Linear Algebra Practice: Systems, Matrices and Vectors

$A$ is $2\times3$, $B$ is $3\times2$, and $\mathbf v$ is $3\times1$. Thus $A+B$ is undefined. The products $AB$, $BA$, and $A\mathbf v$ have sizes $2\times2$, $3\times3$, and $2\times1$, respectively. The sum $AB+BA$ is undefined because its two terms have different sizes.

For example, the top-left entry of $AB$ is $1(2)+(-1)(1)+2(-1)=-1$. Continuing row by column gives

$$AB=\begin{pmatrix}-1&5\\2&2\end{pmatrix},$$

$$BA=\begin{pmatrix}2&1&5\\1&-1&2\\-1&7&0\end{pmatrix}.$$

For the vector product, the two row calculations are $1-2-2=-3$ and $0+6-1=5$. Hence

$$A\mathbf v=\begin{pmatrix}-3\\5\end{pmatrix}.$$

A product can be defined in each order without having the same dimensions in those orders. Matrix multiplication is not an entry-by-entry operation.

Check the result and conditions

All entries are real and no parameter restriction is needed. Independently check the last entry of $BA$: $(-1)(2)+2(1)=0$. Also check that every reported result has the size predicted before calculation.

Common errors

Do not add differently sized matrices, multiply matching positions entry by entry, or assume $AB=BA$. Here the dimensions alone prevent that equality.

Choose what to review

If the dimensions or row-by-column pairing were difficult, revisit multiplication, determinants and inverses. Then redo just one row of each product without looking at the answer.

The augmented matrix is

$$\left[\begin{array}{rrr|r}0&1&1&1\\2&-1&3&9\\1&2&-1&-3\end{array}\right].$$

First use $R_1\leftrightarrow R_3$, then $R_2\leftarrow R_2-2R_1$:

$$\left[\begin{array}{rrr|r}1&2&-1&-3\\0&-5&5&15\\0&1&1&1\end{array}\right].$$

Use $R_2\leftrightarrow R_3$, then $R_3\leftarrow R_3+5R_2$:

$$\left[\begin{array}{rrr|r}1&2&-1&-3\\0&1&1&1\\0&0&10&20\end{array}\right].$$

The last row gives $z=2$. The second gives $y=1-z=-1$. The first then gives $x=-3-2y+z=1$. There is a pivot in each variable column and no contradiction, so the solution is unique:

$$(x,y,z)=(1,-1,2).$$

Check the result and conditions

Substitution gives $-1+2=1$, $2(1)-(-1)+3(2)=9$, and $1+2(-1)-2=-3$. All three equations hold. Every row operation used is reversible; no division by zero occurred.

Common errors

The missing $x$ in the first equation needs a zero entry. Operate on the constant column too. After a swap, row labels refer to the current rows, not their original positions.

Choose what to review

If an operation changed your answer, review why permitted row operations are reversible. If you reached echelon form but could not finish, revisit solving and checking a system.

Expanding along row 1 requires one minor:

$$\det(D)=-2\begin{vmatrix}3&-1\\2&4\end{vmatrix}.$$

The minor is $3(4)-(-1)(2)=14$, so $\det(D)=-28$.

For row reduction, the swap multiplies the determinant by $-1$. Scaling the new first row by $1/3$ multiplies it by $1/3$. Adding a multiple of another row does not change it. After the swap and scaling, the matrix is

$$\begin{pmatrix}1&1/3&-1/3\\0&2&0\\2&0&4\end{pmatrix}.$$

The operation $R_3\leftarrow R_3-2R_1$ makes row 3 equal to $(0,-2/3,14/3)$. Then $R_3\leftarrow R_3+(1/3)R_2$ gives

$$U=\begin{pmatrix}1&1/3&-1/3\\0&2&0\\0&0&14/3\end{pmatrix}.$$

Consequently,

$$\det(U)=1\cdot2\cdot\frac{14}{3}=\frac{28}{3},$$

$$\det(U)=-\frac13\det(D),$$

so $\det(D)=-3\det(U)=-28$. The first method is shorter here because the zeros remove two cofactor terms.

