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SAT II Math II : Matrices

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Example Question #11 : 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:

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

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

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

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

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

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 #12 : Matrices

Solve for :

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

Possible Answers:

N = -7 or N = 7

N = 0

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

N = 12

The equation has no solution.

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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Example Question #91 : Mathematical Relationships

Multiply:

$\begin{bmatrix} x& 8\\ -6& t \end{bmatrix}\begin{bmatrix} t& x \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} tx +6x\\xt-8t \end{bmatrix}$

$\begin{bmatrix} tx +8x&xt-6t \end{bmatrix}$

The matrices cannot be multiplied.

$\begin{bmatrix} tx +6x& xt-8t \end{bmatrix}$

$\begin{bmatrix} tx +8x\\xt-6t \end{bmatrix}$

Correct answer:

The matrices cannot be multiplied.

Explanation:

Two matrices can be multiplied if and only if the number of columns in the first matrix and the number of rows in the second are equal. The first matrix has two columns; the second matrix has one row. This violates the condition, so they cannot be multiplied in this order.

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Example Question #14 : Matrices

Evaluate:

$\begin{vmatrix} x-1& 3\\ 3& x+1 \end{vmatrix}$

Possible Answers:

-6x

$x^{2}-8$

$x^{2}-10$

$x^{2}+8$

Correct answer:

$x^{2}-10$

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 = x-1, B = C = 3,D = x+1 :

$\begin{vmatrix} x-1& 3\\ 3& x+1 \end{vmatrix}$

$= (x-1)(x+1) - 3 \cdot 3$

$=x^{2} -1 - 9$

$=x^{2} -10$

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Example Question #15 : Matrices

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

Give $A^{-1}$ .

Possible Answers:

$A^{-1} = \begin{bmatrix} -3&4\\2 & -2\frac{1}{2} \end{bmatrix}$

$A^{-1} = \begin{bmatrix} -2\frac{1}{2} &4\\2 & -3 \end{bmatrix}$

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

$A^{-1} = \begin{bmatrix} -3&-4\\-2 & -2\frac{1}{2} \end{bmatrix}$

$A^{-1} = \begin{bmatrix} -2\frac{1}{2} &-4\\-2 & -3 \end{bmatrix}$

Correct answer:

$A^{-1} = \begin{bmatrix} -3&4\\2 & -2\frac{1}{2} \end{bmatrix}$

Explanation:

The inverse of a 2 x 2 matrix $\begin{bmatrix} A& B\\ C& D \end{bmatrix}$ , if it exists, is the matrix

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

First, we need to establish that the inverse is defined, which it is if and only if the determinant $AD - BC \neq 0$ .

Set A = 5, B = 8, C = 4, D = 6 , and check:

$AD - BC = 5 \cdot 6 - 8 \cdot 4 =30 - 32 = -2$

The inverse exists.

The process: First, switch the upper-left and lower-right entries, and change the other two entries to their opposites:

$\begin{bmatrix} 5& 8\\ 4 & 6 \end{bmatrix} \rightarrow \begin{bmatrix} 6&- 8\\ -4 & 5 \end{bmatrix}$

Then divide the new matrix elementwise by the determinant of the original matrix, which is .

The inverse is

$\frac{\begin{bmatrix} 6&- 8\\ -4 & 5\end{bmatrix} }{-2}$

$= \begin{bmatrix} \frac{6}{-2}& \frac{-8}{-2}\\ \\ \frac{-4}{-2} & \frac{5}{-2} \end{bmatrix}$

$= \begin{bmatrix} -3&4\\2 & -2\frac{1}{2} \end{bmatrix}$

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Example Question #3 : Matrices

Simplify:

$\begin{bmatrix} 14&22\\ -3 &4 \end{bmatrix}+\begin{bmatrix} -3&12\\ 5 &-10 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 17&10\\ -8 &14 \end{bmatrix}$

$\begin{bmatrix} 11&34\\ 2 &-6 \end{bmatrix}$

$\begin{bmatrix} 45&-4 \end{bmatrix}$

$\begin{bmatrix} 17 & -5 \\ 3 & 1 \end{bmatrix}$

$\begin{bmatrix} 13\\ 28 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 11&34\\ 2 &-6 \end{bmatrix}$

Explanation:

Matrix addition is very easy! All that you need to do is add each correlative member to each other. Think of it like this:

$\begin{bmatrix} 14&22\\ -3 &4 \end{bmatrix}+\begin{bmatrix} -3&12\\ 5 &-10 \end{bmatrix}=\begin{bmatrix} 14-3&22+12\\ -3+5 &4-10 \end{bmatrix}$

Now, just simplify:

$\begin{bmatrix} 11&34\\ 2 &-6 \end{bmatrix}$

There is your answer!

