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  • If A equals 3 -4 1 1 2 0 and B equals 2 1 2 1 2 4 , then verify (BA)2 ≠ B2A2.

\begin{aligned} &A=\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ccc} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right] \end{aligned}  then verify (BA)^{2} \neq B^{2} A^{2}

Answers (1)

We have the following matrices,

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

We need to verify (BA)\textsuperscript{2} \neq B\textsuperscript{2}A\textsuperscript{2}.

Take L.H.S: (BA)\textsuperscript{2}

First, compute BA.

\text { B. } A=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]

We understand what a order of matrix is,

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

Order of matrix B:

Number of rows = 2

⇒ M = 2

Number of columns = 3

⇒ N = 3

Then, order of matrix = M × N

⇒ Order of matrix B = 2 × 3

Order of matrix A:

Number of rows = 3

⇒ M = 3

Number of columns = 2

⇒ N = 2

Then, order of matrix = M × N

⇒ Order of matrix A = 3 × 2

If we have two given matrices A and B which need to be multiplied, then 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.

So, A and B can be multiplied.

\text { B. } A=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]

 

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

\\(2, 1, 2)(3, 1, 2) = (2 \times 3) + (1 \times 1) + (2 \times 2) \\ \Rightarrow (2, 1, 2)(3, 1, 2) = 6 + 1 + 4 \\ \Rightarrow (2, 1, 2)(3, 1, 2) = 11

\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 11 & \\ & \end{bmatrix}

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

\\(2, 1, 2)(-4, 1, 0) = (2 \times -4) + (1 \times 1) + (2 \times 0) \\ \Rightarrow (2, 1, 2)(-4, 1, 0) = -8 + 1 + 0 \\ \Rightarrow (2, 1, 2)(-4, 1, 0) = -7

\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 11 &-7 \\ & \end{bmatrix}

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

\\(1, 2, 4)(3, 1, 2) = (1 \times 3) + (2 \times 1) + (4 \times 2) \\ \Rightarrow (1, 2, 4)(3, 1, 2) = 3 + 2 + 8 \\ \Rightarrow (1, 2, 4)(3, 1, 2) = 13

\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 11 &-7 \\ 13& \end{bmatrix}

Multiply 2nd row of matrix B by matching member of 2nd column of matrix A, then finally end by summing them up.

\\ (1, 2, 4)(-4, 1, 0) = (1 \times -4) + (2 \times 1) + (4 \times 0) \\ \Rightarrow (1, 2, 4)(-4, 1, 0) = -4 + 2 + 0 \\ \Rightarrow (1, 2, 4)(-4, 1, 0) = -2

\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 11 &-7 \\ 13& -2\end{bmatrix}

So,

(BA)\textsuperscript{2} = (BA).(BA)

\begin{aligned} &\Rightarrow(\mathrm{BA})^{2}=\left[\begin{array}{ll} 11 & -7 \\ 13 & -2 \end{array}\right]\left[\begin{array}{ll} 11 & -7 \\ 13 & -2 \end{array}\right]\\ &\text { Similarly, }\\ &\Rightarrow(\mathrm{BA})^{2}=\left[\begin{array}{ll} (11 \times 11+(-7) \times 13) & (11 \times-7+(-7) \times(-2)) \\ (13 \times 11+(-2) \times 13) & (13 \times-7+(-2) \times(-2)) \end{array}\right]\\ &\Rightarrow(\mathrm{BA})^{2}=\left[\begin{array}{cc} 121-91 & -77+14 \\ 143-26 & -91+4 \end{array}\right]\\ &\Rightarrow(\mathrm{BA})^{2}=\left[\begin{array}{cc} 30 & -63 \\ 117 & -87 \end{array}\right] \end{aligned}

Take R.H.S: B\textsuperscript{2}A\textsuperscript{2}

Let us first compute B\textsuperscript{2}.

B\textsuperscript{2} = B.B

\Rightarrow \mathrm{B}^{2}=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]

For multiplication of two matrices, say 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.

Note that in matrix B, number of columns is not equal to the number of rows.

Which means, we can’t find B\textsuperscript{2}.

\Rightarrow L.H.S \neq R.H.S

Thus, we have verified that, (BA)\textsuperscript{2} \neq B\textsuperscript{2}A\textsuperscript{2}.

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