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Q4:
If $A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$, then find $(A + 2B)'$

Solution :

Step 1: Determining Matrix $A$

Given the transpose of matrix $A$, denoted as $A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$, we find $A$ by applying the property of transposition: $(A')' = A$ [Property: The transpose of a transpose is the original matrix].

Thus, transposing the rows and columns of $A'$:

$A = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}$

Step 2: Calculating Matrix $2B$

Given $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$, we perform scalar multiplication by multiplying every element of the matrix by the scalar $2$ [Definition of Scalar Multiplication]:

$2B = 2 \times \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2(-1) & 2(0) \\ 2(1) & 2(2) \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$

Step 3: Calculating $(A + 2B)$

To find the sum of matrices $A$ and $2B$, we add the corresponding elements [Property: Matrix addition is defined for matrices of the same order]:

$A + 2B = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$

$A + 2B = \begin{bmatrix} -2 + (-2) & 1 + 0 \\ 3 + 2 & 2 + 4 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}$

Step 4: Finding the Transpose $(A + 2B)'$

To find the transpose, we interchange the rows and columns of the resulting matrix from Step 3 [Definition: If $C = [c_{ij}]$, then $C' = [c_{ji}]$]:

$(A + 2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$

Matrix Summation Logic: -4 1 5 6 Transpose -4 5 1 6

Final Solution: $(A + 2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$


More Questions from Class 12 Mathematics Matrices EXERCISE 3.3


CBSE Solutions for Class 12 Mathematics Matrices


Chapters in CBSE - Class 12 Mathematics


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