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Exercise B · Q3

Q.Consider A=[134−2123−21]A = \begin{bmatrix} 1 & 3 & 4 \\ -2 & 1 & 2 \\ 3 & -2 & 1 \end{bmatrix}, verify that A.I=I.A=AA.I = I.A = A, where II is the identity matrix of order 3×33\times 3.

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Multiplying AA by the 3×33\times3 identity on either side returns AA itself.

The identity matrix I=[100010001]I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} satisfies AI=IA=AAI=IA=A for every square matrix AA of the same order.

  1. Given A=[134−2123−21]A=\begin{bmatrix}1&3&4\\-2&1&2\\3&-2&1\end{bmatrix} and I=[100010001]I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}.
  2. Compute A⋅IA\cdot I. Each entry (AI)ij=∑kaikIkj(AI)_{ij}=\sum_k a_{ik}I_{kj}; since Ikj=1I_{kj}=1 only when k=jk=j, (AI)ij=aij(AI)_{ij}=a_{ij}. For example, row 1:

[134] ⁣[100]=1,[134] ⁣[010]=3,[134] ⁣[001]=4\begin{bmatrix}1&3&4\end{bmatrix}\!\begin{bmatrix}1\\0\\0\end{bmatrix}=1,\quad \begin{bmatrix}1&3&4\end{bmatrix}\!\begin{bmatrix}0\\1\\0\end{bmatrix}=3,\quad \begin{bmatrix}1&3&4\end{bmatrix}\!\begin{bmatrix}0\\0\\1\end{bmatrix}=4

Doing this for all rows gives A⋅I=[134−2123−21]=AA\cdot I=\begin{bmatrix}1&3&4\\-2&1&2\\3&-2&1\end{bmatrix}=A.

3. Compute I⋅AI\cdot A. Here (IA)ij=∑kIikakj=aij(IA)_{ij}=\sum_k I_{ik}a_{kj}=a_{ij}, since Iik=1I_{ik}=1 only when i=ki=k. Hence

I⋅A=[134−2123−21]=AI\cdot A=\begin{bmatrix}1&3&4\\-2&1&2\\3&-2&1\end{bmatrix}=A

  1. Therefore A⋅I=I⋅A=AA\cdot I=I\cdot A=A, verifying II is the multiplicative identity.
✓Final answer

A⋅I=I⋅A=A=[134−2123−21]A\cdot I=I\cdot A=A=\begin{bmatrix}1&3&4\\-2&1&2\\3&-2&1\end{bmatrix}.

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