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  • If A =[1 2 4 1 5 6 ]then verify that: (i) (2A + B)’ = 2A’ + B’ (ii) (A - B)’ = A’ - B’.

If A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right], B=\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right] then verify that:

(i) (2A + B)’ = 2A’ + B’
(ii) (A - B)’ = A’ - B’.

Answers (1)

We are given the following matrices A and B, such that

A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right], B=\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]

In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as AT or A’.

So, in transpose of a matrix,

The rows of the matrix become the columns of the matrix. .

(i). We need to verify that, (2A + B)’ = 2A’ + B’.

Take L.H.S: (2A + B)’

By substituting the matrices A and B, in (2A + B)’, we get,

\begin{aligned} &(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(2\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right]+\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]\right)^{\prime}\\ &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(\left[\begin{array}{lll} 2 \times 1 & 2 \times 2 \\ 2 \times 4 & 2 \times 1 \\ 2 \times 5 & 2 \times 6 \end{array}\right]+\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]\right)^{\prime}\\ &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(\left[\begin{array}{cc} 2 & 4 \\ 8 & 2 \\ 10 & 12 \end{array}\right]+\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]\right)^{\prime}\\ &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(\left[\begin{array}{cc} 2+1 & 4+2 \\ 8+6 & 2+4 \\ 10+7 & 12+3 \end{array}\right]\right)^{\prime}\\ &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(\left[\begin{array}{cc} 3 & 6 \\ 14 & 6 \\ 17 & 15 \end{array}\right]\right)^{\prime} \end{aligned}

For transpose of (2A + B),

(3, 6), (14, 6) and (17, 15) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.

\begin{aligned} &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}= \begin{bmatrix} 3 &14 &17 \\6 &6 &15 \end{bmatrix} \end{aligned}

Take R.H.S: 2A’ + B’

If A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right]

(1, 2), (4, 1) and (5, 6) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.

\Rightarrow A^{\prime}=\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]
Multiply both sides by 2 we get,
$$ 2 \mathrm{~A}^{\prime}=2\left[\begin{array}{lll} 1 & 4 & 5 \\ 2 & 1 & 6 \end{array}\right]
\\\Rightarrow 2 A^{\prime}=\left[\begin{array}{lll}2 \times 1 & 2 \times 4 & 2 \times 5 \\ 2 \times 2 & 2 \times 1 & 2 \times 6\end{array}\right]$ \\$\Rightarrow 2 \mathrm{~A}^{\prime}=\left[\begin{array}{lll}2 & 8 & 10 \\ 4 & 2 & 12\end{array}\right]
Also,
If
B=\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]

(1, 2), (6, 4) and (7, 3) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.

\begin{aligned} &\Rightarrow \mathrm{B}^{\prime}=\left[\begin{array}{lll} 1 & 6 & 7 \\ 2 & 4 & 3 \end{array}\right]\\ &\text { Now, add } 2 \mathrm{~A}^{\prime} \text { and } \mathrm{B}^{\prime}\\ &2 \mathrm{~A}^{\prime}+\mathrm{B}^{\prime}=\left[\begin{array}{lll} 2 & 8 & 10 \\ 4 & 2 & 12 \end{array}\right]+\left[\begin{array}{lll} 1 & 6 & 7 \\ 2 & 4 & 3 \end{array}\right]\\ &\Rightarrow 2 A^{\prime}+B^{\prime}=\left[\begin{array}{lll} 2+1 & 8+6 & 10+7 \\ 4+2 & 2+4 & 12+3 \end{array}\right]\\ &\Rightarrow 2 A^{\prime}+B^{\prime}=\left[\begin{array}{ccc} 3 & 14 & 17 \\ 6 & 6 & 15 \end{array}\right] \end{aligned}

Since, L.H.S = R.H.S

Thus, (2A + B)’ = 2A’ + B’.

(ii). We need to verify that, (A - B)’ = A’ - B’.

Take L.H.S: (A - B)’

By substituting the matrices A and B in (A - B)’, we get,

\begin{array}{l} (A-B)^{\prime}=\left(\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right]-\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]\right)^{\prime} \\ \Rightarrow(A-B)^{\prime}=\left(\left[\begin{array}{ll} 1-1 & 2-2 \\ 4-6 & 1-4 \\ 5-7 & 6-3 \end{array}\right]\right)^{\prime} \\ \Rightarrow(A-B)^{\prime}=\left(\left[\begin{array}{cc} 0 & 0 \\ -2 & -3 \\ -2 & 3 \end{array}\right]\right)^{\prime} \end{array}

To find transpose of (A - B),

(0, 0), (-2, -3) and (-2, 3) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.

$$ \Rightarrow(A-B)^{\prime}=\left[\begin{array}{ccc} 0 & -2 & -2 \\ 0 & -3 & 3 \end{array}\right]
Take R.H.S: \mathrm{A}^{\prime}-\mathrm{B}^{\prime}
A=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]
(1, 2), (4, 1), \text{ and } (5, 6) \text{ are } 1^{\text{st}}, 2^{\text{nd}}, \text{ and } 3^{\text{rd}} \text{ points, respectively.}

rows respectively, will become 1^{\text{st}}, \quad 2^{\text{nd}}, \quad \text{and} \quad 3^{\text{rd}} columns respectively.
\Rightarrow A^{\prime}=\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]
Also,
B=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]

(1, 2), (6, 4) and (7, 3) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.
\Rightarrow \mathrm{B}^{\prime}=\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right]
When Subtracting \mathrm{B}^{\prime} \text{ from } \mathrm{A}^{\prime}, we get,

A^{\prime}-B^{\prime}=\left[\begin{array}{lll} 1 & 4 & 5 \\ 2 & 1 & 6 \end{array}\right]-\left[\begin{array}{lll} 1 & 6 & 7 \\ 2 & 4 & 3 \end{array}\right]

\begin{array}{l} \Rightarrow A^{\prime}-B^{\prime}=\left[\begin{array}{rrr} 1-1 & 4-6 & 5-7 \\ 2-2 & 1-4 & 6-3 \end{array}\right] \\ \Rightarrow A^{\prime}-B^{\prime}=\left[\begin{array}{rrr} 0 & -2 & -2 \\ 0 & -3 & 3 \end{array}\right] \\ \text { As, L.H.S = R.H.S } \\ \text { Hence proved, }(A-B)^{\prime}=A^{\prime}-B \end{array}

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