## MATH1002 Quizzes

Quiz 10: Determinants
Question 1 Questions
Let $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill -4\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill 5\hfill \end{array}\right]$. Which of the following is true ? Exactly one option must be correct)
 a) $det\left(A\right)=3$; b) $det\left(A\right)=63$; c) $det\left(A\right)=7$; d) $det\left(A\right)$ is undefined; e) none of the above.

Choice (a) is incorrect
Choice (b) is correct!
The determinant of a square matrix is always defined, so (4) is wrong.
Either: $\begin{array}{llll}\hfill det\left(A\right)& =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill -4\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill 5\hfill \end{array}\right|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1\left|\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right|-\left(-3\right)\left|\begin{array}{cc}\hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right|+\left(-4\right)\left|\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill -2\hfill \end{array}\right|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1\left(1×5-0×3\right)-\left(-3\right)\left(2×5-0×3\right)+\left(-4\right)\left(2×\left(-2\right)-1×3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1×5-\left(-3\right)×10+\left(-4\right)×\left(-7\right)=63,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ or, using elementary row operations ${R}_{2}:={R}_{2}-2{R}_{1};\phantom{\rule{2.77695pt}{0ex}}{R}_{3}:={R}_{3}-3{R}_{1}\phantom{\rule{2.77695pt}{0ex}}{R}_{3}:={R}_{3}-{R}_{2}$ (all of which give matrices with the same determinant) to reduce to (upper) triangular form: $det\left(A\right)=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 7\hfill & \hfill 8\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill 5\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 7\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill 7\hfill & \hfill 17\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 7\hfill & \hfill 8\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 9\hfill \end{array}\right|=1×7×9=63.$ Hence the correct answer is (2).
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Let $A=\left[\begin{array}{cccc}\hfill 1\hfill & \hfill -3\hfill & \hfill -4\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill 5\hfill & \hfill 0\hfill \end{array}\right]$. Which of the following is true ? Exactly one option must be correct)
 a) $det\left(A\right)=63$; b) $det\left(A\right)=7$; c) $det\left(A\right)=0$; d) $det\left(A\right)$ is undefined; e) none of the above.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
The determinant of a non-square matrix is undefined, so (4) is correct.
Choice (e) is incorrect
Let $A=\left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill \\ \hfill 7\hfill & \hfill 8\hfill & \hfill 9\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 7\hfill & \hfill 8\hfill & \hfill 9\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$. Which of the following is true ? Exactly one option must be correct)
 a) $det\left(A\right)=3888$; b) $det\left(A\right)=0$; c) $det\left(A\right)=28$; d) $det\left(A\right)$ is undefined; e) none of the above.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is correct!
The determinant of a square matrix is always defined, so (4) is wrong.
$A$ is not triangular (which means all zero on one side of the main top-left to bottom-right diagonal, not the top-right to bottom-left diagonal), so $det\left(A\right)$ is not necessarily the product of the “wrong” diagonal elements, i.e., not necessarily $6\cdot 2\cdot 6\cdot 9\cdot 2\cdot 3=3888$.
The three elementary row operations ${R}_{1}↔{R}_{6};$ ${R}_{2}↔{R}_{5};$ ${R}_{3}↔{R}_{4};$ change the sign of the determinant three times, i.e., $det\left(A\right)=-\left|\begin{array}{cccccc}\hfill 3\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 7\hfill & \hfill 8\hfill & \hfill 9\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 7\hfill & \hfill 8\hfill & \hfill 9\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill \end{array}\right|=-3\cdot 2\cdot 9\cdot 6\cdot 2\cdot 6=-3888.$ So the correct answer is (5).
Let $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right]$. Which of the following statements is correct ? Exactly one option must be correct)
 a) $A$ is invertible since det $A=0$ b) $A$ is not invertible since det $A=0$ c) $A$ is invertible since det $A\ne 0$ d) $A$ is not invertible since det $A\ne 0$

