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SAT II Math II : Mathematical Relationships

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Example Questions

Example Question #2 : Matrices

Define matrix $A = \begin{bmatrix} 4& 5\\ 6&7 \\ 8& 9 \end{bmatrix}$

For which of the following matrix values of is the expression defined?

Possible Answers:

$B = \begin{bmatrix} 3& -5& 7\\ -1& 4& -7 \end{bmatrix}$

$B = \begin{bmatrix} 3& -5\\ 7& -1\\4& -7 \end{bmatrix}$

$B = \begin{bmatrix} 3 \\ 7 \\4 \end{bmatrix}$

The expression is defined for all of the values of given in the other responses.

$B = \begin{bmatrix} 3 \end{bmatrix}$

Correct answer:

$B = \begin{bmatrix} 3& -5& 7\\ -1& 4& -7 \end{bmatrix}$

Explanation:

For the matrix product to be defined, itis necessary and sufficent for the number of columns in to be equal to the number of rows in .

$A = \begin{bmatrix} 4& 5\\ 6&7 \\ 8& 9 \end{bmatrix}$ has two columns. Of the choices, only

$B = \begin{bmatrix} 3& -5& 7\\ -1& 4& -7 \end{bmatrix}$ has two rows, making it the correct choice.

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Example Question #4 : Matrices

$A = \begin{bmatrix} x+8& 7\\ 5& x - 7 \end{bmatrix}$

$B = \begin{bmatrix} 4&x-6 \\ x-7 & 2 \end{bmatrix}$

Calculate: A - B

Possible Answers:

$\begin{bmatrix} x+4& - x+1\\ -x+12 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+4& - x+13\\ -x-2 & x - 5 \end{bmatrix}$

$\begin{bmatrix} x+4& - x+1\\ -x-2 & x - 5 \end{bmatrix}$

$\begin{bmatrix} x+4& - x+1\\ -x-2 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+4& - x+13\\ -x+12 & x - 9 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} x+4& - x+13\\ -x+12 & x - 9 \end{bmatrix}$

Explanation:

To subtract two matrices, subtract the elements in corresponding positions:

$A - B = \begin{bmatrix} x+8& 7\\ 5& x - 7 \end{bmatrix} - \begin{bmatrix} 4&x-6 \\ x-7 & 2 \end{bmatrix}$

$= \begin{bmatrix} x+8 - 4& 7- \left ( x-6 \right )\\ 5-\left ( x - 7 \right ) & x - 7 -2 \end{bmatrix}$

$= \begin{bmatrix} x+4& - x+13\\ -x+12 & x - 9 \end{bmatrix}$

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Example Question #1 : Matrices

Evaluate:

$\begin{vmatrix} y-3&5\\ y+3& 5 \end{vmatrix}$

Possible Answers:

$y^{2} - 34$

10y

$y^{2} + 16$

Correct answer:

Explanation:

The determinant of the matrix $\begin{bmatrix} A& B\\ C& D \end{bmatrix}$ is

$\begin{vmatrix} A& B\\ C& D \end{vmatrix} = AD - BC$ .

Substitute A = y-3, C = y + 3 , B = D = 5 :

$\begin{vmatrix} y-3&5\\ y+3& 5 \end{vmatrix}$

= (y-3) 5 - 5 (y+3)

=(5y - 15) - (5y+15)

=5y - 5y - 15-15

= -30

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Example Question #4 : Matrices

Give the determinant of the matrix $\begin{bmatrix} a&5 \\ a& 7 \end{bmatrix}$

Possible Answers:

$12-a^{2}$

-2a

12a

$a^{2}-12$

Correct answer:

Explanation:

The determinant of the matrix $\begin{bmatrix} A& B\\ C& D \end{bmatrix}$ is

$\begin{vmatrix} A& B\\ C& D \end{vmatrix} = AD - BC$ .

Substitute A = C = a , B = 5, D = 7 :

$\begin{vmatrix} a& 5\\ a& 7 \end{vmatrix} = a \cdot 7 - a \cdot 5 = 7a - 5a = 2a$

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Example Question #3 : Matrices

Multiply:

$\begin{bmatrix} x&5 \\ 6& x \end{bmatrix}\begin{bmatrix} 3\\ 4 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix}3x+20\\ 4x+18 \end{bmatrix}$

$\begin{bmatrix} 3x&15 \\ 24& 4 x \end{bmatrix}$

$\begin{bmatrix}3x+15\\ 4x+24\end{bmatrix}$

$\begin{bmatrix}3x+24 \\ 4x+15\end{bmatrix}$

$\begin{bmatrix} 3x&5 \\ 24& x \end{bmatrix}$

Correct answer:

$\begin{bmatrix}3x+20\\ 4x+18 \end{bmatrix}$

Explanation:

The product of a 2 x 2 matrix and a 2 x 1 matrix is a 2 x 1 matrix.

Multiply each row in the first matrix by the column matrix by multiplying elements in corresponding positions, then adding the products, as follows:

$\begin{bmatrix} x&5 \\ 6& x \end{bmatrix}\begin{bmatrix} 3\\ 4 \end{bmatrix}$

$= \begin{bmatrix} x \cdot 3 + 5 \cdot 4\\ 6 \cdot 3 + x \cdot 4 \end{bmatrix}$

$= \begin{bmatrix}3x+20\\ 4x+18 \end{bmatrix}$

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Example Question #1 : Matrices

Let $A = \begin{bmatrix} 3& 2\\ 4& 3\\ -5& 0 \end{bmatrix}$ .

Give $A^{-1}$ .

