Math, asked by preetishroff8, 5 hours ago

If A=[ 6 2 4 −4 ] and B=[ 1 2 −5 1 ]. Find a matrix X such that 2A+4B-5X=0.

Answers

Answered by MathHacker001
6

Correct Question :-

If \begin{gathered} \sf A = \begin{bmatrix} 6 & 2 \\ 5 &  - 4 \end{bmatrix} \end{gathered} and  \begin{gathered} \sf B = \begin{bmatrix} 1 & 2 \\  - 5 &  1 \end{bmatrix} \end{gathered} Find a matrix X such that 2A + 3B - 5x = 0

Solution :-

According to the question

→ 2A + 3B - 5x = 0

→ 5x = 2A + 3B

Put the values

 \small:\sf\longmapsto{5x =\begin{gathered} \sf 2 \begin{bmatrix} 6 & 2 \\ 5 &  - 4 \end{bmatrix} \end{gathered}   \begin{gathered} \sf +  3  \begin{bmatrix} 1 & 2 \\  - 5 & 1 \end{bmatrix} \end{gathered}}  \: \\  \\  \\  \small:\sf\longmapsto{5x =\begin{gathered} \sf  \begin{bmatrix} 12 & 4 \\ 10 &  - 8 \end{bmatrix} \end{gathered}   \begin{gathered} \sf +    \begin{bmatrix} 3 & 6 \\  - 15 & 3 \end{bmatrix} \end{gathered} } \\  \\  \\ \small:\sf\longmapsto{x  =   \frac{1}{5}\begin{gathered} \sf   \begin{bmatrix} 15 & 10 \\  - 5 &  - 5 \end{bmatrix} \end{gathered} } \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\  \\\small:\bf\longmapsto \red{x = \begin{gathered} \sf  \begin{bmatrix} 3 & 2 \\  - 1 &  - 1 \end{bmatrix} \end{gathered}}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

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Answered by mathdude500
3

\large\underline{\sf{Solution-}}

Given matrices are

\rm :\longmapsto\:A = \bigg[ \begin{matrix}6&2 \\ 4& - 4 \end{matrix} \bigg]

and

\rm :\longmapsto\:B = \bigg[ \begin{matrix}1&2 \\  - 5&1 \end{matrix} \bigg]

Now, it is given that

\red{\rm :\longmapsto\:2A + 4B - 5X = 0}

\rm :\longmapsto\:5X = 2A + 4B

\rm :\longmapsto\:5X = 2\bigg[ \begin{matrix}6&2 \\ 4& - 4 \end{matrix} \bigg] + 4\bigg[ \begin{matrix}1&2 \\  - 5&1 \end{matrix} \bigg]

\rm :\longmapsto\:5X = \bigg[ \begin{matrix}12&4 \\ 8& - 8 \end{matrix} \bigg] + \bigg[ \begin{matrix}4&8 \\  - 20&4 \end{matrix} \bigg]

\rm :\longmapsto\:5X = \bigg[ \begin{matrix}12 + 4&4 + 8 \\ 8 - 20& - 8 + 4 \end{matrix} \bigg]

\rm :\longmapsto\:5X = \bigg[ \begin{matrix}16&12 \\  - 12& - 4 \end{matrix} \bigg]

\rm :\longmapsto\:X =\dfrac{1}{5}  \bigg[ \begin{matrix}16&12 \\  - 12& - 4 \end{matrix} \bigg]

Additional Information :-

1. Matrix addition is possible only when order of both the matrices are same otherwise matrix addition is not defined.

2. Matrix subtraction is possible only when order of both the matrices are same otherwise matrix subtraction is not defined.

3. Matrix multiplication is defined when number of columns of pre multiplier is equals to number of rows of post multiplier, otherwise matrix multiplication is not defined.

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