To find the value of \(\displaystyle n\) when,

\(\displaystyle 6\times 7\equiv n\ (\text{mod } 5)\) first multiply six and seven together.

\(\displaystyle 6\times 7=42\)

Now, recall that mod means the remainder after division occurs.

In this case

\(\displaystyle 6\times 7\equiv n\ (\text{mod } 5)\)

       \(\displaystyle 8\)

\(\displaystyle 5)\overline{42\ }\)

 \(\displaystyle -40\)

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       \(\displaystyle 2\)

Therefore, the remainder is two.

\(\displaystyle n=2\)

This question is testing the matrix operation. Remember to use order of operations and perform the algebraic operation that is inside the parentheses first.

\(\displaystyle (\%\ \#\ \%)\ \$\ \%\)

First look at the # table.

\(\displaystyle \begin{tabular}{ c|cc } \# & \%& * \\ \hline \% & *& \% \\ * & \% & * \\ \end{tabular}\)

Multiplying % by % results in *.

Now go to the $ table and multiply * by %.

\(\displaystyle \begin{tabular}{ c|cc } \$ & * & \% \\ \hline * & \%& * \\ \% & * & * \\ \end{tabular}\)

\(\displaystyle *\ \$\ \%=*\)

Therefore,

\(\displaystyle (\%\ \#\ \%)\ \$\ \%=*\)

To find the value of \(\displaystyle n\) when,

\(\displaystyle 3\times 9\equiv n\ (\text{mod } 4)\) first multiply three and nine together.

\(\displaystyle 3\times 9=27\)

Now, recall that mod means the remainder after division occurs.

In this case

\(\displaystyle 3\times 9\equiv n\ (\text{mod } 4)\)

       \(\displaystyle 6\)

\(\displaystyle 4)\overline{27\ }\)

 \(\displaystyle -24\)

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       \(\displaystyle 3\)

Therefore, the remainder is three.

\(\displaystyle n=3\)

The determinant of a two-by-two matrix

\(\displaystyle A= \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\)

can be found by evaluating the expression

\(\displaystyle \det A = a_{11}a_{22} - a_{12}a_{23}\)

Substitute the corresponding elements to get

\(\displaystyle \det A = 3(4)- (-1)(-2) = 12 - 2 = 10\).

One way to identify whether the matrix is consistent and independent is to form a matrix of its variable coefficients, and calculate its determinant. The matrix is

\(\displaystyle A = \begin{bmatrix} 1 & 8 \\ -4 & 32 \end{bmatrix}\)

The determinant of a two-by-two matrix

\(\displaystyle A= \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\)

can be found by evaluating the expression

\(\displaystyle \det A = a_{11}a_{22} - a_{12}a_{23}\)

Substitute the corresponding elements to get

\(\displaystyle \det A = 1(32)- 8(-4) = 32 -(-32) = 64\)

Since \(\displaystyle \det A \ne 0\),

it follows that the system is consistent and independent.

\(\displaystyle B^{-1}\), the inverse of a two-by-two matrix

 \(\displaystyle B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}\),

can be calculated as follows:

\(\displaystyle B^{-1} = \frac{1}{D} \begin{bmatrix} b_{22} & -b_{12} \\ -b_{21} & b_{11} \end{bmatrix}\),

where \(\displaystyle D = b_{11}b_{22} - b_{12} b_{21}\).

Setting each of the values accordingly,

\(\displaystyle D = 3(3)- 2(4) = 9 - 8 = 1\)

\(\displaystyle B^{-1} = \frac{1}{1} \begin{bmatrix} 3& -2 \\ -4 & 3 \end{bmatrix}\), or

\(\displaystyle B^{-1} = \begin{bmatrix} 3& -2 \\ -4 & 3 \end{bmatrix}\).

For the sum of two matrices to be defined, they must have the same number of rows and columns. \(\displaystyle A\) is a matrix with three columns; since, in this problem, \(\displaystyle I\) refers to the two-by-two identity matrix

\(\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\),

\(\displaystyle I\) has two columns. Since the number of columns differs, \(\displaystyle A+ I\) is undefined.

The product \(\displaystyle AB\) of two matrices \(\displaystyle A\) and \(\displaystyle B\), where \(\displaystyle A\) has \(\displaystyle m\) rows and \(\displaystyle n\) columns and \(\displaystyle B\) has \(\displaystyle n\) rows and \(\displaystyle p\) columns, is a matrix with \(\displaystyle m\) rows and \(\displaystyle p\) columns. It follows that \(\displaystyle AB\) must have the same number of rows as \(\displaystyle A\). Since \(\displaystyle AB\) has three rows, so does \(\displaystyle A\). Nothing can be inferred about the number of rows of \(\displaystyle B\).

The product of any scalar value and a matrix is the matrix formed when each entry in the matrix is multiplied by that scalar. Thus, if

\(\displaystyle K = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix}\)

then

\(\displaystyle 7K = \begin{bmatrix} 7 (1) & 7 ( -2 )& 7(3) \\ 7(-4) & 7(2) & 7(1) \\ 7(2) & 7(1) & 7(3) \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} 7& -14 & 21 \\ -28 & 14 & 7 \\ 14 & 7 & 21 \end{bmatrix}\)

This is not equal to the matrix

\(\displaystyle \begin{bmatrix} 7 & -14 & 21 \\ -4 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix}\),

since the entries in the second and third rows differ. The statement is false.

For two matrices to be equal, two conditions must hold:

1) The matrices must have the same dimension. This can be seen to be the case, since both matrices have two rows and one column.

2) All corresponding entries must be equal. For this to happen, it must hold that

\(\displaystyle x+y=14\)

\(\displaystyle x-y=25\)

This is a system of two equations in two variables, which can be solved as follows:

Add both sides of the equations:

\(\displaystyle \begin{matrix} x+y=14 \\ \underline{x-y=25} \\ 2x\; \; \; \; \; = 39 \end{matrix}\)

It follows that

\(\displaystyle x = 19.5\)

Substitute back:

\(\displaystyle x+y=14\)

\(\displaystyle 19.5+y=14\)

\(\displaystyle y = -5.5\)

Thus, there exists a solution to the system, and, consequently, to the original matrix equation. The statement is therefore false.