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  • If A = [0 -1 2] and , then verify that: (i) (A’)’ = A (ii) (AB)’ = B’A’ (iii) (kA)’ = (kA’)

If A=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right], then verify that:
(i) (A’)’ = A
(ii) (AB)’ = B’A’
(iii) (kA)’ = (kA’)

 

Answers (1)

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

A=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right]

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

Take L.H.S: (A’)’

In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, thatis it switches the row and column indices of the matrix by producing another matrix denoted as A\textsuperscript{T} or A’.

So, in transpose of a matrix,

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

So, If A=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]

(0, -1, 2) and (4, 3, -4) are 1st and 2nd rows respectively, will become 1st and 2nd columns respectively.

Then A^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]
 
Also, if A^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]

Similarly, (0, 4), (-1, 3) and (2, -4) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.

Then \left(A^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]
Note, that

\left(A^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]=A

Thus, verified that \left(A^{\prime}\right)^{\prime}=A

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

Take L.H.S: (AB)’

Compute AB.

AB=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right]

Order of A = 2 × 3

Order of B = 3 × 2

Then, order of AB = 2 × 2

Multiplying 1st row of matrix A by matching members of 1st column of matrix B, then finally sum them up.

(0, -1, 2)(4, 1, 2) = (0 × 4) + (-1 × 1) + (2 × 2)

⇒ (0, -1, 2)(4, 1, 2) = 0 - 1 + 4

⇒ (0, -1, 2)(4, 1, 2) = 3

\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right] = \begin{bmatrix} 3 & \\ & \end{bmatrix}

Similarly, repeat the steps to fill for the other elements.

\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right] =\left[\begin{array}{cc} 3 & (0 \times 0)+(-1 \times 3)+(2 \times 6) \\ (4 \times 4)+(3 \times 1)+(-4 \times 2) & (4 \times 0)+(3 \times 3)+(-4 \times 6) \end{array}\right]

\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]\left[\begin{array}{cc} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right]=\left[\begin{array}{cc} 3 & 0-3+12 \\ 16+3-8 & 0+9-24 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]\left[\begin{array}{cc} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right]=\left[\begin{array}{cc} 3 & 9 \\ 11 & -15 \end{array}\right] \\ \Rightarrow \mathrm{AB}=\left[\begin{array}{cc} 3 & 9 \\ 11 & -15 \end{array}\right] \end{array}

Transpose of AB is (AB)’.

(3, 9) and (11, -15) are 1st and 2nd rows respectively, will become 1st and 2nd columns respectively.

$$ (\mathrm{AB})^{\prime}=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right]
Take R.H.S:
\mathrm{B}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]
(4, 0), (1, 3) and (2, 6) are 1st, 2nd and 3rd rows of matrix B respectively, will become 1st, 2nd and 3rd columns respectively.
\Rightarrow \mathrm{B}^{\prime}=\left[\begin{array}{lll}4 & 1 & 2 \\ 0 & 3 & 6\end{array}\right]
Also, if A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]
(0, -1, 2) and (4, 3, -4) are 1st and 2nd rows respectively, will become 1st and 2nd columns respectively.
\Rightarrow A^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]
By multiplyingB^{\prime} by A^{\prime}we get,

\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right]

Order of B’ = 2 × 3

Order of A’ = 3 × 2

Then, order of B’A’ = 2 × 2

Multiply 1st row of matrix B’ by matching members of 1st column of matrix A’, then finally sum them up.

(4, 1, 2)(0, -1, 2) = (4 × 0) + (1 × -1) + (2 × 2)

⇒ (4, 1, 2)(0, -1, 2) = 0 - 1 + 4

⇒ (4, 1, 2)(0, -1, 2) = 3

\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right] = \begin{bmatrix} 3 & \\ & \end{bmatrix}

Similarly, repeat the same steps to fill the rest of the elements.

\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right] = \left[\begin{array}{cc} 3 & (4 \times 4)+(1 \times 3)+(2 \times-4) \\ (0 \times 0)+(3 \times-1)+(6 \times 2) & (0 \times 4)+(3 \times 3)+(6 \times-4) \end{array}\right]

\begin{array}{l} \Rightarrow\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 & 16+3-8 \\ 0-3+12 & 0+9-24 \end{array}\right] \\ \Rightarrow\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right] \\ \Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right] \\ \text { since, } \mathrm{L.H.S}=\mathrm{R.H.S} \\ \text { Thus, }(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime} \end{array}

(iii). We need to verify that, (kA)’ = kA’.

Take L.H.S: (kA)’

We know that,

A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]

By Multiplying k on both sides, we get, (k is a scalar quantity)

\begin{array}{l} k A=k \times\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \\ \Rightarrow k A=\left[\begin{array}{ccc} k \times 0 & k \times-1 & k \times 2 \\ k \times 4 & k \times 3 & k \times-4 \end{array}\right] \\ \Rightarrow k A=\left[\begin{array}{ccc} 0 & -k & 2 k \\ 4 k & 3 k & -4 k \end{array}\right] \end{array}

Now, to find transpose of kA,

(0, -k, 2k) and (4k, 3k, -4k) are 1st and 2nd rows of matrix kA respectively, will become 1st and 2nd columns respectively.

\begin{aligned} &\Rightarrow(\mathrm{kA})^{\prime}=\left[\begin{array}{cc} 0 & 4 \mathrm{k} \\ -\mathrm{k} & 3 \mathrm{k} \\ 2 \mathrm{k} & -4 \mathrm{k} \end{array}\right]\\ &\text { Take R.H.S: kA }\\ &A=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \end{aligned}

Then, for transpose of A,

(0, -1, 2) and (4, 3, -4) are 1st and 2nd rows of matrix A respectively, will become 1st and 2nd columns respectively.

A'=\begin{bmatrix} 0 &4 \\-1 &3 \\2 &-4 \end{bmatrix}

By Multiplying k on both sides, we get,

\\ \mathrm{kA}^{\prime}=\mathrm{k}\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]$ \\$\Rightarrow \mathrm{kA}^{\prime}=\left[\begin{array}{cc}\mathrm{k} \times 0 & \mathrm{k} \times 4 \\ \mathrm{k} \times-1 & \mathrm{k} \times 3 \\ \mathrm{k} \times 2 & \mathrm{k} \times-4\end{array}\right]$ \\$\Rightarrow \mathrm{kA}^{\prime}=\left[\begin{array}{cc}0 & 4 \mathrm{k} \\ -\mathrm{k} & 3 \mathrm{k} \\ 2 \mathrm{k} & -4 \mathrm{k}\end{array}\right]
As, L.H.S = R.H.S.
Hence proved, (kA)' = kA'.

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