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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Show that: $ii) \qquad \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \: \neq \: \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} $

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Toolbox:
  • Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
  • If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B:
  • If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
  • If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
We need to prove that $LHS: \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \: \neq \: RHS: \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$.
$LHS: Multiplying \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}$:
$\Rightarrow LHS: \begin{bmatrix} 1\times(-1)+2\times 0+3\times 2 & 1\times 1+2(-1)+3\times 3 & 1\times 0+2\times 1+3\times 4 \\ 0\times(-1)+1\times 0+0\times 2 & 0\times 1+1\times (-1)+0\times 3 & 0\times 0+1\times 1+0\times 4 \\ 1\times (-1)+1\times 0+0\times 2 & 1\times 1+1\times (-1)+0\times 4 & 1\times 0+1\times 1+0\times 4 \end{bmatrix} $
$\Rightarrow LHS: \begin{bmatrix}-1+0+6 & 1-2+9 & 0+2+12\\0+0+0 & 0-1+0 & 0+1+0\\-1+0+0 & 1-1+0 & 0+1+0\end{bmatrix}$
$\Rightarrow LHS: \begin{bmatrix}5 & 8 & 14\\0 & -1 & 1\\-1& 0 &1 \end{bmatrix}$
$RHS: Multiplying$ $ \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} $:
$\Rightarrow RHS:$ $ \begin{bmatrix} -1\times1+1\times 0+0\times 1 & -1\times 2+1\times 1+0\times 1 & -1\times 3+1\times 0+0\times 0 \\ 0\times(1)+-1\times 0+1\times 1 & 0\times 2+-1\times (1)+1\times 1 & 0\times 3+(-1)\times 0+1\times 0 \\ 2\times (1)+2\times 0+2\times 2 & 2\times 2+3\times (1)+4\times 1 & 2\times 3+3\times 0+4\times 0 \end{bmatrix} $
$\Rightarrow RHS: \begin{bmatrix}-1+0+0 & -2+1+0 & -3+0+0\\0-0+1 & 0-1+1 & 0+0+0\\2+0+4 & 4+3+4 & 6+0+0\end{bmatrix}$
$\Rightarrow RHS: \begin{bmatrix}-1 & -1 & -3\\1& 0 & 0\\6& 11 &6 \end{bmatrix}$
We can see that $LHS \neq RHS$
answered Feb 27, 2013 by balaji.thirumalai
 

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