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Suppose matrix A = \(\begin{bmatrix} 7 & 2 & 3 \\ 4 & 8 & 1 \\ 5 & 12 & 4 \end{bmatrix}\) is multiplied by the identity matrix. What you will get ?

  • A.\(\begin{bmatrix} 7 & 2 & 3 \\ 4 & 8 & 1 \\ 5 & 12 & 4 \end{bmatrix}\)
  • B.\(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
  • C.\(\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)
  • D.\(\begin{bmatrix} 7 & 4 & 5 \\ 2 & 8 & 12 \\ 3 & 1 & 4 \end{bmatrix}\)

Answer: A

The product is the matrix A itself.

The identity matrix \(I\) has 1s on the diagonal and 0s elsewhere. It is the multiplicative identity for matrices: \(AI = IA = A\) for any square matrix \(A\) of the same order, just as multiplying a number by 1 leaves it unchanged.

The identity itself would result only if \(A\) were the identity. The zero matrix comes from multiplying by the zero matrix, and the last option is the transpose of \(A\), which multiplication by \(I\) does not produce.

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