bmatrix}
$$

Key concept: Symmetric Matrix

c) $$ A’ = A $$
[Solution Description] A matrix A is symmetric if and only if its transpose is equal to itself, i.e., $$A’ = A$$.

Key concept: Definition Understanding

c) The number of columns in the first matrix must equal the number of rows in the second matrix.
[Solution Description] According to the definition of matrix multiplication, for two matrices A and B, the product AB is defined only if the number of columns in A equals the number of rows in B.

Key concept: Scalar Distribution Over Matrix Addition

a) $$\begin{bmatrix} 12 & 9 \\ 15 & -12 \end{bmatrix}$$
[Solution Description] Use distribution to calculate $$3(P+Q)$$ by first finding $$P+Q$$ and then multiplying the result by 3.

Step 1: Calculate $$P + Q$$ by adding corresponding elements:
$$
P + Q = \begin{bmatrix} 1+3 & -1+4 \\ 0+5 & 2+(-6) \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 5 & -4 \end{bmatrix}
$$

Step 2: Multiply each element of $$P+Q$$ by 3:

First row: $$\begin{bmatrix} 4 \cdot 3 & 3 \cdot 3 \end{bmatrix} = \begin{bmatrix} 12 & 9 \end{bmatrix}$$

Second row: $$\begin{bmatrix} 5 \cdot 3 & -4 \cdot 3 \end{bmatrix} = \begin{bmatrix} 15 & -12 \end{bmatrix}$$

Thus, $$3(P+Q) = \begin{bmatrix} 12 & 9 \\ 15 & -12 \end{bmatrix}$$.

Key concept: Commutative Property

a) Yes
[Solution Description] Matrix addition is commutative, meaning $$P + Q = Q + P$$. Hence, it holds true that $$\begin{bmatrix} 8 & 10 \\ 12 & 14 \end{bmatrix} = \begin{bmatrix} 8 & 10 \\ 12 & 14 \end{bmatrix}$$.

Key concept: Zero Matrix and Matrix Inversion, Zero Matrix and Linear Transformations

c) $$D^{-1}$$ does not exist
[Solution Description] If $$D^{2} = O$$, then multiplying D by itself results in a zero matrix, meaning D is not invertible. Generally, only non-singular matrices have inverses, and having $$D^{2} = O$$ implies D is singular. Therefore, $$D^{-1}$$ does not exist.

Key concept: Matrix Representation

b) $$\begin{bmatrix} -5 \\ 8 \end{bmatrix}$$
[Solution Description]
To represent the point (-5, 8) as a column matrix, we need to place these coordinates into a single-column format. In this structure, each coordinate will occupy one row in a two-row column matrix.

The correct representation is:
$$
\begin{bmatrix} -5 \\ 8 \end{bmatrix}
$$
Option b) matches this matrix.

Key concept: Element Identification

a) 0
[Solution Description]
In an identity matrix, only the diagonal elements are 1, and all other elements are 0. Hence, the element in the third row and second column is 0.

Key concept: Zero Element Recognition

b) 0, 0, 0
[Solution Description] Non-diagonal elements in a diagonal matrix are all zeroes. So, here they are all zero.

Key concept: Advanced Diagonal Matrix Problem, Complex Matrix Operations

b), D is an identity matrix
[Solution Description]
In this scenario, multiplication of D and S should result in S itself, i.e., $$DS = S$$ and similarly $$SD = S$$.

Since D is diagonal and S is a scalar matrix, the product maintains form only if D equals the identity matrix multiplied by some constant value affecting S.

It ensures $$d_{ii}s_{ii} = s_{ii}$$ for non-zero scalar elements across identical indices matching unity. Essentially $$d_{ii}$$ must equal 1 and thus make D the identity matrix to fulfill these multiplications returning S unchanged.

Key concept: Element Count

c) 5
[Solution Description] To find the number of elements in the row matrix $$[7, 11, 5, 8, 10]$$, we simply count each element within the matrix:
– Elements: $$7, 11, 5, 8, 10$$
Total count of elements = 5

Key concept: Order of Matrix

a) 2 × 3
[Solution Description]
The matrix B has 2 rows and 3 columns. Thus, the order of the matrix is 2 × 3.

