The third midterm

Practice problems with solutions

Computations with matrices

a. Matrix products: Homework 5, problem 4 & problem 5

b. Consider the following matrices:

$A= \mat 1 & 3 & -1 \\ 2 & 0 & 1\rix, \qquad B= \mat 3 & 3\rix$

Compute each of the following matrix expressions, if they are defined:

1. $AB$, $BA$, $AA^\top$, $A^\top A$

$\begin{aligned} AB& \text{ can't multiply $2\times 3$ and $1\times 2$} \\ BA&= \mat 9 & 9 & 0\rix\\ AA^\top&= \mat 11 & 1 \\ 1 & 5\rix\\ A^\top A&= \mat 5 & 3 & 1 \\ 3 & 9 & -3\\ 1 & -3 & 2\rix \end{aligned}$

2. $BB^\top$, $B^\top B$

$BB^\top = \mat 18\rix, \qquad B^\top B= \mat 9 & 9 \\ 9& 9 \rix$

3. $A^2$ is not defined because $A$ is not square
4. $(AA^\top)^2 = \tmat 122 & 16 \\ 16 & 26\trix$
5. $A^\top B^\top B A = \tmat 81&81&0\\81&81&0\\ 0&0&0 \trix$
6. $AA^\top+BB^\top$ is undefined because $AA^\top$ and $BB^\top$ do not have the same size
7. $AA^\top+B^\top B = \tmat 20&10\\10&14\trix$

Compute the matrix of $T:V\to V$ with respect to a given basis of $V$

a. $V=\cP_4(\R)$, the standard basis $\{1,x,x^2,x^3,x^4\}$, $Tf(x) = f''(x)-xf'(x)$

$\left. \begin{aligned} T(1) &= 0\\ T(x) &=-x\\ T(x^2)&=2-2x^2\\ T(x^3)&=6x-3x^3\\ T(x^4)&=12x^2-4x^4 \end{aligned}\right\}\implies [T]_{1, x, x^2, x^3, x^4}=\mat 0& 0 & 2 & 0 & 0\\ 0&-1 & 0 & 6 & 0\\ 0& 0 &-2 & 0 &12\\ 0& 0 & 0 &-3 & 0\\ 0& 0 & 0 & 0 &-4 \rix$

b. $V=\R^2$ and $T:\R^2\to\R^2$ is given by $T(x_1, x_2) = (2x_1, x_2)$.

• Find the matrix of $T$ with respect to the standard basis $\{e_1, e_2\}$ of $\R^2$

$\left. \begin{gathered} Te_1=2e_1 \\ Te_2=e_2 \end{gathered}\right\}\implies [T]_{e_1, e_2}= \mat 2 & 0 \\ 0 & 1\rix$

• Find the matrix of $T$ with respect to the basis $\{u,v\}$ where $u=\binom{1}{2}$, $v=\binom{2}{1}$.
  Compute $Tu, Tv$:

$Tu = \binom 22= au+bv, \qquad Tv=\binom41 = cu+dv$

Solve for $a,b,c,d$:

$\left. \begin{gathered} au+bv=\binom{a+2b}{2a+b} = \binom{2}{2}\implies a=b=\frac23 \\ cu+dv=\binom{c+2d}{2c+d}= \binom 41\implies c=-\frac23, d=\frac{7}{3} \end{gathered}\right\} \implies [T]_{u,v}=\mat 2/3 & -2/3 \\ 2/3 & 7/3\rix$

c. $V$ is the vector space

$V=\{(x_1,x_2,x_3)\in\R^3 \mid x_1+x_2+x_3=0\},$

with basis $\{u,v\}$ where $u = \tmat 1\\-1\\0\trix$, $v=\tmat 0\\1\\-1 \trix$, and $T:V\to V$ is the linear transformation $T(x_1, x_2, x_3) = (x_3,x_1,x_2).$

The vector space $V$ is a linear subspace of $\R^3$. It contains the two vectors $u,v$ and we are given they form a basis for $V$, so $V$ is two dimensional: $V$ is a plane.

Compute $Tu$ and $Tv$ and write them as linear combinations of $u$ and $v$.

$Tu= \tmat 0 \\ 1\\ -1\trix = v,\quad Tv= \tmat -1\\0\\1\trix=-u-v\implies [T]_{u,v} = \mat 0 & -1 \\ 1 & -1\rix$

Compute some determinants

a. From the book, page 237,Problem 4

b. For which value of $a\in\R$ is $\tmat 1 &a &1\\a & 1 & 1 \\ 1 & 1 & 1 \trix$ invertible?