Choice (a) is incorrect
Choice (b) is correct!
$\begin{array}{rcll}det\phantom{\rule{1em}{0ex}}A& =& 1\left|\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 4\hfill & \hfill 5\hfill \end{array}\right|-2\left|\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill 5\hfill \end{array}\right|+3\left|\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right|& \text{}\\ & =& 4-16+12=0.& \text{}\end{array}$ It is not invertible since det $A=0$.
Choice (c) is incorrect
Choice (d) is incorrect
What is the determinant of the matrix $\left[\begin{array}{cccc}\hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right]$ ?

Correct!
$\begin{array}{llll}\hfill \left|\begin{array}{cccc}\hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right|& =-\left|\begin{array}{cccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\left(-1\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right|+1\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right|\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\left(-\left|\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 5\hfill \end{array}\right|+2\left|\begin{array}{cc}\hfill 3\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 5\hfill \end{array}\right|-3\left|\begin{array}{cc}\hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 4\hfill \end{array}\right|+\left|\begin{array}{cc}\hfill 2\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right|-\left|\begin{array}{cc}\hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 4\hfill \end{array}\right|+2\left|\begin{array}{cc}\hfill 3\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right|\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\left(+13+22-42+11-14+10\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Try expanding the determinant along the third row.
Let $A=\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill i\hfill \end{array}\right]$. Which of the following is true ? Exactly one option must be correct)
 a) $det\left(A\right)=aei+afh+bdi+bgf+cdh+ceg$; b) $det\left(A\right)=aei-afh-bdi+bgf+cdh-ceg$; c) $det\left(A\right)=aei+afh+bdi-bgf-cdh-ceg$; d) $det\left(A\right)=aei-afh+bdi-bgf+cdh-ceg$; e) none of the above.

Choice (a) is incorrect
Choice (b) is correct!

$\begin{array}{llll}\hfill det\left(A\right)& =\left|\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill i\hfill \end{array}\right|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =a\left|\begin{array}{cc}\hfill e\hfill & \hfill f\hfill \\ \hfill h\hfill & \hfill i\hfill \end{array}\right|-b\left|\begin{array}{cc}\hfill d\hfill & \hfill f\hfill \\ \hfill g\hfill & \hfill i\hfill \end{array}\right|+c\left|\begin{array}{cc}\hfill d\hfill & \hfill e\hfill \\ \hfill g\hfill & \hfill h\hfill \end{array}\right|\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =aei-afh+bdi-bgf+cdh-ceg.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Hence the correct answer is (4).
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
The matrix $A$ has the elementary row operations
${R}_{2}:={R}_{2}-{R}_{1},\phantom{\rule{2em}{0ex}}{R}_{2}↔{R}_{3},\phantom{\rule{2em}{0ex}}{R}_{3}:={R}_{3}-3{R}_{2},$
applied to it to transform it to the matrix $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 6\hfill \end{array}\right]$. What is the determinant of $A$ ? Exactly one option must be correct)
 a) -6 b) 6 c) 18 d) -18 e) None of the above

Choice (a) is correct!
The determinant of $A$ has the opposite sign of the determinant of the matrix
$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 6\hfill \end{array}\right]$
since a row swap has been applied. The determinant of a triangular matrix is the product of the diagonal entries hence $detA=-1×1×6=-6$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
The matrix $A$ has the elementary row operations
${R}_{2}:={R}_{2}-3{R}_{1},{R}_{3}:={R}_{3}+4{R}_{1},{R}_{2}↔{R}_{3},{R}_{2}:=\frac{1}{2}{R}_{2},{R}_{4}:={R}_{4}-2{R}_{2},$
applied to it to transform it to the matrix $\left[\begin{array}{cccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$. What is the determinant of $A$ ? Exactly one option must be correct)
 a) -3 b) 3 c) 12 d) -12 e) None of the above

