Precalculus Quiz: Zero And Identity Matrices And Determinants

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Zero And Identity Matrices And DeterminantsQuestion 1 of 20

Let $A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . If $C = A \cdot B - B \cdot A$ , what is the determinant of matrix $C$ ?

0

10

-10

1

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Precalculus Quiz

Precalculus Quiz: Zero And Identity Matrices And Determinants

Practice Zero And Identity Matrices And Determinants in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Zero And Identity Matrices And Determinants, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let $A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . If $C = A \cdot B - B \cdot A$ , what is the determinant of matrix $C$ ?

$0$ (correct answer)
$10$
$-10$
$1$

Explanation: Since

B

is the identity matrix,

A \cdot B = A

and

B \cdot A = A

. Therefore,

C = A - A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}

, which is the zero matrix. The determinant of the zero matrix is

0

. Choice B is the determinant of

A

(

2 \cdot 3 - (-1) \cdot 4 = 10

). Choice C reflects a sign error in computing the determinant of

A

. Choice D assumes

C

equals the identity matrix.

Question 2

Matrix $M$ is a $3 \times 3$ matrix such that $M \cdot I_3 = M$ and $\det(M) = 0$ . Which statement about matrix $M$ must be true?

Matrix $M$ has a multiplicative inverse and can be used to solve linear systems uniquely.
Matrix $M$ has no multiplicative inverse and represents a transformation that collapses dimension. (correct answer)
Matrix $M$ equals the zero matrix since its determinant is zero.
Matrix $M$ equals the identity matrix since it satisfies the identity property.

Explanation: Since

\det(M) = 0

, matrix

M

has no multiplicative inverse and represents a transformation that reduces dimension (collapses space). The condition

M \cdot I_3 = M

is satisfied by any

3 \times 3

matrix since multiplying by the identity always returns the original matrix. Choice A contradicts the fundamental theorem about determinants and inverses. Choice C is incorrect because many non-zero matrices have zero determinant. Choice D misinterprets the identity property.

Question 3

Consider the matrix $W = \begin{pmatrix} p & 2 \\ -1 & q \end{pmatrix}$ . If $W \cdot 0_{2 \times 2} = 0_{2 \times 2}$ and $\det(W) = 7$ , what is the value of $p + q$ ?

The value cannot be determined since multiplying by zero gives no information about $W$ .
$p + q = 7$ since the determinant equals the trace when multiplied by zero.
$p + q$ can be any value since the zero matrix property is satisfied by all matrices. (correct answer)
$p + q = \sqrt{7}$ since this balances the determinant condition with the zero matrix property.

Explanation: The condition

W \cdot 0_{2 \times 2} = 0_{2 \times 2}

is always true for any matrix

W

since any matrix multiplied by the zero matrix equals the zero matrix. This gives us no constraint on

p

and

q

. The determinant condition

\det(W) = pq - (2)(-1) = pq + 2 = 7

gives us

pq = 5

, but this doesn't uniquely determine

p + q

. For example,

p = 5, q = 1

gives

p + q = 6

, while

p = \frac{5}{2}, q = 2

gives

p + q = \frac{9}{2}

. Choices A, B, and D all provide incorrect reasoning about the relationship between these conditions.

Question 4

Given the $2\times2$ zero matrix $O=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ , what is $\det(O)$ ?

$1$
$0$ (correct answer)
$-1$
$2$

Explanation: This question tests understanding of zero matrices and their determinants. The zero matrix O has all entries equal to zero and serves as the additive identity for matrices: A + O = O + A = A for any matrix A of the same dimensions, analogous to adding 0 to a number. The determinant of the zero matrix is always 0, which can be verified using the formula det([0 0; 0 0]) = (0)(0) - (0)(0) = 0 - 0 = 0. This means the zero matrix is singular (has no inverse), which makes sense since no matrix multiplied by O can give the identity matrix. Choice B is correct because det(O) = 0 for any zero matrix, reflecting that it's always singular (non-invertible). Choice A incorrectly gives 1, confusing the zero matrix with the identity matrix. Key insight: det(O) = 0 means the zero matrix is never invertible, while det(I) = 1 means the identity matrix is always invertible—these are fundamental properties of these special matrices.

Question 5

Let $I_3=\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$ be the $3\times 3$ identity matrix. What is $\det(I_3)$ ?