For a system, the same reversible operation is applied to an entire augmented row and preserves the shared solutions. A determinant is a scalar attached to a square coefficient matrix; swaps and nonzero row scalings change that scalar in the stated ways. Preserving solutions and preserving a determinant are different claims.

Check the result and conditions

The two methods agree on $-28$. Since it is nonzero, $D$ is invertible. The factor $1/3$ scales one row, not the entire matrix; it therefore contributes one factor of $1/3$, not three.

Common errors

Watch the negative cofactor sign at position $(1,2)$, the subtraction in $12-(-2)$, and the direction of the final multiplier: $\det(U)=-\det(D)/3$, so multiply $\det(U)$ by $-3$ to recover the original.

Choose what to review

If “same solutions” led you to keep the determinant unchanged, revisit what row operations preserve, then redo this question with a two-column log of operation and determinant multiplier. For the nonzero-determinant/inverse connection, use the matrix fundamentals lesson.

Start with $[L\mid C]=[M\mid I_3]$. We display the left block $L$ and right block $C$ separately so every entry stays readable. Each stated operation acts on both blocks.

After $R_3\leftarrow R_3-R_1$, followed by $R_3\leftarrow R_3+2R_2$, the blocks are

$$L=\begin{pmatrix}1&2&0\\0&1&1\\0&0&3\end{pmatrix},$$

$$C=\begin{pmatrix}1&0&0\\0&1&0\\-1&2&1\end{pmatrix}.$$

Now divide row 3 by 3. Use the resulting row 3 to clear the third entry of row 2 with $R_2\leftarrow R_2-R_3$. This gives

$$L=\begin{pmatrix}1&2&0\\0&1&0\\0&0&1\end{pmatrix},$$

$$C=\frac13\begin{pmatrix}3&0&0\\1&1&-1\\-1&2&1\end{pmatrix}.$$

Finally use $R_1\leftarrow R_1-2R_2$. The left block is now $I_3$, and the right block is the inverse:

$$M^{-1}=\frac13\begin{pmatrix}1&-2&2\\1&1&-1\\-1&2&1\end{pmatrix}.$$

For the first operation alone, applying $R_3\leftarrow R_3-R_1$ to $I_3$ gives

$$E=\begin{pmatrix}1&0&0\\0&1&0\\-1&0&1\end{pmatrix}.$$

Left multiplication by $E$ performs that row operation: $EM$ is the new left block and $EI_3=E$ is the new right block. One operation does not yet give the inverse. The full sequence sends $M$ to $I_3$ and sends $I_3$ to $M^{-1}$.

Now compute

$$\mathbf x=M^{-1}\mathbf b=\frac13\begin{pmatrix}6\\-3\\9\end{pmatrix}.$$

$$\mathbf x=\begin{pmatrix}2\\-1\\3\end{pmatrix}.$$

Check the result and conditions

Writing $N=3M^{-1}$, direct multiplication gives $MN=NM=3I_3$, so both inverse identities hold. Also, $\det(M)=3\ne0$, and the row reduction never divides by zero. The forward check is $M(2,-1,3)^{\mathsf T}=(0,2,5)^{\mathsf T}$, since the three row results are $2-2=0$, $-1+3=2$, and $2+3=5$.

Common errors

Changing only the left block loses the inverse calculation. Also distinguish $EM$, which performs a row operation, from $ME$, which generally has a different effect. The inverse belongs to the matrix and can be reused for a different right-hand side.

Choose what to review

If the meaning of an inverse is unclear, revisit the introductory inverse example. If a block went out of sync, return to whole-row operations, then repeat this three-by-three construction.