Fact: a matrix is invertible if and only if its determinant is not zero. We compute (subtract the 1st column from the 2nd and 3rd columns)

$\deter1 &a &1\\a & 1 & 1 \\ 1 & 1 & 1 \minant= \deter1 &a-1 &0\\a & 1-a & 1-a \\ 1 & 0 & 0\minant= -(a-1)^2.$

So the matrix is always invertible, except when $a=1$.

c. Subtract the first column from each of the other columns:

$\deter 1& 1 & 1& 1& 1\\ 1& 2 & 3& 4& 5\\ 1& 3 & 5& 7& 9\\ 1& 4 & 7& 10& 13\\ 1& 5 & 9& 13& 17 \minant= \deter 1& 0 & 0& 0& 0\\ 1& 1 & 2& 3& 4\\ 1& 2 & 4& 6& 9\\ 1& 3 & 6& 9& 12\\ 1& 4 & 8& 12& 16 \minant =0 .$

The last determinant is zero because the third column is twice the second column.

d. Let $A=\tmat 1& 2 & 1\\ 2& 0 &-1 \\ -5& 0 &3 \trix$, and let $B=A^{-1}$. Compute the cofactor matrix of $A$. Then use the cofactor formula to compute $b_{11}$, $b_{12}$, and $b_{13}$.

e. Let $A=\tmat 1& 2 & 1\\ 2& 0 &-1 \\ -5& 0 &3 \trix$. Use row reduction (or “Gaussian elimination”) to compute $A^{-1}$.

f. For which values of $t\in\R$ are the vectors

$u=\mat 1 \\ 2 \\ t\rix,\quad v=\mat 2 \\ 1 \\ t\rix,\quad w=\mat 1 \\ t \\ 0\rix$

a basis of $\R^3$?

Answer. $\{u,v,w\}$ is a basis if and only if

$0\neq\det(u,v,w)= \deter 1 & 2 & 1\\ 2 & 1 & t\\ t & t & 0 \minant= \deter 1 & 1 & 1\\ 2 & -1 & t\\ t & 0 & 0 \minant=t\deter1&1\\-1&t\minant=t(t+1)$

So $\{u,v,w\}$ is a basis if and only if $t\neq0$ and $t\neq-1$.

True/False?

1. For any $k\times \ell$ matrix $A$, the matrix product $AA^\top$ is always defined, no matter what $k$ and $\ell$ are. True: the sizes match.

2. If $A= \tmat1 & 3 & -1 \\ 2 & 0 & 1\trix$, then $\det(AA^\top) = \det (A) \det (A^\top)$ $A$ is not a square matrix so its determinant is not defined: False!

3. If $A$ is a square matrix then $\det(A^\top A) \geq 0$ $\det(A^\top A) = \det (A^\top)\det(A)=\det(A)\det(A)=(\det A)^2\geq$, so True.

4. If $A$ is a square matrix with $A=-A^\top$, then $\det A^\top = \det A$.

   For an $n\times n$ matrix $A$ one always has $\det(cA)=c^n\det A$. Therefore $\det(-A)=(-1)^n\det(A)$. If $A^\top=-A$ then we have $\det(A)=\det(A^\top)=(-1)^n\det A$.

   If $n$ is even then this just says $\det(A^\top)=\det(A)$ so the statement is true. If $n$ is odd then we get $\det(A)=-\det(A)$ so $\det(A)=0$. Thus S\det(A^\top)=\det(A)$ because both are zero.

5. If $A$ is a $3\times 3$ square matrix with $A^\top=-A$ then $\det A=0$ True; see the previous question with $n=3$.

6. If $A$ is a square matrix with $A^2=O$ then $A=O$. Example: $A=\tmat 0&1\\0&0\trix$. This matrix satisfies $A^2=O$.

7. If $A$ is an $n\times n$ matrix and if there is a nonzero vector $x\in\F^n$ with $Ax=0$ then $\det A = 0$.True. If $Ax=0$ and $x\neq 0$ then $x\in N(A)$ so $N(A)\neq \{0\}$. Hence $A$ is not injective, and hence $\det A=0$.

8. If $A$ is a square matrix with $A^2=O$ then $\det A=0$. True. If $A^2= O$ then $\det(A)^2=\det(A)\det(A)=\det(A\cdot A)=\det(A^2)=det(O)=0$. Therefore $\det(A)=0$.

9. From the book: §4.4, page 236, problem 1—if a T/F question claims some equation, and you think it is wrong, say how the equation can be fixed.