$0$
$3$
$-1$
$1$ (correct answer)

A \cdot I = I \cdot A = A

, just as multiplying a number by 1 leaves it unchanged. The determinant has the property that

det(AB) = det(A) \cdot det(B)

for square matrices A and B,

det(I) = 1

for the identity matrix, and

det(O) = 0

for the zero matrix, with the non-zero determinant criterion determining invertibility. For the 3×3 identity matrix

I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

det(I_3) = 1

. Choice D is correct because it correctly states the identity property. Choice A claims

det(I) = 0

, but the determinant of the identity matrix is always 1. Determinant properties:

det(I) = 1

(identity always invertible),

det(O) = 0

(zero matrix never invertible), and

det(AB) = det(A) \cdot det(B)

(determinants multiply, not add). Remember the roles of special matrices: the identity matrix I is the multiplicative identity (

AI = IA = A

), while the zero matrix O is the additive identity (

A + O = A

) and absorbs under multiplication (

A \cdot O = O

Question 6

Let $P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $ad - bc = 5$ . If $Q = P + kI_2$ for some scalar $k$ , what is $\det(Q)$ in terms of $a$ , $d$ , and $k$ ?

$5 + k$
$(a + k)(d + k) - bc$ (correct answer)
$ad - bc + k^2$
$5k + k^2$

Explanation:

$$. Therefore, $$\det(Q) = (a+k)(d+k) - bc$$. Choice A incorrectly adds $$k$$ to the original determinant. Choice C incorrectly adds $$k^2$$ to $$ad - bc$$. Choice D substitutes the given value $$ad - bc = 5$$ but uses an incorrect formula for the determinant.

Question 7

Let $A$ be a $3\times 3$ matrix and let $I_3=\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$ . For matrix $A$ , which statement correctly describes the identity matrix property under multiplication?

$A\cdot I_3 = I_3$
$A\cdot I_3 = O_3$
$A\cdot I_3 = A$ (correct answer)
$A\cdot I_3 \ne A$ in general because matrix multiplication is not commutative.

Explanation: This question tests understanding of identity matrices and their properties. The identity matrix I is a square matrix with 1s on the main diagonal and 0s elsewhere, serving as the multiplicative identity: for any matrix A of compatible dimensions, A·I = I·A = A, just as multiplying a number by 1 leaves it unchanged. The identity matrix for 3×3 is I = [[1,0,0],[0,1,0],[0,0,1]]. When we compute A·I, the result is A, confirming the identity property. Choice C is correct because it correctly states the identity property. Choice A states that A·I = I, but the identity property is A·I = A—multiplying by the identity matrix leaves A unchanged, not replaces it with I. Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 8

Let $A$ be any $2\times2$ matrix and let $O$ be the $2\times2$ zero matrix $O=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ . What property does the zero matrix have under multiplication, i.e., what is $A\cdot O$ ?

$A\cdot O=O$ (correct answer)
$A\cdot O=A$
$A\cdot O=I_2$
$A\cdot O=\det(A)$

Explanation: This question tests understanding of zero matrices and their properties under multiplication. The zero matrix O has all entries equal to zero and serves as the additive identity for matrices: A + O = O + A = A for any matrix A of the same dimensions, analogous to adding 0 to a number. For multiplication, when any matrix A is multiplied by the zero matrix O, the result is always the zero matrix: A·O = O. This is because each entry in the product is computed as a sum of products, and since all entries of O are zero, every sum equals zero. Choice A is correct because A·O = O for any matrix A—the zero matrix 'absorbs' under multiplication. Choice B incorrectly states A·O = A, confusing the multiplication property with the addition property (A + O = A). Remember that while the zero matrix is the additive identity (A + O = A), it acts as an absorbing element under multiplication (A·O = O·A = O), similar to how multiplying any number by 0 gives 0.

Question 9

Matrix $T$ represents a linear transformation that maps the unit square to a parallelogram with area $6$ . If $S = T + 0 \cdot I_2$ , what is the relationship between $\det(S)$ and the invertibility of $S$ ?

$\det(S) = 6$ and $S$ is invertible since the transformation preserves shape.
$\det(S) = 1$ and $S$ is invertible since the identity component ensures invertibility.
$\det(S) = 0$ and $S$ is not invertible since adding zero changes invertibility.
$\det(S) = 6$ and $S$ is invertible since the determinant is nonzero. (correct answer)

Explanation: When you encounter questions about linear transformations and matrices, focus on how the determinant connects geometric properties to algebraic ones. The key insight is that a matrix's determinant tells you both the area scaling factor of the transformation and whether the matrix is invertible. Since matrix

T

maps the unit square (area = 1) to a parallelogram with area 6, the transformation scales areas by a factor of 6. This means

