Symmetric Matrices, Inverses and Mixed Practice — Worked Solutions

Ques­tion 1: Show­ing a matrix is skew-sym­met­ric#

The prob­lem#

Show that $A = \begin{bmatrix} 0 & 3 & -2 \\ -3 & 0 & 5 \\ 2 & -5 & 0 \end{bmatrix}$ is skew-sym­met­ric.

Under­stand­ing the prob­lem#

A square matrix is skew-sym­met­ric when $A^T = -A$, that is, $a_{ji} = -a_{ij}$ for every $i, j$. In par­tic­u­lar, every diag­o­nal entry must be $0$.

The idea#

Write down $A^T$ (rows become columns) and $-A$ (change every sign), and com­pare them.

Step-by-step solu­tion#

Step 1. Trans­pose: the first row of $A^T$ is the first col­umn of $A$, and so on.

$$A^T = \begin{bmatrix} 0 & -3 & 2 \\ 3 & 0 & -5 \\ -2 & 5 & 0 \end{bmatrix}$$

Step 2. Neg­a­tive of $A$.

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

Step 3. The two matri­ces agree entry by entry, so $A^T = -A$ and $A$ is skew-sym­met­ric.

Check­ing the answer#

The diag­o­nal is all zeros, and each mir­ror pair has oppo­site signs: $(3, -3)$, $(-2, 2)$, $(5, -5)$. ✓

Answer#

$A^T = -A$, so $A$ is skew-sym­met­ric.

Ques­tion 2: Sym­met­ric plus skew-sym­met­ric (2 × 2)#

The prob­lem#

Write $\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a sym­met­ric and a skew-sym­met­ric matrix.

Under­stand­ing the prob­lem#

Every square matrix $A$ splits as $A = P + Q$ with $P$ sym­met­ric ($P^T = P$) and $Q$ skew-sym­met­ric ($Q^T = -Q$).

The idea#

Use $\displaystyle P = \tfrac12(A + A^T)$ and $\displaystyle Q = \tfrac12(A - A^T)$.

Step-by-step solu­tion#

Step 1. Trans­pose.

$$A^T = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$$

Step 2. Sym­met­ric part.

$$\displaystyle P = \frac12\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$$

Step 3. Skew-sym­met­ric part.

$$\displaystyle Q = \frac12\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$

Check­ing the answer#

$P + Q = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = A$ ✓; $P$ is sym­met­ric and $Q$ has zero diag­o­nal with oppo­site off-diag­o­nal entries. ✓

Answer#

$\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Ques­tion 3: Sym­met­ric plus skew-sym­met­ric (3 × 3)#

The prob­lem#

Write $\begin{bmatrix} 1 & 4 & -2 \\ 0 & 3 & 6 \\ 2 & -2 & 5 \end{bmatrix}$ as a sum of sym­met­ric and skew-sym­met­ric parts.

Under­stand­ing the prob­lem#

Same task as Ques­tion 2, for a $3 \times 3$ matrix.

The idea#

$\displaystyle P = \tfrac12(A + A^T)$ and $\displaystyle Q = \tfrac12(A - A^T)$.

Step-by-step solu­tion#

Step 1. Trans­pose.

$$A^T = \begin{bmatrix} 1 & 0 & 2 \\ 4 & 3 & -2 \\ -2 & 6 & 5 \end{bmatrix}$$

Step 2. Add and halve.

$$A + A^T = \begin{bmatrix} 2 & 4 & 0 \\ 4 & 6 & 4 \\ 0 & 4 & 10 \end{bmatrix}, \qquad P = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & 2 \\ 0 & 2 & 5 \end{bmatrix}$$

Step 3. Sub­tract and halve.

$$A - A^T = \begin{bmatrix} 0 & 4 & -4 \\ -4 & 0 & 8 \\ 4 & -8 & 0 \end{bmatrix}, \qquad Q = \begin{bmatrix} 0 & 2 & -2 \\ -2 & 0 & 4 \\ 2 & -4 & 0 \end{bmatrix}$$

Check­ing the answer#

$P + Q = \begin{bmatrix} 1 & 4 & -2 \\ 0 & 3 & 6 \\ 2 & -2 & 5 \end{bmatrix} = A$. ✓

Answer#

$P = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & 2 \\ 0 & 2 & 5 \end{bmatrix}$ (sym­met­ric), $Q = \begin{bmatrix} 0 & 2 & -2 \\ -2 & 0 & 4 \\ 2 & -4 & 0 \end{bmatrix}$ (skew-sym­met­ric), $A = P + Q$.

Ques­tion 4: A Aᵀ is always sym­met­ric#

The prob­lem#

Show that for any square $A$, $AA^T$ is sym­met­ric.

Under­stand­ing the prob­lem#

This is a proof. A matrix $M$ is sym­met­ric when $M^T = M$. So you must show $(AA^T)^T = AA^T$.

The idea#

Use two rules of trans­poses: $(XY)^T = Y^TX^T$ (order reverses) and $(X^T)^T = X$.

