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Engineering

Can someone explain - Algebra - BITSAT-4

The root of the question is?

\begin{bmatrix} x &1&2 \end{bmatrix}\begin{bmatrix} 0 &1&1 \\1&0&1\\ 1&1&0\end{bmatrix}\begin{bmatrix} x \\-1\\1 \end{bmatrix}= 0

  • Option 1)

    \frac{1}{3}

  • Option 2)

    \frac{-1}{3}

  • Option 3)

    0

  • Option 4)

    1

 
Answers (2)
108 Views
A Aadil Khan

As learnt in

Multiplication of matrices -

-

 

 

\begin{bmatrix} x & 1 &2 \end{bmatrix} \begin{bmatrix} 0 &1 &1 \\ 1 &0 & 1\\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} x\\-1 \\1 \end{bmatrix}=0

\begin{bmatrix} 3 & x+2 & x+1 \end{bmatrix}\begin{bmatrix} x\\-1 \\1 \end{bmatrix}=0

3x-x-2+x+1=0

x=\frac{1}{3}

 


Option 1)

\frac{1}{3}

This option is correct.

Option 2)

\frac{-1}{3}

This option is incorrect.

Option 3)

0

This option is incorrect.

Option 4)

1

This option is incorrect.

A Aadil Khan

As learnt in

Multiplication of matrices -

-

 

 

\begin{bmatrix} x & 1 &2 \end{bmatrix} \begin{bmatrix} 0 &1 &1 \\ 1 &0 & 1\\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} x\\-1 \\1 \end{bmatrix}=0

\begin{bmatrix} 3 & x+2 & x+1 \end{bmatrix}\begin{bmatrix} x\\-1 \\1 \end{bmatrix}=0

3x-x-2+x+1=0

x=\frac{1}{3}

 


Option 1)

\frac{1}{3}

This option is correct.

Option 2)

\frac{-1}{3}

This option is incorrect.

Option 3)

0

This option is incorrect.

Option 4)

1

This option is incorrect.

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