\det(T) = 6

. Now,

S = T + 0 \cdot I_2

simplifies to

S = T + 0 = T

, so

S

and

T

are identical matrices. Therefore,

\det(S) = \det(T) = 6

. Since the determinant is nonzero,

S

is invertible. Choice A incorrectly claims the transformation "preserves shape." Linear transformations that map squares to parallelograms change shape unless the parallelogram is also a square. The reasoning conflates area scaling with shape preservation. Choice B contains a fundamental error: it states

\det(S) = 1

, but we know from the area scaling that the determinant must be 6. The phrase "identity component ensures invertibility" is meaningless since we're adding

0 \cdot I_2 = 0

. Choice C makes two mistakes: it claims

\det(S) = 0

(contradicting the given area information) and suggests that "adding zero changes invertibility," which is nonsensical since adding zero doesn't change anything. Remember: the determinant of a linear transformation matrix equals the area scaling factor. If this factor is nonzero, the matrix is invertible. Always check what operations are actually being performed on the matrix—adding zero matrices changes nothing.

Question 10

Let $A=\begin{bmatrix}5&2\\-1&3\end{bmatrix}$ and let $O$ be the $2\times 2$ zero matrix $O=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ . For matrix $A$ , what is $A+O$ ?

$\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\begin{bmatrix}5&2\\-1&3\end{bmatrix}$ (correct answer)
$\begin{bmatrix}5&0\\0&3\end{bmatrix}$
$\begin{bmatrix}5&2\\1&3\end{bmatrix}$

Explanation: This question tests understanding of zero matrices and their properties. The zero matrix O has all entries equal to zero and serves as the additive identity for matrices: A + O = O + A = A for any matrix A of the same dimensions, analogous to adding 0 to a number. When we add A + O = [5 2; -1 3] + [0 0; 0 0], we add corresponding entries: (5+0)=5, (2+0)=2, (-1+0)=-1, and (3+0)=3, giving us [5 2; -1 3], which equals A. Choice B is correct because it properly shows that A + O = A = [5 2; -1 3], confirming that O is the additive identity. Choice A incorrectly gives the zero matrix as the answer, claiming A + O = O, but the zero matrix is the additive identity, so A + O = A (adding zero doesn't change A). Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 11

Let $A=\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ be a $2\times 2$ matrix and let $I_2=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$ be the $2\times 2$ identity matrix. For matrix $A$ , what is $A\cdot I_2$ ?

$\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$
$\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}$
$\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ (correct answer)
$\begin{bmatrix}2 & 1\\ -3 & 4\end{bmatrix}$

Explanation: This question tests understanding of identity matrices and their properties. The identity matrix I is a square matrix with 1s on the main diagonal and 0s elsewhere, serving as the multiplicative identity: for any matrix A of compatible dimensions, A·I = I·A = A, just as multiplying a number by 1 leaves it unchanged. The identity matrix for 2×2 is I = [[1,0],[0,1]]. When we compute A·I = [[2,-1],[3,4]] · [[1,0],[0,1]] = [[21 + (-1)0, 20 + (-1)1], [31 + 40, 30 + 41]] = [[2,-1],[3,4]], we get back A, confirming the identity property. Choice C is correct because it correctly states the identity property by showing A·I2 equals A. Choice A states that A·I = I, but the identity property is A·I = A—multiplying by the identity matrix leaves A unchanged, not replaces it with I. Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 12

Let $A=\begin{bmatrix}2&3\\1&4\end{bmatrix}$ (a $2\times2$ matrix). Based on the determinant formula $\det\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=ad-bc$ , what is the determinant of matrix $A$ ?

$-5$
$11$
$5$ (correct answer)
$8$

Explanation: This question tests understanding of determinant calculation for 2×2 matrices. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix A = [2 3; 1 4], we compute det(A) = ad - bc = (2)(4) - (3)(1) = 8 - 3 = 5. Choice C is correct because it properly calculates det(A) = ad - bc = 8 - 3 = 5. Choice A gives -5, which would result from reversing the subtraction (3 - 8 instead of 8 - 3), while choice B incorrectly adds the products (8 + 3 = 11) instead of subtracting. Key to determinants: for a 2×2 matrix [a b; c d], always use det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det = 0 means the matrix has no inverse (singular) while det ≠ 0 means it's invertible.

Question 13

Let $A$ be any $2\times2$ matrix and let $O_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ . What property does the zero matrix have under multiplication (with compatible dimensions), i.e., what is $A\cdot O_2$ ?