Step-by-step solu­tion#

Step 1. Apply the prod­uct rule with $X = A$, $Y = A^T$.

$$(AA^T)^T = (A^T)^T A^T$$

Step 2. Trans­pos­ing twice gives back the orig­i­nal: $(A^T)^T = A$.

$$(AA^T)^T = AA^T$$

Step 3. So $AA^T$ equals its own trans­pose, which means it is sym­met­ric.

Check­ing the answer#

Take $A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$: $AA^T = \begin{bmatrix} 5 & 6 \\ 6 & 9 \end{bmatrix}$, which is sym­met­ric. ✓

Answer#

$(AA^T)^T = (A^T)^TA^T = AA^T$, so $AA^T$ is sym­met­ric.

Ques­tion 5: Inverses by ele­men­tary row oper­a­tions#

The prob­lem#

Find the inverse by row oper­a­tions: (a) $\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$; (b) $\begin{bmatrix} 4 & -3 \\ 1 & -1 \end{bmatrix}$.

Under­stand­ing the prob­lem#

Write the aug­mented matrix $[A \mid I]$. Apply row oper­a­tions to the whole row until the left half becomes $I$; the right half is then $A^{-1}$.

The idea#

Aim to cre­ate a $1$ in the top-left, clear the entry below it, then clear the entry above the sec­ond $1$. Choose oper­a­tions that avoid frac­tions where pos­si­ble.

Step-by-step solu­tion#

Part (a)

Step 1. Start.

$$\left[\begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 5 & 2 & 0 & 1 \end{array}\right]$$

Step 2. $R_2 \to R_2 - R_1$.

$$\left[\begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 2 & 1 & -1 & 1 \end{array}\right]$$

Step 3. $R_1 \to R_1 - R_2$ gives a lead­ing $1$.

$$\left[\begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 2 & 1 & -1 & 1 \end{array}\right]$$

Step 4. $R_2 \to R_2 - 2R_1$.

$$\left[\begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 0 & 1 & -5 & 3 \end{array}\right]$$

The left half is $I$, so $A^{-1} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}$.

Part (b)

Step 1. Start, then $R_1 \leftrightarrow R_2$ to bring a $1$ to the top.

$$\left[\begin{array}{cc|cc} 4 & -3 & 1 & 0 \\ 1 & -1 & 0 & 1 \end{array}\right] \to \left[\begin{array}{cc|cc} 1 & -1 & 0 & 1 \\ 4 & -3 & 1 & 0 \end{array}\right]$$

Step 2. $R_2 \to R_2 - 4R_1$.

$$\left[\begin{array}{cc|cc} 1 & -1 & 0 & 1 \\ 0 & 1 & 1 & -4 \end{array}\right]$$

Step 3. $R_1 \to R_1 + R_2$.

$$\left[\begin{array}{cc|cc} 1 & 0 & 1 & -3 \\ 0 & 1 & 1 & -4 \end{array}\right]$$

So $A^{-1} = \begin{bmatrix} 1 & -3 \\ 1 & -4 \end{bmatrix}$.

Check­ing the answer#

(a) $\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} 6 - 5 & -3 + 3 \\ 10 - 10 & -5 + 6 \end{bmatrix} = I$. ✓
(b) $\begin{bmatrix} 4 & -3 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 1 & -3 \\ 1 & -4 \end{bmatrix} = \begin{bmatrix} 4 - 3 & -12 + 12 \\ 1 - 1 & -3 + 4 \end{bmatrix} = I$. ✓

Answer#

(a) $\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}$; (b) $\begin{bmatrix} 1 & -3 \\ 1 & -4 \end{bmatrix}$.

Com­mon mis­take to avoid#

Apply­ing an oper­a­tion to the left half only. Every oper­a­tion must be applied to the right half too.

Ques­tion 6: A matrix with no inverse#

The prob­lem#

Show that $\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$ has no inverse.

Under­stand­ing the prob­lem#

With row oper­a­tions, a matrix has no inverse when, at some stage, a row of zeros appears on the left-hand side: then the left side can never become $I$.

The idea#

Notice row 1 is $-3$ times row 2. So adding $3R_2$ to $R_1$ should wipe out row 1.

Step-by-step solu­tion#

Step 1. Write $[A \mid I]$.

$$\left[\begin{array}{cc|cc} 6 & -3 & 1 & 0 \\ -2 & 1 & 0 & 1 \end{array}\right]$$

Step 2. $R_1 \to R_1 + 3R_2$.

$$\left[\begin{array}{cc|cc} 0 & 0 & 1 & 3 \\ -2 & 1 & 0 & 1 \end{array}\right]$$

Step 3. The first row on the left is all zeros. No fur­ther row oper­a­tions can turn it into a row of $I$, so $A^{-1}$ does not exist.

Check­ing the answer#

Row 1 of $A$ is $-3$ times row 2, so for any matrix $C$, row 1 of $AC$ is $-3$ times row 2 of $AC$. If $AC$ were $I$, you would need $(1, 0) = -3(0, 1)$, which is false. So no inverse can exist. ✓

Answer#

$R_1 \to R_1 + 3R_2$ gives a zero row, so the matrix has no inverse.