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Example Question #4 : Matrices

Simplify:

$\begin{bmatrix} -5&7\\ 102 & -12 \end{bmatrix}+\begin{bmatrix} 3&107\\ -10 & -9 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 90\\ 93 \end{bmatrix}$

$\begin{bmatrix} 112\\ 71 \end{bmatrix}$

$\begin{bmatrix} -8&-100\\ 112 & -3 \end{bmatrix}$

$\begin{bmatrix} 112 & 71\end{bmatrix}$

$\begin{bmatrix} -2&114\\ 92 & -21 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -2&114\\ 92 & -21 \end{bmatrix}$

Explanation:

Matrix addition is really easy—don't overthink it! All you need to do is combine the two matrices in a one-to-one manner for each index:

$\begin{bmatrix} -5&7\\ 102 & -12 \end{bmatrix}+\begin{bmatrix} 3&107\\ -10 & -9 \end{bmatrix}=\begin{bmatrix} -5+3&7+107\\ 102-10 & -12-9 \end{bmatrix}$

Then, just simplify all of those simple additions and subtractions:

$\begin{bmatrix} -5+3&7+107\\ 102-10 & -12-9 \end{bmatrix}=\begin{bmatrix} -2&114\\ 92 & -21 \end{bmatrix}$

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Example Question #8 : Find The Product Of A Matrix And A Scalar

Evaluate: $-3\begin{bmatrix} 2& 3 & 4\\ 4&-5 & 10 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} -12&-6 \\ 15& -9\\ -30&-12 \end{bmatrix}$

$\begin{bmatrix} -6& -9& -12\\ -12& 15& -30 \end{bmatrix}$

$\begin{bmatrix} -1& 0& 1\\ -12& 15&-30 \end{bmatrix}$

$\begin{bmatrix} -12& -1\\ 15& 0\\ -30& 1 \end{bmatrix}$

$\begin{bmatrix} -18& 6& -42 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -6& -9& -12\\ -12& 15& -30 \end{bmatrix}$

Explanation:

This problem involves a scalar multiplication with a matrix. Simply distribute the negative three and multiply this value with every number in the 2 by 3 matrix. The rows and columns will not change.

$-3\begin{bmatrix} 2& 3 & 4\\ 4&-5 & 10 \end{bmatrix}=\begin{bmatrix} -6&-9 &-12 \\ -12& 15&-30 \end{bmatrix}$

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Example Question #9 : Find The Product Of A Matrix And A Scalar

Simplify:

$\begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix} + 4\begin{bmatrix} 5 & 3\\ 11 &22 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 49\\ 206\end{bmatrix}$

$\begin{bmatrix} 33 & 16\\ 115 & 91\end{bmatrix}$

$\begin{bmatrix} 49 & 206\end{bmatrix}$

$\begin{bmatrix} 31 & 93\\ 3 &21 \end{bmatrix}$

$\begin{bmatrix} 20 & 12\\ 44 &88 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 33 & 16\\ 115 & 91\end{bmatrix}$

Explanation:

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first:

$\begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix} + 4\begin{bmatrix} 5 & 3\\ 11 &22 \end{bmatrix}= \begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix}+\begin{bmatrix} 20 & 12\\ 44 &88 \end{bmatrix}$

The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

$\begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix} + \begin{bmatrix} 20 & 12\\ 44 &88 \end{bmatrix} = \begin{bmatrix} 13 +20 & 4+12\\ 71+44 &3 +88\end{bmatrix}$

$\begin{bmatrix} 13 +20 & 4+12\\ 71+44 &3 +88\end{bmatrix}= \begin{bmatrix} 33 & 16\\ 115 & 91\end{bmatrix}$

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Example Question #10 : Find The Product Of A Matrix And A Scalar

$5\begin{bmatrix} a \\ b \end{bmatrix} = 14\begin{bmatrix} 20 \\ 12 \end{bmatrix}$

What is a+b ?

Possible Answers:

$\frac{191}{4}$

$\frac{148}{5}$

$\frac{41}{5}$

$\frac{448}{5}$

Correct answer:

$\frac{448}{5}$

Explanation:

You can begin by treating this equation just like it was:

5x=14y

That is, you can divide both sides by :

$\begin{bmatrix} a \\ b \end{bmatrix} = \frac{14}{5}\begin{bmatrix} 20 \\ 12 \end{bmatrix}$

Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component:

$\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \frac{14}{5}*20 \\ \frac{14}{5}*12 \end{bmatrix}$

Then, simplify:

$\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 56 \\ \frac{168}{5} \end{bmatrix}$

Therefore, $a+b=56+\frac{168}{5}=\frac{280}{5}+\frac{168}{5}=\frac{448}{5}$

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SAT II Math II : Matrices

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

All SAT II Math II Resources

Example Questions

Example Question #11 : Matrices

Example Question #12 : Matrices

Example Question #91 : Mathematical Relationships

Example Question #14 : Matrices

Example Question #15 : Matrices

Example Question #3 : Matrices

Example Question #4 : Matrices

Example Question #8 : Find The Product Of A Matrix And A Scalar

Example Question #9 : Find The Product Of A Matrix And A Scalar

Example Question #10 : Find The Product Of A Matrix And A Scalar

All SAT II Math II Resources

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