The coefficient matrix and its replacement for $y$ are

$$H=\begin{pmatrix}2&-1&1\\1&1&0\\0&2&1\end{pmatrix},$$

$$H_y=\begin{pmatrix}2&6&1\\1&1&0\\0&-1&1\end{pmatrix}.$$

Expand each determinant along row 2, where the third entry is zero. For the original matrix,

$$\det(H)=-1(-1-2)+1(2)=5.$$

Because $\det(H)\ne0$, the system has a unique solution and Cramer’s rule applies. For the replacement matrix,

$$\begin{aligned}&\det(H_y)\\&=-1(6+1)+1(2)\\&=-5.\end{aligned}$$

Therefore $y=\det(H_y)/\det(H)=-1$. The second equation gives $x=1-y=2$, and the third gives $z=-1-2y=1$.

$$(x,y,z)=(2,-1,1).$$

Finding just the requested determinant and then substituting avoids two unnecessary replacement determinants. For a large system, elimination is usually more efficient than computing a determinant for every unknown.

Check the result and conditions

The original left sides are $4+1+1=6$, $2-1=1$, and $-2+1=-1$. The denominator is 5, so no zero division is hidden. If the original determinant had been zero, these quotients would not classify consistency; row reduction would be needed.

Common errors

Replace a column, not a row; use the constants in their original equation order. A negative replacement determinant is allowed. It is the original nonzero determinant that permits division.

Choose what to review

If you replaced the wrong entries, review the difference between coefficients and constants. If the determinant condition was missed, revisit when an inverse exists, then redo the first two steps here.

The operations $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-R_1$ give the equivalent triangular system

$$x+y+z=2,$$

$$(a-1)y+z=1,$$

$$(a-1)(a-2)z=a-2.$$

Let $P_a$ denote the coefficient matrix of the original system. Only row additions were used, so its determinant is the product of these diagonal coefficients:

$$\det(P_a)=(a-1)^2(a-2).$$

Case 1: $a\ne1,2$. Both factors are nonzero. The last equation gives $z=1/(a-1)$. The second then gives

$$y=\frac{1-z}{a-1}=\frac{a-2}{(a-1)^2}.$$

Finally,

$$x=\frac{2a^2-6a+5}{(a-1)^2}.$$

This is the unique solution because all three pivots are nonzero.

Case 2: $a=1$. The last reduced equation becomes $0=-1$. There is no solution. In the original system, the first and third equations have the same left side but constants 2 and 1, which confirms the contradiction.

Case 3: $a=2$. The last reduced equation is $0=0$, while the second is $y+z=1$. Set $z=t$, with $t\in\mathbb R$; then $y=1-t$ and $x=1$. Every solution is

$$(x,y,z)=(1,1-t,t),\qquad t\in\mathbb R.$$

The coefficient matrix is singular exactly at $a=1$ and $a=2$. These two singular cases have different consistency outcomes.

Check the result and conditions

For $a\ne1,2$, substitute the formulas into the reduced equations: $(a-1)y+z=1$ and $(a-1)(a-2)z=a-2$; the chosen $x$ gives $x+y+z=2$. Reversing the row additions verifies the original system. At $a=2$, the family gives original left sides $2,3,2$ for every real $t$. Substituting $a=2$ into the generic formulas produces only $(1,0,1)$, one member of that family, and misses the others.

Common errors

Do not cancel $a-2$ without recording $a\ne2$, or divide by $a-1$ before checking $a=1$. A zero determinant does not by itself mean “no solution.” A zero row must be read together with its augmented entry and the remaining equations.

Choose what to review

If the exceptional cases were lost, revisit reading final rows and free variables. Then redo this question by listing the excluded values before writing any fractions.

