Answer
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Hint:Start by computing the values for $(f - g)(x)$. Then one by one substitute the values of the terms $f(x)$ and $g(x)$. Then further open the brackets and combine all the like together. Finally evaluate the value of the function $(f - g)(3)$.
Complete step by step answer:
First we will start off by mentioning both the equations separately
$7x - 5y + z = 13$….. (1)
$\,\,\,\,\,19x = 8z = 10$ …… (2)
Now here, we will split equation (2) into two equations.
$7x - 5y + z = 13$….. (1)
$\,\,\,\,\,19x\,\,\,\,\,\,\,\, = 10$ ….... (2)
$ + 8x = 10$…… (3)
Now we will write the augmented matrix.
\[\left[ {\begin{array}{*{20}{c}}
7&{ - 5}&{1|\,\,\,13} \\
{19}&0&{0|\,\,\,10} \\
0&0&{8|\,\,\,10}
\end{array}} \right]\]
We will start by applying the operation ${R_3} \div 8$ and ${R_2} \div 19$.
Hence, the matrix will become,
\[\left[ {\begin{array}{*{20}{c}}
7&{ - 5}&{1|\,\,\,13} \\
1&0&{0|\,\,\,\dfrac{{10}}{{19}}} \\
0&0&{1|\,\,\,\dfrac{{10}}{8}}
\end{array}} \right]\]
We will now apply the operation ${R_1} - 7{R_2}$.
Hence, the matrix will become,
\[\left[ {\begin{array}{*{20}{c}}
0&{ - 5}&{1\,\,\,\,\,|\,\,\dfrac{{237}}{{19}}} \\
1&0&{0|\,\,\,\dfrac{{10}}{{19}}} \\
0&0&{1|\,\,\,\dfrac{{10}}{8}}
\end{array}} \right]\]
We will now apply the operation ${R_1} \div ( - 5)$.
Hence, the matrix will become,
\[\left[ {\begin{array}{*{20}{c}}
0&1&{\dfrac{{ - 1}}{5}\,\,\,\,\,|\,\,\dfrac{{ - 237}}{{95}}} \\
1&0&{0\,\,\,\,\,|\,\,\,\dfrac{{10}}{{19}}} \\
0&0&{1\,\,\,\,\,|\,\,\,\dfrac{{10}}{8}}
\end{array}} \right]\]
We will now apply the operation ${R_1} + \dfrac{1}{5}{R_3}$.
Hence, the matrix will become,
\[\left[ {\begin{array}{*{20}{c}}
0&1&{0\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{ - 667}}{{380}}} \\
1&0&{0\,\,\,\,\,|\,\,\,\dfrac{{10}}{{19}}} \\
0&0&{1\,\,\,\,\,|\,\,\,\dfrac{{10}}{8}}
\end{array}} \right]\]
\[\left[ {\begin{array}{*{20}{c}}
1&0&{0\,\,\,|\,\,\,\dfrac{{10}}{{19}}} \\
0&1&{\,\,\,0\,\,\,|\,\,\,\dfrac{{ - 667}}{{380}}} \\
0&0&{1\,\,\,|\,\,\,\dfrac{5}{4}}
\end{array}} \right]\]
Additional Information:
An augmented matrix is a matrix obtained by appending the columns of two given matrices, usually for the purpose of performing the same elementary row operations on each of the given matrices. For a given number of unknowns, the number of solutions to a system of linear equations depends only on the rank of the matrix representing the system and the rank of the corresponding augmented matrix. An augmented matrix may also be used to find the inverse of a matrix by combining it with the identity matrix.
Note:While solving such type of matrix related questions very carefully. Split the equations step by step. Now apply different conditions and arithmetic operations in order to convert the set of linear equations to augmented matrix.
Complete step by step answer:
First we will start off by mentioning both the equations separately
$7x - 5y + z = 13$….. (1)
$\,\,\,\,\,19x = 8z = 10$ …… (2)
Now here, we will split equation (2) into two equations.
$7x - 5y + z = 13$….. (1)
$\,\,\,\,\,19x\,\,\,\,\,\,\,\, = 10$ ….... (2)
$ + 8x = 10$…… (3)
Now we will write the augmented matrix.
\[\left[ {\begin{array}{*{20}{c}}
7&{ - 5}&{1|\,\,\,13} \\
{19}&0&{0|\,\,\,10} \\
0&0&{8|\,\,\,10}
\end{array}} \right]\]
We will start by applying the operation ${R_3} \div 8$ and ${R_2} \div 19$.
Hence, the matrix will become,
\[\left[ {\begin{array}{*{20}{c}}
7&{ - 5}&{1|\,\,\,13} \\
1&0&{0|\,\,\,\dfrac{{10}}{{19}}} \\
0&0&{1|\,\,\,\dfrac{{10}}{8}}
\end{array}} \right]\]
We will now apply the operation ${R_1} - 7{R_2}$.
Hence, the matrix will become,
\[\left[ {\begin{array}{*{20}{c}}
0&{ - 5}&{1\,\,\,\,\,|\,\,\dfrac{{237}}{{19}}} \\
1&0&{0|\,\,\,\dfrac{{10}}{{19}}} \\
0&0&{1|\,\,\,\dfrac{{10}}{8}}
\end{array}} \right]\]
We will now apply the operation ${R_1} \div ( - 5)$.
Hence, the matrix will become,
\[\left[ {\begin{array}{*{20}{c}}
0&1&{\dfrac{{ - 1}}{5}\,\,\,\,\,|\,\,\dfrac{{ - 237}}{{95}}} \\
1&0&{0\,\,\,\,\,|\,\,\,\dfrac{{10}}{{19}}} \\
0&0&{1\,\,\,\,\,|\,\,\,\dfrac{{10}}{8}}
\end{array}} \right]\]
We will now apply the operation ${R_1} + \dfrac{1}{5}{R_3}$.
Hence, the matrix will become,
\[\left[ {\begin{array}{*{20}{c}}
0&1&{0\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{ - 667}}{{380}}} \\
1&0&{0\,\,\,\,\,|\,\,\,\dfrac{{10}}{{19}}} \\
0&0&{1\,\,\,\,\,|\,\,\,\dfrac{{10}}{8}}
\end{array}} \right]\]
\[\left[ {\begin{array}{*{20}{c}}
1&0&{0\,\,\,|\,\,\,\dfrac{{10}}{{19}}} \\
0&1&{\,\,\,0\,\,\,|\,\,\,\dfrac{{ - 667}}{{380}}} \\
0&0&{1\,\,\,|\,\,\,\dfrac{5}{4}}
\end{array}} \right]\]
Additional Information:
An augmented matrix is a matrix obtained by appending the columns of two given matrices, usually for the purpose of performing the same elementary row operations on each of the given matrices. For a given number of unknowns, the number of solutions to a system of linear equations depends only on the rank of the matrix representing the system and the rank of the corresponding augmented matrix. An augmented matrix may also be used to find the inverse of a matrix by combining it with the identity matrix.
Note:While solving such type of matrix related questions very carefully. Split the equations step by step. Now apply different conditions and arithmetic operations in order to convert the set of linear equations to augmented matrix.
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