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Call the new matrix ${A}^{\prime }$. Now $det{A}^{\prime }=-\frac{1}{2}\phantom{\rule{1em}{0ex}}detA$ since ${A}^{\prime }$ has had a row swap to change the sign and ${A}^{\prime }$ has had a row divided by 2. Now $det{A}^{\prime }=6$ since it is triangular. Thus $detA=-2×6=-12$.
Choice (e) is incorrect
Let $A=\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill i\hfill \end{array}\right]$, and suppose $det\left(A\right)=0$. Consider the system $\begin{array}{llll}\hfill ax+by+cz& =5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill dx+ey+fz& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill gx+hy+iz& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Which of the following is necessarily true ? Exactly one option must be correct)
 a) the system has no solutions; b) the system has many solutions; c) the system has a unique solution ; d) none of the above.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Since $det\left(A\right)=0$, $A$ is not invertible, so we cannot solve the system by left multiplying by ${A}^{-1}$ both sides of $A\mathbf{{x}^{T}=\mathbf{{b}^{}}T}$ ($\mathbf{x}=\left[x\phantom{\rule{2.77695pt}{0ex}}y\phantom{\rule{2.77695pt}{0ex}}z\right]$, $\mathbf{b}=\left[5\phantom{\rule{2.77695pt}{0ex}}1\phantom{\rule{2.77695pt}{0ex}}4\right]$) to get a unique solution $\mathbf{{x}^{T}={A}^{-1}\mathbf{{b}^{}}T}$.
In fact, if we solve by reducing the augmented matrix $\left[A|\mathbf{{b}^{T}}\right]$, the reduced form will have third row $\left[0\phantom{\rule{2.77695pt}{0ex}}0\phantom{\rule{2.77695pt}{0ex}}0\phantom{\rule{2.77695pt}{0ex}}|\phantom{\rule{2.77695pt}{0ex}}\ast \right]$. Hence, at least one column will be missing a leading 1, so (c) cannot possibly be true.
If $\ast \ne 0$, (a) will be true. If $\ast =0$, (b) will be true.
However we do not have enough information to decide if $\ast =0$ or not. So neither (a) nor (b) is necessarily true. Hence the correct answer is (d)!!
Which of the following statements about determinants are true ?

Exactly one option must be correct)
 a) $det{A}^{-1}=-detA$ b) $det{A}^{-1}=\frac{1}{detA}$ c) $\left|\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \\ \hfill g\hfill & \hfill h\hfill & \hfill i\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill a\hfill & \hfill d\hfill & \hfill g\hfill \\ \hfill b\hfill & \hfill e\hfill & \hfill h\hfill \\ \hfill c\hfill & \hfill f\hfill & \hfill i\hfill \end{array}\right|$ d) $\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 4\hfill & \hfill 6\hfill \\ \hfill 8\hfill & \hfill 10\hfill & \hfill 12\hfill \\ \hfill 14\hfill & \hfill 16\hfill & \hfill 18\hfill \end{array}\right|=2\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill \\ \hfill 7\hfill & \hfill 8\hfill & \hfill 9\hfill \end{array}\right|$

Choice (a) is incorrect
$1=detI=det\left(A{A}^{-1}\right)=detAdet{A}^{-1}$.
Choice (b) is correct!
$1=detI=det\left(A{A}^{-1}\right)=detAdet{A}^{-1}$.
Choice (c) is correct!
This is correct because $det{A}^{T}=detA$.
Choice (d) is incorrect
As $\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 4\hfill & \hfill 6\hfill \\ \hfill 8\hfill & \hfill 10\hfill & \hfill 12\hfill \\ \hfill 14\hfill & \hfill 16\hfill & \hfill 18\hfill \end{array}\right|={2}^{3}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill \\ \hfill 7\hfill & \hfill 8\hfill & \hfill 9\hfill \end{array}\right|$, this is incorrect.