Q.Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

We are given two $3 \times 3$ matrices $A$ and $B$ and their product $AB$. The entries of $A$ and $B$ contain the unknowns $a, b, c$. Instead of solving a full system of nine equations, we can multiply $A$ and $B$ symbolically and then match the result to the given matrix. This yields equations that determine $a, b, c$ directly.

1. Write down the matrices clearly

We know:

2. Compute the product $AB$ entry by entry

The $(i,j)$ entry of $AB$ is the dot product of row $i$ of $A$ with column $j$ of $B$.

• Row 1 of $A$: $[b, a, 0]$

  • Column 1 of $B$: $[0, b, b]^T$

    $(AB)_{11} = b \cdot 0 + a \cdot b + 0 \cdot b = ab$

  • Column 2 of $B$: $[a, 0, a]^T$

    $(AB)_{12} = b \cdot a + a \cdot 0 + 0 \cdot a = ab$

  • Column 3 of $B$: $[b, c, a]^T$

    $(AB)_{13} = b \cdot b + a \cdot c + 0 \cdot a = b^2 + ac$

• Row 2 of $A$: $[c, 0, b]$

  • Column 1: $c \cdot 0 + 0 \cdot b + b \cdot b = b^2$

  • Column 2: $c \cdot a + 0 \cdot 0 + b \cdot a = ac + ab$

  • Column 3: $c \cdot b + 0 \cdot c + b \cdot a = bc + ab$

• Row 3 of $A$: $[a, a, b]$

  • Column 1: $a \cdot 0 + a \cdot b + b \cdot b = ab + b^2$

  • Column 2: $a \cdot a + a \cdot 0 + b \cdot a = a^2 + ab$

  • Column 3: $a \cdot b + a \cdot c + b \cdot a = ab + ac + ab = 2ab + ac$

So the symbolic product is:

3. Equate to the given matrix

We match each entry:

4. Solve systematically

From $(2,1)$: $b^2 = 1 \Rightarrow b = \pm 1$.

From $(1,1)$: $ab = 2$.

If $b = 1$, then $a = 2$.

If $b = -1$, then $a = -2$.

Check $(3,1)$: $ab + b^2 = 2 + 1 = 3$ works for $b=1, a=2$.

For $b=-1, a=-2$: $ab = 2$ (since $(-2)(-1)=2$), $b^2=1$, sum = 3 — also works. So both sign pairs are possible so far.

Now use $(2,2)$: $ab + ac = 8$. With $ab=2$, we get $2 + ac = 8 \Rightarrow ac = 6$. …

The rank of a matrix tells you the dimension of the vector space spanned by its rows (or columns). For a $3 \times 4$ matrix, the maximum possible rank is 3, but it could be less if rows are linearly dependent. The cleanest way to find rank is to row-reduce to echelon form and count the number of non-zero rows — each non-zero row corresponds to a vector that cannot be written as a combination of the others.

Let’s work through it step by step.

1. Write the matrix and start row reduction. The given matrix is

We want to get zeros below the first pivot. The first pivot is $-4$ in row 1, column 1. To eliminate the first column entries in rows 2 and 3, we use row operations.

2. Eliminate below the first pivot.

   Replace row 2 with $4R_2 - 3R_1$ (this avoids fractions for now):

   $4R_2 = \begin{bmatrix} -12 & 0 & 8 & 12 \end{bmatrix}$,

   $3R_1 = \begin{bmatrix} -12 & -3 & 3 & 12 \end{bmatrix}$,

   so $4R_2 - 3R_1 = \begin{bmatrix} 0 & 3 & 5 & 0 \end{bmatrix}$.

   Thus the new row 2 is $\begin{bmatrix} 0 & 3 & 5 & 0 \end{bmatrix}$.

   Replace row 3 with $2R_3 - R_1$:

   $2R_3 = \begin{bmatrix} -4 & -2 & 0 & -8 \end{bmatrix}$,

   $R_1 = \begin{bmatrix} -4 & -1 & 1 & 4 \end{bmatrix}$,

   so $2R_3 - R_1 = \begin{bmatrix} 0 & -1 & -1 & -12 \end{bmatrix}$.

   The matrix now looks like:

3. Eliminate below the second pivot. The second pivot is $3$ in row 2, column 2. To clear the entry below it, replace row 3 with $3R_3 + R_2$: $3R_3 = \begin{bmatrix} 0 & -3 & -3 & -36 \end{bmatrix}$, $R_2 = \begin{bmatrix} 0 & 3 & 5 & 0 \end{bmatrix}$, so $3R_3 + R_2 = \begin{bmatrix} 0 & 0 & 2 & -36 \end{bmatrix}$. …