Ques­tion 7: Inverse of an upper tri­an­gu­lar matrix#

The prob­lem#

Find $A^{-1}$ for $A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}$.

Under­stand­ing the prob­lem#

$A$ already has $1$s on the diag­o­nal and zeros below it, so you only need to clear the entries above the diag­o­nal.

The idea#

Work from the bot­tom up: use $R_3$ to clear col­umn 3, then $R_2$ to clear col­umn 2.

Step-by-step solu­tion#

Step 1. Write $[A \mid I]$.

$$\left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & 3 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]$$

Step 2. $R_2 \to R_2 - 3R_3$ clears the $3$.

$$\left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -3 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]$$

Step 3. $R_1 \to R_1 - 2R_2$ clears the $2$.

$$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -2 & 6 \\ 0 & 1 & 0 & 0 & 1 & -3 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]$$

Check­ing the answer#

$$\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & -2 & 6 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 + 2 & 6 - 6 \\ 0 & 1 & -3 + 3 \\ 0 & 0 & 1 \end{bmatrix} = I$$

Answer#

$A^{-1} = \begin{bmatrix} 1 & -2 & 6 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{bmatrix}$

Ques­tion 8: The rota­tion matrix is orthog­o­nal#

The prob­lem#

If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$, show $A^TA = I$ and hence $A^{-1} = A^T$.

Under­stand­ing the prob­lem#

First com­pute the prod­uct $A^TA$ and show it is the iden­tity. Then explain why that makes $A^T$ the inverse of $A$.

The idea#

Mul­ti­ply and use $\cos^2\alpha + \sin^2\alpha = 1$.

Step-by-step solu­tion#

Step 1. Trans­pose.

$$A^T = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$$

Step 2. Mul­ti­ply $A^TA$.

$$A^TA = \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & -\cos\alpha\sin\alpha + \sin\alpha\cos\alpha \\ -\sin\alpha\cos\alpha + \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha \end{bmatrix}$$

Step 3. Sim­plify: each diag­o­nal entry is $1$, each off-diag­o­nal entry is $0$.

$$A^TA = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$

Step 4. In the same way, $AA^T = \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & \cos\alpha\sin\alpha - \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha - \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha \end{bmatrix} = I$. Since $A^T$ mul­ti­plied by $A$ on either side gives $I$, $A^T$ is the inverse of $A$: $A^{-1} = A^T$.

Check­ing the answer#

Take $\alpha = 90^\circ$: $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$, $A^T = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$, and $A^TA = I$. ✓

Answer#

$A^TA = AA^T = I$, so $A^{-1} = A^T$.

Ques­tion 9: The rever­sal law for inverses#

The prob­lem#

If $A$ and $B$ are invert­ible, ver­ify $(AB)^{-1} = B^{-1}A^{-1}$ by mul­ti­ply­ing.

Under­stand­ing the prob­lem#

A matrix $C$ is the inverse of $AB$ when $(AB)C = I$ and $C(AB) = I$. So take $C = B^{-1}A^{-1}$ and check both prod­ucts. ($A$ and $B$ are square of the same order.)

The idea#

Use asso­cia­tiv­ity to regroup the brack­ets so that $BB^{-1}$ or $A^{-1}A$ sit next to each other.

Step-by-step solu­tion#

Step 1. Mul­ti­ply on the right.

$$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$$

Step 2. Mul­ti­ply on the left.

$$(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I$$

Step 3. Both prod­ucts are $I$, so $B^{-1}A^{-1}$ is the inverse of $AB$.

Check­ing the answer#

Note the order: in $A^{-1}B^{-1}(AB)$ noth­ing can­cels, because $B^{-1}$ and $A$ are not next to each other; that is why the order must reverse.

Answer#

$(AB)(B^{-1}A^{-1}) = (B^{-1}A^{-1})(AB) = I$, so $(AB)^{-1} = B^{-1}A^{-1}$.

Ques­tion 10: Find­ing k in a matrix equa­tion#

The prob­lem#

Find $k$ if $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ sat­is­fies $A^2 - 5A - kI = O$.

Under­stand­ing the prob­lem#

$O$ is the zero matrix. You need the num­ber $k$ that makes $A^2 - 5A$ equal to $kI$.

The idea#

Com­pute $A^2 - 5A$; if it comes out as a mul­ti­ple of $I$, that mul­ti­ple is $k$.

Step-by-step solu­tion#

Step 1. Square $A$.

$$A^2 = \begin{bmatrix} 1 + 6 & 2 + 8 \\ 3 + 12 & 6 + 16 \end{bmatrix} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$$

Step 2. Sub­tract $5A$.

$$A^2 - 5A = \begin{bmatrix} 7 - 5 & 10 - 10 \\ 15 - 15 & 22 - 20 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2I$$

Step 3. So $A^2 - 5A - kI = (2 - k)I = O$ requires $k = 2$.

Check­ing the answer#

With $k = 2$: $A^2 - 5A - 2I = 2I - 2I = O$. ✓

Answer#

$k = 2$