The dot product is a scalar:

$$\mathbf u\cdot\mathbf v=2(1)+(-1)(2)+2(0)=0.$$

The lengths are

$$\lVert\mathbf u\rVert=\sqrt{4+1+4}=3,$$

$$\lVert\mathbf v\rVert=\sqrt{1+4}=\sqrt5.$$

Both vectors are nonzero, so the angle formula applies:

$$\cos\theta=\frac{\mathbf u\cdot\mathbf v}{\lVert\mathbf u\rVert\lVert\mathbf v\rVert}=0.$$

With $0\le\theta\le\pi$, this gives $\theta=\pi/2$, or $90^\circ$.

For distance, $\mathbf u-\mathbf v=(1,-3,2)$, giving

$$d=\sqrt{1+9+4}=\sqrt{14}.$$

For the parameter condition,

$$\mathbf u\cdot\mathbf w(s)=2s-1-2=2s-3.$$

It vanishes exactly when $s=3/2$. The vector $\mathbf w(3/2)=(3/2,1,-1)$ is nonzero, so the perpendicular-angle interpretation is valid too.

Algebraically, the zero vector is orthogonal to every vector under the definition “dot product equals zero.” However, it has no direction, and the angle formula would divide by its zero length. The angle with the zero vector is undefined; it is not assigned $90^\circ$ by that formula.

Check the result and conditions

At $s=3/2$, the dot product is $3-1-2=0$. Also, $d^2=14=\lVert\mathbf u\rVert^2+\lVert\mathbf v\rVert^2$, which agrees with the zero dot product. Square roots denote nonnegative lengths; the angle calculation requires both vectors to be nonzero.

Common errors

A dot product returns a number, not a vector. Do not write $d=\lVert\mathbf u\rVert-\lVert\mathbf v\rVert$, forget the square root in a norm, or infer an angle for a zero vector from its dot product.

Choose what to review

If a scalar and a vector got mixed up, revisit dot products and orthogonality. Then label each requested answer here as a scalar, length, angle, or parameter before calculating.

Subtracting row 1 and then row 2 from row 3 makes a zero coefficient row. Row additions preserve the determinant, and a matrix with a zero row has determinant zero. Thus $\det(S)=0$, so $S^{-1}$ does not exist and Cramer quotients would have zero denominator.

On the augmented matrix, the same operations give the final row

$$\left[\begin{array}{rrr|c}0&0&0&k-5\end{array}\right].$$

If $k\ne5$, this is a contradiction and there is no solution. If $k=5$, the remaining equations are $x+y=2$ and $y+z=3$. Let $z=t$, with $t\in\mathbb R$. Then

$$\mathbf x(t)=(t-1,3-t,t).$$

Every real $t$ gives a solution. The additional orthogonality condition is

$$\mathbf x(t)\cdot\mathbf n=(t-1)-(3-t)+t.$$

$$\mathbf x(t)\cdot\mathbf n=3t-4.$$

It vanishes exactly when $t=4/3$, selecting

$$\mathbf x_*=(1/3,5/3,4/3).$$

The extra condition selects one member because the equation $3t-4=0$ has one solution. It does not change the original matrix or its determinant. The original system with $k=5$ still has infinitely many solutions, and $S$ is still singular.

Check the result and conditions

The three row results at $\mathbf x_*$ are $1/3+5/3=2$, $5/3+4/3=3$, and $1/3+10/3+4/3=5$. Its dot product with $\mathbf n$ is $(1-5+4)/3=0$. As a further check, $S\mathbf n=\mathbf0$; changing $t$ by a real amount changes the solution by that amount times $\mathbf n$ without changing the right-hand side. However, the dot product with $\mathbf n$ changes by three times that amount.

Common errors

Do not conclude “no solution” from singularity alone, stop after finding one convenient solution, or say the extra condition creates an inverse. Also, check the constant row: coefficient dependence requires the same relation among the constants for consistency.

Choose what to review

If the consistency test or family was missing, return to contradictions and free variables. If the family was correct but the extra condition was not, revisit the dot-product test for orthogonality, then redo only part 3.