Matrices: questions with answers

Multiply each quantity by its matching unit price, then add the three costs. This is the row-by-column matrix product, giving 225 rupees.

In $a_{23}$ the first subscript is the row and the second is the column, so the substitution is that row number for $i$ and that column number for $j$, giving 10. Feeding them in the other order gives 5. Subscripts start at 1, not 0, so shifting both down by one gives 7 -- a real entry of $A$, but not this one.

Do the scaling first, then the addition, and only then equate corresponding entries. The coefficient 4 multiplies the whole entry, bracket and all, so the bottom-right position gives $4(y - 4) = 8$ and hence $y = 6$; applying 4 to the $y$ alone and not to the $-3$ gives a different value. Striking the coefficient out of the equation altogether gives $x = 8$, $y = 12$, which satisfies no position at all.

The figure has 4 corners and each one needs two numbers. One column per corner fixes which dimension is 4, and the order is written rows first, so it is $2 \times 4$. The reversed shape $4 \times 2$ is what the other convention produces. Writing all 8 coordinates in a single line would be a $1 \times 8$ matrix, in which no corner is a unit at all.

Reading the four corners off the numbered axes gives $A$, $B$, $C$ and $D$ in turn. With one column per vertex each column is a point written vertically, so the first row holds the four $x$-coordinates and the second row the four $y$-coordinates: the matrix is \begin{bmatrix} 1 & 3 & 1 & -1 \\ 0 & 2 & 3 & 2 \end{bmatrix}, of order $2 \times 4$. Putting one vertex per row instead gives \begin{bmatrix} 1 & 0 \\ 3 & 2 \\ 1 & 3 \\ -1 & 2 \end{bmatrix} -- the same numbers in a $4 \times 2$ shape, which is not what was asked for.

Subtraction is entrywise and $A - B$ means each entry of $B$ taken from the entry of $A$ in the same position: $A - B = \begin{bmatrix} 3 & -6 & 4 \\ -5 & 4 & -8 \end{bmatrix}$. The reverse, $B - A$, is \begin{bmatrix} -3 & 6 & -4 \\ 5 & -4 & 8 \end{bmatrix}; these two share not a single entry, because $A$ and $B$ were chosen so that no difference is its own negative. Adding instead of subtracting gives \begin{bmatrix} 7 & 2 & 2 \\ 7 & 10 & 8 \end{bmatrix}, and leaving one row unsubtracted gives \begin{bmatrix} 5 & -2 & 3 \\ -5 & 4 & -8 \end{bmatrix}.

Row 1, column 1 gives $x - 2$ against its partner and row 2, column 1 gives $4x$ against its own. Both yield $x = 7$, and every remaining position is a number matching itself, so the equation really is satisfied -- by that one value. Solving one entry is not enough to conclude it; the agreement of the second is what makes the answer safe.

Being scalar forces $c = 0$ and $a = b = -3$, so $2a + 3b = 5 \times -3 = -15$. Reading the condition as merely 'diagonal' forces $c$ alone and leaves the two diagonal letters free, so it settles nothing and any pair of unequal values -- $a = 3$, $b = 8$, giving 30 -- would do. Reading it as 'the identity' pins the diagonal at 1 and gives 5.

That entry is $a_{21}a_{11} + a_{22}a_{21}$ -- row 2 of $A$ taken against column 1 of $A$, both products added -- and it comes to 24. Squaring the entry that already sits in that position gives 9, which is the entrywise reading of a power. Doubling it gives 6, the corresponding entry of $2A$. And keeping only the first of the two products gives 15: every entry of a product is a whole sum, never one term of it.

The reason is a true statement. The assertion is not: matrix multiplication is not commutative, and a single pair with $AB \ne BA$ settles it. So the correct verdict is that the assertion is false while the reason is true. Rating the assertion true and the reason merely unrelated accepts the very claim being tested -- it is the commuting assumption in the form of a verdict.

$A'$ is $3 \times 2$, so $A'A$ is defined and has order $3 \times 3$ -- but it cannot be an identity matrix that undoes $A$, because $A$ is not square and has no inverse at all. The product exists; what fails is the claim that it is $I$ in any useful sense. And the product certainly is defined: reading $A'$ as though it kept the order $2 \times 3$ would make $A'A$ meaningless, which is a mistake about the transpose, not about this matrix.

$AB$ needs $B$ to have $n$ rows, and $BA$ needs $B$ to have $m$ columns. Together that pins $B$ at $n \times m$ -- the two dimensions of $A$ in the other order. Demanding that $A$ and $B$ have the same order is far too strong: it forces $m = n$, and rules out the perfectly good case of a $2 \times 3$ with a $3 \times 2$. Demanding that both be square is also sufficient without being necessary, and $AB$ existing on its own settles nothing about $BA$.

The skew part is $\frac{1}{2}(A - A')$: the difference is what makes mirrored entries opposite, and the half is what makes the two parts add back to $A$ rather than to $2A$. $\frac{1}{2}(A + A')$ is the symmetric part -- the sum and the difference are not interchangeable. $A - A'$ has the right shape but no half, so the decomposition fails. And $\frac{1}{2}(A' - A)$ is the negative of the skew part.

No. Skewness says $a_{ji} = -a_{ij}$ in every position, and on the diagonal that reads $k = -k$, which forces the entry to be zero. Here the diagonal entry is $k$, which is not zero, so the matrix cannot be skew. It is symmetric -- every such matrix is -- and being symmetric is no help: the two conditions overlap only at the zero matrix.

No. An inverse $B$ would have to satisfy both $AB = I$ and $BA = I$. For the products to exist at all, $B$ must be $3 \times 2$; then $AB$ is $2 \times 2$ and $BA$ is $3 \times 3$, so the two identity matrices would have different orders -- and in fact $BA$ can never be the identity of that order. Only a square matrix has an inverse. Reciprocating the entries does not produce one either; that is not what inverting means.