Prove: If A and B are n×n diagonal matrices, then AB = BA.

Question

Prove: If A and B are n×n diagonal matrices, then  AB = BA.

Asma Vang · Accepted Answer

Step 1  Given, A and B are n×n diagonal matrices  Step 2  Suppose  A=[a110…0ca22…0⋮⋮⋱000…ann],B=[b110…00b22…0⋮⋮⋱000…bnn]  Now  L.H.S.=AB  =[a110…0ca22…0⋮⋮⋱000…ann][b110…00b22…0⋮⋮⋱000…bnn]  =[a11b110…00a22b22…0⋮⋮⋱000…annbnn]  Now  R.H.S.=BA  =[b110…00b22…0⋮⋮⋱000…bnn][a110…0ca22…0⋮⋮⋱000…ann]  =[b11a110…00b22a22…0⋮⋮⋱000…bnnann]  =[a11b110…00a22b22…0⋮⋮⋱000…annbnn]  Thus AB=BA

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

nick1337 · Accepted Answer

Result:AB = BASolution:A=[a110…00a22…0⋮⋮⋱⋮00…ann]B=[b110…00b22…0⋮⋮⋱⋮00…bnn]Now, we can compute the product AB:AB=[a110…00a22…0⋮⋮⋱⋮00…ann][b110…00b22…0⋮⋮⋱⋮00…bnn]When we multiply these matrices, we perform the multiplication of corresponding entries in each row of A with the corresponding entries in each column of B. Since both A and B are diagonal matrices, all the off-diagonal entries are zero. Therefore, the only non-zero multiplications occur when the row index is equal to the column index.So, the product AB is:AB=[a11b110…00a22b22…0⋮⋮⋱⋮00…annbnn]Similarly, let&#039;s compute the product BA:BA=[b110…00b22…0⋮⋮⋱⋮00…bnn][a110…00a22…0⋮⋮⋱⋮00…ann]Again, the only non-zero multiplications occur when the row index is equal to the column index. Thus, the product BA is:BA=[b11a110…00b22a22…0⋮⋮⋱⋮00…bnnann]By comparing the matrices AB and BA, we can see that they have the same diagonal entries a11b11,a22b22,…,annbnn. Since the diagonal entries uniquely determine a matrix, we conclude that AB = BA.Therefore, we have proved that if A and B are n × n diagonal matrices, then AB = BA.

RizerMix · Accepted Answer

To prove the statement &#039;&#039;If&amp;nbsp;A&amp;nbsp;and&amp;nbsp;B&amp;nbsp;are&amp;nbsp;n×n&amp;nbsp;diagonal&amp;nbsp;matrices,&amp;nbsp;then&amp;nbsp;AB=BA,&#039;&#039; we can use the definition of diagonal matrices and basic matrix multiplication properties.Let A and B be given as diagonal matrices:A=[a110⋯00a22⋯0⋮⋮⋱⋮00⋯ann]B=[b110⋯00b22⋯0⋮⋮⋱⋮00⋯bnn]where aij and bij denote the elements of matrices A and B, respectively.Now, let&#039;s compute the product AB:AB=[a11b110⋯00a22b22⋯0⋮⋮⋱⋮00⋯annbnn]Similarly, let&#039;s compute the product BA:BA=[a11b110⋯00a22b22⋯0⋮⋮⋱⋮00⋯annbnn]By comparing AB and BA, we observe that they have the same elements in each corresponding position. Therefore, AB=BA, which proves the given statement.

Vasquez · Accepted Answer

Let&#039;s denote the diagonal entries of matrix A as aii, where 1≤i≤n, and the diagonal entries of matrix B as bii, where 1≤i≤n. Since A and B are diagonal matrices, all off-diagonal entries are zero.The product AB can be expressed as:AB=[a110…00a22…0⋮⋮⋱⋮00…ann][b110…00b22…0⋮⋮⋱⋮00…bnn].Performing the matrix multiplication, we get:AB=[a11b110…00a22b22…0⋮⋮⋱⋮00…annbnn].Similarly, let&#039;s calculate the product BA:BA=[b110…00b22…0⋮⋮⋱⋮00…bnn][a110…00a22…0⋮⋮⋱⋮00…ann].Performing the matrix multiplication, we have:BA=[b11a110…00b22a22…0⋮⋮⋱⋮00…bnnann].Comparing the matrices AB and BA, we can observe that they have the same diagonal entries. Since matrices AB and BA are both diagonal matrices with the same diagonal entries, we conclude that AB=BA.Therefore, we have proved that if A and B are n×n diagonal matrices, then AB=BA.

Prove: If A and B are n times n diagonal matrices, then AB = BA.

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