Possible Answers:

$A = \begin{bmatrix} -3&-4 &5 \\ -2 & -3 & 0 \end{bmatrix}$

$A = \begin{bmatrix} -2& -3\\ -3& -4\\ 0& 5 \end{bmatrix}$

$A = \begin{bmatrix} -3& -2\\ -4& -3\\ 5& 0 \end{bmatrix}$

$A^{-1}$ is not defined.

$A = \begin{bmatrix} 5&-4 &-3 \\ 0 & -3 & -2 \end{bmatrix}$

Correct answer:

$A^{-1}$ is not defined.

Explanation:

has three rows and two columns; since the number of rows is not equal to the number of columns, is not a square matrix, and, therefore, it does not have an inverse.

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Example Question #1 : Matrices

Define matrix $A = \begin{bmatrix} 4& 5\\ 6&7 \\ 8& 9 \end{bmatrix}$ .

For which of the following matrix values of is the expression A+B defined?

I: $B = \begin{bmatrix} 4& 5\\ 6&7 \\ 8& 9 \end{bmatrix}$

II: $B = \begin{bmatrix} 4& 5\\ 6&7 \end{bmatrix}$

III: $B = \begin{bmatrix} 3 & 4& 5\\ 5 & 6&7 \\ 7 & 8& 9 \end{bmatrix}$

Possible Answers:

I only

I and II only

II and II only

I and III only

I, II, and III

Correct answer:

I only

Explanation:

For the matrix sum A+B to be defined, it is necessary and sufficent for and to have the same number of rows and the same number of columns. has three rows and two columns; of the three choices, only (I) has the same dimensions.

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Example Question #7 : Matrices

Let $A = \begin{bmatrix} 5& -4\\ -2& -7 \end{bmatrix}$ and be the 2 x 2 identity matrix.

Let A - B = I .

Which of the following is equal to ?

Possible Answers:

$B = \begin{bmatrix} 4& -5 \\ -3& -8\end{bmatrix}$

$B = \begin{bmatrix} 4& -3 \\ -1& -8\end{bmatrix}$

$B = \begin{bmatrix} 3& -5 \\ -3& -9\end{bmatrix}$

$B = \begin{bmatrix} 5& -5 \\ -3& -7\end{bmatrix}$

$B = \begin{bmatrix} 4& -4 \\ -2& -8\end{bmatrix}$

Correct answer:

$B = \begin{bmatrix} 4& -4 \\ -2& -8\end{bmatrix}$

Explanation:

The 2 x 2 identity matrix is $I =\begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}$ .

A - B = I , or, equivalently,

B = A - I ,

$B = \begin{bmatrix} 5& -4\\ -2& -7 \end{bmatrix} - \begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}$

Subtract the elements in the corresponding positions:

$B = \begin{bmatrix} 5-1& -4-0\\ -2-0& -7 -1\end{bmatrix}$

$B = \begin{bmatrix} 4& -4 \\ -2& -8\end{bmatrix}$

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Example Question #81 : Mathematical Relationships

$A = \begin{bmatrix} x+8& 7\\ 5& x - 7 \end{bmatrix}$

$B = \begin{bmatrix} 4&x-6 \\ x-7 & 2 \end{bmatrix}$

Calculate: A+B

Possible Answers:

$\begin{bmatrix} x+12& x+1\\ x-12 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+12& x+1\\ x-2 & x - 5 \end{bmatrix}$

$\begin{bmatrix} x+12& x+1\\ x-2 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+12& x+1\\ x-12 & x - 9 \end{bmatrix}$

$\begin{bmatrix} x+12& x-13\\ x-12 & x - 9 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} x+12& x+1\\ x-2 & x - 5 \end{bmatrix}$

Explanation:

To add two matrices, add the elements in corresponding positions:

$A + B = \begin{bmatrix} x+8& 7\\ 5& x - 7 \end{bmatrix} + \begin{bmatrix} 4&x-6 \\ x-7 & 2 \end{bmatrix}$

$= \begin{bmatrix} x+8 + 4& 7+ \left ( x-6 \right )\\ 5+\left ( x - 7 \right ) & x - 7 +2 \end{bmatrix}$

$= \begin{bmatrix} x+12& x+1\\ x-2 & x - 5 \end{bmatrix}$

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Example Question #82 : Mathematical Relationships

Solve for :

$\begin{vmatrix} N + 1& 6 \\ 6& N - 1 \end{vmatrix} = 12$

Possible Answers:

N = -7 or N = 7

N = 12

N = 0

The equation has no solution.

$N = -\sqrt{47}$ or $N = \sqrt{47}$

Correct answer:

N = -7 or N = 7

Explanation:

The determinant of a matrix can be evaluated as follows:

$\begin{vmatrix} A&B \\ C&D \end{vmatrix} = AD - BC$

Therefore, the equation can be rewritten:

$\begin{vmatrix} N + 1& 6 \\ 6& N - 1 \end{vmatrix} = 12$

$\left ( N + 1 \right )\left (N - 1 \right ) - 6 \cdot 6= 12$

$N^{2} - 1 -3 6 = 12$

$N^{2} - 37= 12$

$N^{2} - 37+ 37 = 12 + 37$

$N^{2} = 49$

$N = \pm \sqrt{49} = \pm 7$

The solution set is

N = -7 or N = 7 .

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SAT II Math II : Mathematical Relationships

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #2 : Matrices

Example Question #4 : Matrices

Example Question #1 : Matrices

Example Question #4 : Matrices

Example Question #3 : Matrices

Example Question #1 : Matrices

Example Question #1 : Matrices

Example Question #7 : Matrices

Example Question #81 : Mathematical Relationships

Example Question #82 : Mathematical Relationships

All SAT II Math II Resources

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