# If the column matrix $A = \left[ {\begin{array}{*{20}{c}}

{ - 2} \\

4 \\

5

\end{array}} \right]$ and row matrix $B = \left[ {\begin{array}{*{20}{c}}

1&3&{ - 6}

\end{array}} \right]$ then verify that ${\left( {AB} \right)^\prime } = B'A'$.

Last updated date: 28th Mar 2023

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Answer

Verified

308.7k+ views

Hint: In this question apply the property of transpose (i.e. rows changed into column and column changed into rows) later on apply the property of matrix multiplication, so use these concepts to reach the solution of the question.

Given matrix is

$A = \left[ {\begin{array}{*{20}{c}}

{ - 2} \\

4 \\

5

\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}

1&3&{ - 6}

\end{array}} \right]$

Now A’ is the transpose of matrix (i.e. rows changed into column and column changed into rows) so apply transpose of A we have,

$A' = \left[ {\begin{array}{*{20}{c}}

{ - 2}&4&5

\end{array}} \right]$

And the transpose of B is

$B' = \left[ {\begin{array}{*{20}{c}}

1 \\

3 \\

{ - 6}

\end{array}} \right]$

Now the given equation is

${\left( {AB} \right)^\prime } = B'A'$

Now, consider L.H.S

$ = {\left( {AB} \right)^\prime }$

First calculate AB we have

$AB = \left[ {\begin{array}{*{20}{c}}

{ - 2} \\

4 \\

5

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

1&3&{ - 6}

\end{array}} \right]$

Now apply matrix multiplication we have

$AB = \left[ {\begin{array}{*{20}{c}}

{ - 2 \times 1}&{ - 2 \times 3}&{ - 2 \times \left( { - 6} \right)} \\

{4 \times 1}&{4 \times 3}&{4 \times \left( { - 6} \right)} \\

{5 \times 1}&{5 \times 3}&{5 \times - 6}

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{ - 2}&{ - 6}&{12} \\

4&{12}&{ - 24} \\

5&{15}&{ - 30}

\end{array}} \right]$

Now take transpose of above matrix we have

${\left( {AB} \right)^\prime } = \left[ {\begin{array}{*{20}{c}}

{ - 2}&4&5 \\

{ - 6}&{12}&{15} \\

{12}&{ - 24}&{ - 30}

\end{array}} \right]$………… (1)

Now consider R.H.S

$ = B'A'$

$ \Rightarrow B'A' = \left[ {\begin{array}{*{20}{c}}

1 \\

3 \\

{ - 6}

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

{ - 2}&4&5

\end{array}} \right]$

Now apply matrix multiplication we have

$ \Rightarrow B'A' = \left[ {\begin{array}{*{20}{c}}

{1 \times \left( { - 2} \right)}&{1 \times 4}&{1 \times 5} \\

{3 \times \left( { - 2} \right)}&{3 \times 4}&{3 \times 5} \\

{ - 6 \times \left( { - 2} \right)}&{ - 6 \times 4}&{ - 6 \times 5}

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{ - 2}&4&5 \\

{ - 6}&{12}&{15} \\

{12}&{ - 24}&{ - 30}

\end{array}} \right]$……………… (2)

Now from equation (1) and (2)

L.H.S = R.H.S

Hence verified.

Note: In such types of question always remember the key concept that we have to remember is that always recall the property of transpose which is stated above so, use this property and calculate the transpose of A and B, then apply matrix multiplication on given equation and calculate L.H.S and R.H.S separately and check whether they are equal or not if yes then the given condition is satisfied which is the required answer.

Given matrix is

$A = \left[ {\begin{array}{*{20}{c}}

{ - 2} \\

4 \\

5

\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}

1&3&{ - 6}

\end{array}} \right]$

Now A’ is the transpose of matrix (i.e. rows changed into column and column changed into rows) so apply transpose of A we have,

$A' = \left[ {\begin{array}{*{20}{c}}

{ - 2}&4&5

\end{array}} \right]$

And the transpose of B is

$B' = \left[ {\begin{array}{*{20}{c}}

1 \\

3 \\

{ - 6}

\end{array}} \right]$

Now the given equation is

${\left( {AB} \right)^\prime } = B'A'$

Now, consider L.H.S

$ = {\left( {AB} \right)^\prime }$

First calculate AB we have

$AB = \left[ {\begin{array}{*{20}{c}}

{ - 2} \\

4 \\

5

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

1&3&{ - 6}

\end{array}} \right]$

Now apply matrix multiplication we have

$AB = \left[ {\begin{array}{*{20}{c}}

{ - 2 \times 1}&{ - 2 \times 3}&{ - 2 \times \left( { - 6} \right)} \\

{4 \times 1}&{4 \times 3}&{4 \times \left( { - 6} \right)} \\

{5 \times 1}&{5 \times 3}&{5 \times - 6}

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{ - 2}&{ - 6}&{12} \\

4&{12}&{ - 24} \\

5&{15}&{ - 30}

\end{array}} \right]$

Now take transpose of above matrix we have

${\left( {AB} \right)^\prime } = \left[ {\begin{array}{*{20}{c}}

{ - 2}&4&5 \\

{ - 6}&{12}&{15} \\

{12}&{ - 24}&{ - 30}

\end{array}} \right]$………… (1)

Now consider R.H.S

$ = B'A'$

$ \Rightarrow B'A' = \left[ {\begin{array}{*{20}{c}}

1 \\

3 \\

{ - 6}

\end{array}} \right]\left[ {\begin{array}{*{20}{c}}

{ - 2}&4&5

\end{array}} \right]$

Now apply matrix multiplication we have

$ \Rightarrow B'A' = \left[ {\begin{array}{*{20}{c}}

{1 \times \left( { - 2} \right)}&{1 \times 4}&{1 \times 5} \\

{3 \times \left( { - 2} \right)}&{3 \times 4}&{3 \times 5} \\

{ - 6 \times \left( { - 2} \right)}&{ - 6 \times 4}&{ - 6 \times 5}

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

{ - 2}&4&5 \\

{ - 6}&{12}&{15} \\

{12}&{ - 24}&{ - 30}

\end{array}} \right]$……………… (2)

Now from equation (1) and (2)

L.H.S = R.H.S

Hence verified.

Note: In such types of question always remember the key concept that we have to remember is that always recall the property of transpose which is stated above so, use this property and calculate the transpose of A and B, then apply matrix multiplication on given equation and calculate L.H.S and R.H.S separately and check whether they are equal or not if yes then the given condition is satisfied which is the required answer.

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