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Q5(i):
For the matrices A and B, verify that $(AB)' = B'A'$, where (i) $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$, $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$

Solution :

Step 1: Given Matrices and Dimensional Analysis

We are given the matrices:

$A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$ (Order: $3 \times 1$)

$B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$ (Order: $1 \times 3$)

[Per the definition of matrix multiplication, the product $AB$ is defined if the number of columns in $A$ equals the number of rows in $B$. Since $A$ is $3 \times 1$ and $B$ is $1 \times 3$, the resulting matrix $AB$ will be of order $3 \times 3$.]

Step 2: Calculating the Product $AB$

Applying the row-by-column multiplication rule:

$AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} (1)(-1) & (1)(2) & (1)(1) \\ (-4)(-1) & (-4)(2) & (-4)(1) \\ (3)(-1) & (3)(2) & (3)(1) \end{bmatrix}$

$AB = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$

Step 3: Finding the Transpose of the Product $(AB)'$

To find the transpose, we interchange rows and columns [Per the definition of Transpose Matrix $(A^T)_{ij} = A_{ji}$]:

$(AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Step 4: Finding Transposes of $A$ and $B$ individually

For $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$, the transpose $A' = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$.

For $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$, the transpose $B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$.

Step 5: Calculating the Product $B'A'$

We now compute the product of $B' (3 \times 1)$ and $A' (1 \times 3)$:

$B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} (-1)(1) & (-1)(-4) & (-1)(3) \\ (2)(1) & (2)(-4) & (2)(3) \\ (1)(1) & (1)(-4) & (1)(3) \end{bmatrix}$

$B'A' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Step 6: Verification

Expression Result
$(AB)'$ $\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$
$B'A'$ $\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Comparing the matrices from Step 3 and Step 5, we observe that $(AB)' = B'A'$.

Verification Result: $(AB)' = B'A' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Final Solution: The property $(AB)' = B'A'$ is verified as both sides yield the matrix $\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$.


More Questions from Class 12 Mathematics Matrices EXERCISE 3.3


CBSE Solutions for Class 12 Mathematics Matrices


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