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  • Given an example of matrices A, B and C such that AB = AC, where A is non-zero matrix, but B ≠ C.

Given an example of matrices A, B and C such that AB = AC, where A is non-zero matrix but B ≠ C,.

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We need to form matrices A, B and C such that AB = AC, where A is a non-zero matrix but B ≠ C.

Take,

\begin{aligned} &A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\\ &B=\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]\\ &C=\left[\begin{array}{ll} 1 & 3 \\ 2 & 2 \end{array}\right]\\ &\text { First, compute AB. }\\ &A B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right] \end{aligned}

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

\\(1, 0)(1, 2) = (1 $ \times $ 1) + (0 $ \times $ 2) \\$ \Rightarrow $ (1, 0)(1, 2) = 1 + 0 \\$ \Rightarrow $ (1, 0)(1, 2) = 1

\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 1 & \\ & \end{bmatrix}

Similarly, let us do the same for other elements.

\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & (1 \times 3)+(0 \times 0) \\ (0 \times 1)+(0 \times 2) & (0 \times 3)+(0 \times 0) \end{array}\right]

AB=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 1 &3 \\0 & 0\end{bmatrix}

Now, let us compute AC.

AC=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 2 \end{array}\right]

Multiply 1st row of matrix A by matching members of 1st column of matrix C, then finally sum them up.

\\(1, 0)(1, 2) = (1 $ \times $ 1) + (0 $ \times $ 2) \\$ \Rightarrow $ (1, 0)(1, 2) = 1 + 0 \\$ \Rightarrow $ (1, 0)(1, 2) = 1

\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 2 \end{array}\right]=\begin{bmatrix} 1 & \\ & \end{bmatrix}

Similarly, let us do the same for other elements.

\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 & (1 \times 3)+(0 \times 2) \\ (0 \times 1)+(0 \times 2) & (0 \times 3)+(0 \times 2) \end{array}\right]

AC=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 1 &3 \\0 & 0\end{bmatrix}

Clearly, AB = AC, but  B ≠ C.

Hence, we have found an example which fulfils the required criteria.

Posted by

infoexpert22

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