Matrices - SAT Subject Test in Math II

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Question

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

Answer

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 , :

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

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Question

Multiply:

$\begin{bmatrix} x&7 \ y& 3 \end{bmatrix}\begin{bmatrix} 2\ 7 \end{bmatrix}$

Answer

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:

$\begin{bmatrix} x&7 \ y& 3 \end{bmatrix}\begin{bmatrix} 2\ 7 \end{bmatrix}$

$=\begin{bmatrix} x \cdot 2+ 7 \cdot 7 \ y \cdot 2 + 3 \cdot 7 \end{bmatrix}$ \

$=\begin{bmatrix} 2x + 49 \ 2 y+ 21 \end{bmatrix}$

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Question

Multiply:

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

Answer

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:

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

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

$= \begin{bmatrix} 6x+5y \ 3y-24 \end{bmatrix}$

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Question

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?

Answer

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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Question

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

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

Calculate:

Answer

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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Question

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

Give $A^{-1}$ .

Answer

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 .

Set , 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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Question

Evaluate:

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

Answer

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 :

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

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Question

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

Give $A^{-1}$ .

Answer

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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Question

Multiply:

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

Answer

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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Question

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?

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

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

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

Answer

For the matrix sum 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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Question

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

Let .

Which of the following is equal to ?

Answer

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

, or, equivalently,

,

so

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

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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Question

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

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

Calculate:

Answer

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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Question

Evaluate:

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

Answer

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 :

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

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

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

$=x^{2} -10$

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Question

Solve for :

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

Answer

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

or .

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Question

Multiply:

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

Answer

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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Question

Given the following matrices, what is the product of and ?

$\begin{bmatrix} 1&-4 \ 2&x \end{bmatrix} - \begin{bmatrix} y&4 \ 8&1 \end{bmatrix} = \begin{bmatrix} -8&-8 \ -6&4 \end{bmatrix}$

Answer

When subtracting matrices, you want to subtract each corresponding cell.

$\begin{bmatrix} 1-y&-4-4 \ 2-8&x-1 \end{bmatrix}= \begin{bmatrix} -8&-8 \ -6&4 \end{bmatrix}$

Now solve for and

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Question

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

Answer

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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Question

Simplify:

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

Answer

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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Question

Simplify:

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

Answer

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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Question

Simplify:

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

Answer

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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