Key concept: Matrix Notation

b) A matrix of order $$ m \times n $$
[Solution Description]
The notation $$ A = [a_{ij}]_{m \times n} $$ represents a matrix of order $$ m \times n $$, where  m  is the number of rows and  n  is the number of columns. Each element in the matrix is denoted by $$ a_{ij} $$, with  i  indicating the row and  j  indicating the column.

Key concept: Simple Matrix Application

a) $$\begin{pmatrix} 10 & 5 \end{pmatrix}$$
[Solution Description] The possession of 10 pencils and 5 erasers by Radha can be represented as a $$1 \times 2$$ matrix: $$\begin{pmatrix} 10 & 5 \end{pmatrix}$$.

Key concept: Scalar Matrix and Identity

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that an identity matrix is a type of scalar matrix where each diagonal element is precisely 1. This statement is true as the definition of an identity matrix aligns perfectly with that of a scalar matrix when k = 1.

The reason provided explains another property of a scalar matrix, which is that all non-diagonal elements are zero. This is also true for both scalar matrices and identity matrices since having zero non-diagonal elements is integral to their definitions.

However, while both statements are true, the reason does not directly explain why an identity matrix is specifically a scalar matrix when k = 1. Therefore, the reason is not the explanation of the assertion.

Hence, the correct option is (b).

Key Concept: Symmetric and Skew Symmetric Decomposition

a) $$\begin{bmatrix} 4 & 4 4 & -3 \end{bmatrix} + \begin{bmatrix} 0 & 3 -3 & 0 \end{bmatrix}$$

[Solution Description] According to the theorem, any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix using $$A = \frac{1}{2}(A + A’) + \frac{1}{2}(A – A’)$$. For $$A = \begin{bmatrix}
4 & 7 \\
1 & -3
\end{bmatrix},$$
calculate $$A’$$:

$$A’ = \begin{bmatrix} 4 & 1 \\ 7 & -3 \end{bmatrix}$$

Now compute $$A + A’$$ and $$A – A’$$:

$$A + A’ = \begin{bmatrix} 4 & 7 \\ 1 & -3 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 7 & -3 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 8 & -6 \end{bmatrix}$$

$$A – A’ = \begin{bmatrix} 4 & 7 \\ 1 & -3 \end{bmatrix} – \begin{bmatrix} 4 & 1 \\ 7 & -3 \end{bmatrix} = \begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix}$$

Therefore, the symmetric component is $$\frac{1}{2} \begin{bmatrix} 8 & 8 \\ 8 & -6 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 4 & -3 \end{bmatrix}$$
and the skew symmetric component is $$\frac{1}{2} \begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$$

Key Concept: Symmetric Matrix

a) $\begin{bmatrix} 1 & 2 \\  2 & 1 \end{bmatrix}$

[Solution Description]

For a matrix to be symmetric, it must satisfy $A’ = A$. Let’s evaluate each option:

Option a):

$$\begin{bmatrix}
1 & 2 \\
2 & 1
\end{bmatrix}’ = \begin{bmatrix}
1 & 2 \\
2 & 1
\end{bmatrix}$$

Option b):
$$\begin{bmatrix} 3 & 4 \\
5 & 6 \end{bmatrix}^T = \begin{bmatrix} 3 & 5 \\
4 & 6 \end{bmatrix}$$

Option c): $\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}’ = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Option d):

$$\begin{bmatrix}
0 & 1 \\
-1 & 0
\end{bmatrix}^T =
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}$$

Options a) and c) are symmetric because their transposes are equal to the original matrices.

Key Concept: Scalar Addition

b) $\begin{pmatrix} 20 & 25 \\ 30 & 35 \end{pmatrix}$

[Solution Description]

Here, we will use the property of scalar addition for matrices. According to the given property: $(k+l)A = kA + lA$.