$A\cdot O_2=A$
$A\cdot O_2=I_2$
$A\cdot O_2=O_2$ (correct answer)
$A\cdot O_2=\det(A)$

Explanation: This question tests understanding of zero matrices and their properties. The zero matrix O has all entries equal to zero and serves as the additive identity for matrices: A + O = O + A = A for any matrix A of the same dimensions, analogous to adding 0 to a number. For multiplication, A·O = [[show result is O]], showing that the zero matrix 'absorbs' under multiplication. Choice C is correct because it correctly states the zero matrix property under multiplication. Choice A claims A·O = A, but the zero matrix absorbs under multiplication, so A·O = O. Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 14

Consider matrices $A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} x & 0 \\ 0 & y \end{pmatrix}$ . If $AB = BA$ and $\det(B) = 0$ , which values of $x$ and $y$ satisfy these conditions?

$x = 3, y = 4$ since these make $B$ diagonal with $A$ 's eigenvalues.
$x = 0, y = 1$ since this makes $\det(B) = 0$ and satisfies commutativity.
$x = 1, y = 0$ since this makes $\det(B) = 0$ and satisfies commutativity.
$x = 0, y = 0$ since the zero matrix commutes with all matrices. (correct answer)

Explanation: When you encounter matrix problems involving commutativity (

AB = BA

) and determinant conditions, you need to systematically check both requirements. Let's verify the correct answer by testing

AB = BA

for each option. For matrices to commute, their product must be the same regardless of order. With

A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}

and $$B = \begin{pmatrix} x & 0 \ 0 & y \end{pmatrix}

$$AB = \begin{pmatrix} 3x & 2y \\ x & 4y \end{pmatrix} $$ and $$BA = \begin{pmatrix} 3x & 2x \\ y & 4y \end{pmatrix}

For

AB = BA

, we need

2y = 2x

and

x = y

, which means

x = y

. Now checking the determinant condition

\det(B) = xy = 0

: since

x = y

, we need

x^2 = 0

, so

x = y = 0

. Option A gives

x = 3, y = 4

, but

\det(B) = 12 \neq 0

, violating the determinant condition. Option B has

x = 0, y = 1

, which satisfies

\det(B) = 0

, but from our commutativity analysis, we need

x = y

, and

0 \neq 1

. Option C has

x = 1, y = 0

, which satisfies

\det(B) = 0

, but again

1 \neq 0

, violating commutativity. Option D gives

x = 0, y = 0

, satisfying both

x = y

(commutativity) and $$\det(B) = 0$. Study tip: When checking matrix commutativity, compute both products explicitly and set corresponding entries equal. Don't assume diagonal matrices automatically commute with all matrices—they only commute when they're scalar multiples of the identity matrix.

Question 15

Let $A=\begin{bmatrix}5 & 0 & -2\\ 1 & 3 & 4\end{bmatrix}$ be a $2\times 3$ matrix and let $O$ be the $2\times 3$ zero matrix $\begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$ . For matrix $A$ , what is $A+O$ ?

$\begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$
$\begin{bmatrix}5 & 0 & -2\\ 1 & 3 & 4\end{bmatrix}$ (correct answer)
$\begin{bmatrix}5 & 0\\ 1 & 3\\ -2 & 4\end{bmatrix}$
$\begin{bmatrix}5 & 0 & 2\\ 1 & 3 & -4\end{bmatrix}$

Explanation: This question tests understanding of zero matrices and their properties. The zero matrix O has all entries equal to zero and serves as the additive identity for matrices: A + O = O + A = A for any matrix A of the same dimensions, analogous to adding 0 to a number. The zero matrix O = [[0,0,0],[0,0,0]]. When we add A + O = [[5+0,0+0,-2+0],[1+0,3+0,4+0]] = [[5,0,-2],[1,3,4]], each entry of A is added to 0, giving result equals A, confirming that O is the additive identity. For multiplication, A·O would result in the zero matrix if dimensions allow, showing that the zero matrix 'absorbs' under multiplication. Choice B is correct because it correctly states zero matrix property under addition. Choice A claims A + O = O, but the zero matrix is the additive identity, so A + O = A (adding zero doesn't change A). Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 16

Let $A=\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ and let $I_2=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$ be the $2\times 2$ identity matrix. For matrix $A$ , what is $A\cdot I_2$ (matrix multiplication)?

