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  • If possible, find BA and AB, where A =2 1 2 1 2 4

If possible, find BA and AB, where

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

Answers (1)

We are given matrices A and B, such that

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

We are required to find BA and AB, if possible.

To carry out the multiplication of matrices A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Let us check for BA.

\mathrm{BA}=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]

If a matrix has M rows and N columns, the order of matrix is M × N.

Order of B:

Number of rows = 3

⇒ M = 3

Number of columns = 2

⇒ N = 2

Then, order of matrix B = M × N

⇒ Order of matrix B = 3 × 2

Order of A:

Number of rows = 2

⇒ M = 2

Number of columns =3

⇒ N = 3

Then, order of matrix A = M × N

⇒ Order of matrix A = 2 × 3

Here,

Number of columns in matrix B = Number of rows in matrix A = 2

So, BA is possible.

Let us check for AB.

\mathrm{AB}=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]

Here,

Number of columns in matrix A = Number of rows in matrix B = 3

So, AB is also possible.

Let us find out BA.

\mathrm{BA}=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]

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

\\(4, 1).(2, 1) = (4 \times 2) + (1 \times 1) \\ \Rightarrow (4, 1).(2, 1) = 8 + 1 \\ \Rightarrow (4, 1).(2, 1) = 9

\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]= \begin{bmatrix} 9 & \\ & \\ & \end{bmatrix}

Multiply 1st row of matrix B by matching members of 2nd column of matrix A, then finally end by summing them up
\\(4, 1).(1, 2) = (4 \times 1) + (1 \times 2) \\ \Rightarrow (4, 1).(1, 2) = 4 + 2 \\ \Rightarrow (4, 1).(1, 2) = 6

\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]= \begin{bmatrix} 9 & 6 \\ & \\ & \end{bmatrix}

Similarly, let us calculate in the matrix itself.

\begin{array}{l} {\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]} \\ \Rightarrow\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]=\left[\begin{array}{ccc} 9 & 6 & 8+4 \\ 4+3 & 2+6 & 4+12 \\ 2+2 & 1+4 & 2+8 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]=\left[\begin{array}{lll} 9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10 \end{array}\right] \end{array}

Now, let us find out AB.

A B=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]

Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally end by summing it up.

\\(2, 1, 2).(4, 2, 1) = (2 \times 4) + (1 \times 2) + (2 \times 1) \\ \Rightarrow (2, 1, 2).(4, 2, 1) = 8 + 2 + 2 \\ \Rightarrow (2, 1, 2).(4, 2, 1) = 12

\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]= \begin{bmatrix} 12 & \\ & \\ & \end{bmatrix}

Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then finally end by summing it up.

\\(2, 1, 2).(1, 3, 2) = (2 \times 1) + (1 \times 3) + (2 \times 2) \\ \Rightarrow (2, 1, 2).(1, 3, 2) = 2 + 3 + 4 \\ \Rightarrow (2, 1, 2).(1, 3, 2) = 9

\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]= \begin{bmatrix} 12 & 9\\ & \\ & \end{bmatrix}

Similarly, let us calculate in the matrix itself.

\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]=\begin{bmatrix} 12 &9 \\ (1 \times 4)+(2 \times 2)+(4 \times 1) & (1 \times 1)+(2 \times 3)+(4 \times 2) \end{bmatrix}

\begin{aligned} &\Rightarrow\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 12 & 9 \\ 4+4+4 & 1+6+8 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 12 & 9 \\ 12 & 15 \end{array}\right]\\ &A B=\left[\begin{array}{cc} 12 & 9 \\ 12 & 15 \end{array}\right]_{\text {and }} B A=\left[\begin{array}{ccc} 9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10 \end{array}\right] \end{aligned}

 

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