First, calculate $k + l$: $k + l = 2 + 3 = 5$

Next, compute $(k+l)A$ as follows:

$$5A = 5 \times \begin{pmatrix} 4 & 5 \\ 6 & 7 \end{pmatrix} = \begin{pmatrix} 20 & 25 \\ 30 & 35 \end{pmatrix}$$

Therefore, $((k+l)A = \begin{pmatrix} 20 & 25 \\ 30 & 35 \end{pmatrix}\).$

Key Concept: Element Calculation

c) 7

[Solution Description] The element in the second row, first column of $A + B$ is calculated by adding the elements in the second row, first column of matrices A and B: $1 + 6 = 7$

Key Concept: Matrix Transpose

a) $\begin{bmatrix} 7 & 9 \\ 8 & 10 \end{bmatrix}$

[Solution Description] The transpose of matrix $D$ is found by swapping its rows with columns:

\[
D^T = \begin{bmatrix} 7 & 89 & 10 \end{bmatrix}^T = \begin{bmatrix} 7 \\ 89 \\ 10 \end{bmatrix}
\]

Thus, the correct answer is $\begin{bmatrix} 7 \\ 89 \\ 10 \end{bmatrix}.$

Key Concept: Order and Element Check

a) Matrices $E$ and $F$

[Solution Description] To determine equality, both order and elements must match:

– Matrices $E$ and $F$ have the same order and identical elements.

– Matrices $G$ and $H$ differ by one element ($8$ and $7$).

– Matrices $I$ and $J$ differ by one element ($5$ and $6$).

– Matrices $K$ and $L$ have different orders (one is $1 \times 2$, other is $2 \times 1$).

Thus, only matrices $E$ and $F$ are equal.

Key Concept: Matrix Properties, Complex Matrix Identification

c) All elements are equal to each other

[Solution Description]

Since the matrix is given to be both a diagonal and a scalar matrix, it must satisfy the properties of both. A diagonal matrix has non-diagonal elements as zero, which is satisfied here. For a scalar matrix, all diagonal elements must be equal. Thus, we have: $x = y = z$

Therefore, the elements $x$, $y$, and $z$ must be equal.

Key Concept:  Matrix Type Identification

d) Square matrix

[Solution Description] A square matrix is defined as one where the number of rows equals the number of columns ($m = n$). Since both the number of rows and columns is 4 in this case, it is a square matrix.

Key Concept: Matrix in Science, Complex Matrix Problem

a) $\begin{bmatrix} 0.55 & 0.45 \\ 0.48 & 0.52 \end{bmatrix}$

[Solution Description] The application of the mutation process involves multiplying the original genetic matrix $G$ with the mutation matrix $M$. Thus, the new matrix $N$ is calculated as:

\[
N = GM = \begin{bmatrix} p & q \\ r & s \end{bmatrix} \begin{bmatrix} 0.9 & 0.1 \\ 0.2 & 0.8 \end{bmatrix}
\]

Performing the multiplication gives us:

\[
\begin{align*}
N_{11} &= p \cdot 0.9 + q \cdot 0.2 = 0.9p + 0.2q \\
N_{12} &= p \cdot 0.1 + q \cdot 0.8 = 0.1p + 0.8q \\
N_{21} &= r \cdot 0.9 + s \cdot 0.2 = 0.9r + 0.2s \\
N_{22} &= r \cdot 0.1 + s \cdot 0.8 = 0.1r + 0.8s
\end{align*}
\]

Assuming (p=0.5), (q=0.5), (r=0.4), (s=0.6) for an example calculation, the new matrix becomes:

\[
N = \begin{bmatrix} 0.9(0.5) + 0.2(0.5) & 0.1(0.5) + 0.8(0.5) \\ 0.9(0.4) + 0.2(0.6) & 0.1(0.4) + 0.8(0.6) \end{bmatrix} = \begin{bmatrix} 0.55 & 0.45 \\ 0.48 & 0.52 \end{bmatrix}
\]

Therefore, the new matrix representing allele frequencies post-mutation is:$\begin{bmatrix} 0.55 & 0.45 \\ 0.48 & 0.52 \end{bmatrix}$