Question

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).

Short answer

The products are\(AD = \left( {\begin{aligned}{*{20}{c}}2&3&5\\2&6&{15}\\2&{12}&{25}\end{aligned}} \right)\), and\(DA = \left( {\begin{aligned}{*{20}{c}}2&2&2\\3&6&9\\5&{20}&{25}\end{aligned}} \right)\).

Each column of matrix A is multiplied by the appropriate diagonal entry of matrix D using right-multiplication by the diagonal matrix D. Each row of matrix A is multiplied by the appropriate diagonal entry of D when left-multiplication by D is used.

It is proved that \(AB = BA\).

Step-by-step solution

Definition of matrix multiplication

Consider two matrices P and Q of order\(m \times n\)and\(n \times p\), respectively. The order of the product PQ matrix is\(m \times p\).

Let\({{\bf{q}}_1},{{\bf{q}}_2},...,{{\bf{q}}_n}\)be the columns of the matrix Q. Then, the product PQ is obtained as shown below:

\(\begin{aligned}{c}PQ = P\left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}& \cdots &{{{\bf{q}}_n}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{P{{\bf{q}}_1}}&{P{{\bf{q}}_2}}& \cdots &{P{{\bf{q}}_n}}\end{aligned}} \right)\end{aligned}\)

Obtain the product AD

Consider the matrices\(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\)and\(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\).

Obtain the product of matrices\(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\)and\(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\)as shown below:

\(\begin{aligned}{c}AD = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{2 + 0 + 0}&{0 + 3 + 0}&{0 + 0 + 5}\\{2 + 0 + 0}&{0 + 6 + 0}&{0 + 0 + 15}\\{2 + 0 + 0}&{0 + 12 + 0}&{0 + 0 + 25}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}2&3&5\\2&6&{15}\\2&{12}&{25}\end{aligned}} \right)\end{aligned}\)

Thus, \(AD = \left( {\begin{aligned}{*{20}{c}}2&3&5\\2&6&{15}\\2&{12}&{25}\end{aligned}} \right)\).

Obtain the product AD

Consider the matrices\(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\)and\(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\).

Obtain the product of matrices\(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\)and\(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\)as shown below:

\(\begin{aligned}{c}DA = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{2 + 0 + 0}&{2 + 0 + 0}&{2 + 0 + 0}\\{0 + 3 + 0}&{0 + 6 + 0}&{0 + 9 + 0}\\{0 + 0 + 5}&{0 + 0 + 20}&{0 + 0 + 25}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}2&2&2\\3&6&9\\5&{20}&{25}\end{aligned}} \right)\end{aligned}\)

Thus, \(DA = \left( {\begin{aligned}{*{20}{c}}2&2&2\\3&6&9\\5&{20}&{25}\end{aligned}} \right)\). So, \(AD \ne DA\).

Right-multiplication and left-multiplication

Each column of matrix A, that is\(\left( {\begin{aligned}{*{20}{c}}1\\1\\1\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}1\\2\\4\end{aligned}} \right)\), and \(\left( {\begin{aligned}{*{20}{c}}1\\3\\5\end{aligned}} \right)\)is multiplied by the appropriate diagonal entry of matrix D using right-multiplication by the diagonal matrix D. Each row of matrix A is multiplied by the appropriate diagonal entry of D when left-multiplication by D is used.

Construct a \(3 \times 3\) matrix, and obtain the product AB

The matrix\(B = \left( {\begin{aligned}{*{20}{c}}{ - 2}&0&0\\0&{ - 2}&0\\0&0&{ - 2}\end{aligned}} \right)\).

Obtain the product of matrices\(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\)and\(B = \left( {\begin{aligned}{*{20}{c}}{ - 2}&0&0\\0&{ - 2}&0\\0&0&{ - 2}\end{aligned}} \right)\)as shown below:

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 2}&0&0\\0&{ - 2}&0\\0&0&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2 + 0 + 0}&{0 - 2 + 0}&{0 + 0 - 2}\\{ - 2 + 0 + 0}&{0 - 4 + 0}&{0 + 0 - 6}\\{ - 2 + 0 + 0}&{0 - 8 + 0}&{0 + 0 - 10}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2}&{ - 2}&{ - 2}\\{ - 2}&{ - 4}&{ - 6}\\{ - 2}&{ - 8}&{ - 10}\end{aligned}} \right)\end{aligned}\)

Thus, \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 2}&{ - 2}&{ - 2}\\{ - 2}&{ - 4}&{ - 6}\\{ - 2}&{ - 8}&{ - 10}\end{aligned}} \right)\).

Obtain the product BA

Obtain the product of matrices\(B = \left( {\begin{aligned}{*{20}{c}}{ - 2}&0&0\\0&{ - 2}&0\\0&0&{ - 2}\end{aligned}} \right)\)and\(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\)as shown below:

\(\begin{aligned}{c}BA = \left( {\begin{aligned}{*{20}{c}}{ - 2}&0&0\\0&{ - 2}&0\\0&0&{ - 2}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2 + 0 + 0}&{ - 2 + 0 + 0}&{ - 2 + 0 + 0}\\{0 - 2 + 0}&{0 - 4 + 0}&{0 - 6 + 0}\\{0 + 0 - 2}&{0 + 0 - 8}&{0 + 0 - 10}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 2}&{ - 2}&{ - 2}\\{ - 2}&{ - 4}&{ - 6}\\{ - 2}&{ - 8}&{ - 10}\end{aligned}} \right)\end{aligned}\)

Thus,\(BA = \left( {\begin{aligned}{*{20}{c}}{ - 2}&{ - 2}&{ - 2}\\{ - 2}&{ - 4}&{ - 6}\\{ - 2}&{ - 8}&{ - 10}\end{aligned}} \right)\). So,\(AB = BA\).

Therefore, \(AD = \left( {\begin{aligned}{*{20}{c}}2&3&5\\2&6&{15}\\2&{12}&{25}\end{aligned}} \right)\), and \(DA = \left( {\begin{aligned}{*{20}{c}}2&2&2\\3&6&9\\5&{20}&{25}\end{aligned}} \right)\). And it is proved that \(AB = BA\).