$\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$
$\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}$
$\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ (correct answer)
$\begin{bmatrix}4 & -2\\ 6 & 8\end{bmatrix}$

Explanation: This question tests understanding of identity matrices and their properties. The identity matrix I is a square matrix with 1s on the main diagonal and 0s elsewhere, serving as the multiplicative identity: for any matrix A of compatible dimensions, A·I = I·A = A, just as multiplying a number by 1 leaves it unchanged. For matrix A = [[2, -1], [3, 4]] and I2 = [[1, 0], [0, 1]], when we compute A·I2, the result is [[(21 + -10), (20 + -11)], [(31 + 40), (30 + 41)]] = [[2, -1], [3, 4]], confirming the identity property. Choice C is correct because it correctly states the identity property by showing A·I2 = A. Choice A states that A·I2 = I2, but the identity property is A·I2 = A—multiplying by the identity matrix leaves A unchanged, not replaces it with I2. Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 17

Let $A=\begin{bmatrix}5 & 2\\ -1 & 3\end{bmatrix}$ and let $O_2=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}$ be the $2\times 2$ zero matrix. For matrix $A$ , what is $A+O_2$ (matrix addition)?

$\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}$
$\begin{bmatrix}5 & 2\\ -1 & 3\end{bmatrix}$ (correct answer)
$\begin{bmatrix}5 & 0\\ 0 & 3\end{bmatrix}$
$\begin{bmatrix}10 & 4\\ -2 & 6\end{bmatrix}$

Explanation: This question tests understanding of zero matrices and their properties. The zero matrix O has all entries equal to zero and serves as the additive identity for matrices: A + O = O + A = A for any matrix A of the same dimensions, analogous to adding 0 to a number. For matrix A = [[5, 2], [-1, 3]] and O2 = [[0, 0], [0, 0]], when we add A + O2, each entry of A is added to 0, giving [[5+0, 2+0], [-1+0, 3+0]] = [[5, 2], [-1, 3]], confirming that O is the additive identity. Choice B is correct because it correctly states the zero matrix property by showing A + O2 = A. Choice A claims A + O2 = O2, but the zero matrix is the additive identity, so A + O2 = A (adding zero doesn't change A). Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 18

Given $M=\begin{bmatrix}2 & 4\\ 1 & 2\end{bmatrix}$ , does matrix $M$ have an inverse? Why or why not? (Use $\det\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=ad-bc$ .)

Yes, because $\det(M)=2\cdot2+4\cdot1=8\neq0$ .
No, because $\det(M)=2\cdot2-4\cdot1=0$ . (correct answer)
Yes, because $\det(M)=4\cdot1-2\cdot2=0$ .
No, because $\det(M)=2\cdot2-4\cdot1=4\neq0$ .

Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix M = [[2,4],[1,2]], we compute det(M) = 22 - 41 = 4 - 4 = 0. Since det(M) = 0, and this is equal to zero, matrix M is singular and has no inverse. Choice B is correct because it correctly applies the invertibility criterion. Choice A reverses the subtraction in the determinant formula, computing 22 + 41 instead of 22 - 41, which gives the wrong value. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test. Key to determinants: for a 2×2 matrix [a b; c d], always use det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det = 0 means the matrix has no inverse (singular) while det ≠ 0 means it's invertible.

Question 19

Let $A=\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ and let the $2\times2$ identity matrix be $I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ . For matrix $A$ , what is $A\cdot I_2$ ?

$\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ (correct answer)
$\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\begin{bmatrix}2 & 1\\ -3 & 4\end{bmatrix}$

Explanation: This question tests understanding of identity matrices and their properties. The identity matrix I is a square matrix with 1s on the main diagonal and 0s elsewhere, serving as the multiplicative identity: for any matrix A of compatible dimensions, A·I = I·A = A, just as multiplying a number by 1 leaves it unchanged. The identity matrix for 2×2 is I = [[1,0],[0,1]]. When we compute A·I = [[2,-1],[3,4]] · [[1,0],[0,1]] = [[2, -1],[3, 4]], we get the result equals A, confirming the identity property. Choice B is correct because it correctly applies the identity property that A·I = A. Choice A states that A·I = I, but the identity property is A·I = A—multiplying by the identity matrix leaves A unchanged, not replaces it with I. Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 20

Let $A$ be a $2\times2$ matrix with $\det(A)=-3$ . Based on the determinant, is $A$ invertible?

Yes, because $\det(A)\neq 0$ . (correct answer)
No, because $\det(A)<0$ .
No, because $\det(A)\neq 1$ .
Yes, because $\det(A)=0$ .

Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). Since det(A) = -3, and this is not equal to zero, matrix A is non-singular and invertible. Choice A is correct because it correctly applies the invertibility criterion. Choice B incorrectly claims the matrix is not invertible when det(A) = -3, but the criterion is that det(A) = 0 means singular (no inverse) and det(A) ≠ 0 means invertible. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test. Key to determinants: for a 2×2 matrix [a b; c d], always use det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det = 0 means the matrix has no inverse (singular) while det ≠ 0 means it's invertible.