We will cover following topics

Row and Column Reduction

The basic elementary operations are:

Interchange one row of the matrix with another of the matrix
Multiply one row of the matrix by a nonzero scalar constant
Replace the one row with the one row plus a constant times another row of the matrix

A system of linear equations can be solved by reducing its augmented matrix into reduced echelon form.

Echelon Form

Echelon Form: A matrix is said to be in echelon form if:

All non-zero rows are above any rows with all zeros, and
The leading coefficient of a non-zero row is always strictly to the right of the leading coefficient of the row above it

For example, the matrix $\left[ \begin{array}{lllll}{1} & {2} & {1} & {7} & {6} \\ {0} & {0} & {2} & {5} & {8} \\ {0} & {0} & {0} & {1} & {7}\end{array}\right]$ is presented in its Echelon form.

Reduced Echelon Form: A matrix is said to be in reduced echelon form if:

It is in row echelon form,
The leading entry in each nonzero row is a 1, and
Each column containing a leading 1 has zeros everywhere else

Example 1: Find the reduced row echelon form of the following matrix: $\begin{bmatrix} 1 & 2 & 1 \\ 2 & 2 & 2 \\ 1 & 0 & 1 \\ \end{bmatrix}$

Sol 1:

\[\begin{bmatrix}1 & 2 & 1 \\ 2 & 2 & 2 \\ 1 & 0 & 1\end{bmatrix}\] \[\rightarrow \begin{bmatrix}1 & 2 & 1\\0 & -2 & 0\\0 & -2 & 0\end{bmatrix}\] \[\rightarrow \begin{bmatrix}1 & 2 & 1\\0 & -2 & 0\\0 & 0 & 0\end{bmatrix}\] \[\rightarrow \begin{bmatrix}1 & 0 & 1\\0 & -2 & 0\\0 & 0 & 0\end{bmatrix}\] \[\rightarrow \begin{bmatrix}1 & 0 & 1\\0 & 1 & 0\\0 & 0 & 0\end{bmatrix}\]

Example 2: Find the reduced row echelon form of the following matrix: $\begin{bmatrix}1 & 1 & 1 & 1\\1 & 1 & 1 & 1\\0 & 1 & 2 & 3\\0 & 1 & 2 & 3\end{bmatrix}$
Sol 2:

\[\begin{bmatrix}1 & 1 & 1 & 1\\1 & 1 & 1 & 1\\0 & 1 & 2 & 3\\0 & 1 & 2 & 3\end{bmatrix}\] \[\rightarrow \begin{bmatrix}1 & 1 & 1 & 1\\0 & 0 & 0 & 0\\0 & 1 & 2 & 3\\0 & 1 & 2 & 3\end{bmatrix}\] \[\rightarrow \begin{bmatrix}1 & 1 & 1 & 1\\0 & 1 & 2 & 3\\0 & 0 & 0 & 0\\0 & 1 & 2 & 3\end{bmatrix}\] \[\rightarrow \begin{bmatrix}1 & 1 & 1 & 1\\0 & 1 & 2 & 3\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}\]

Congruence and Similarity

wo square matrices $A$ and $B$ over a field are called congruent if there exists an invertible matrix $P$ over the same field such that $P^{T} A P=B$.
Two $n \times n$ matrices $A$ and $B$ are called similar if there exists an invertible $n \times n$ matrix $P$ such that $B=P^{-1} A P$.

Inverse of a Matrix

A square matrix $A$ is said to be invertible if there exists a matrix $B$ such that $AB=BA=I$. Such a matrix $B$ is called the inverse of the matrix $A$, and written as $B=A^{-1}$.

For a non-singular matrix $A$, $A^{-1}$ is equal to $Adj(A)/ det(A)$.

Example 1: Determine whether there exists a nonsingular matrix $A$ if $A^{2}=AB+2A$, where $B$ is the following matrix.

\[B=\begin{bmatrix}-1 & 1 & -1\\0 & -1 & 0\\1 & 2 & -2\end{bmatrix}\]

If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{−1}$.

Sol: Suppose that a nonsingular matrix $A$ satisfying $A^{2}=AB+2A$ exists.
Then $A$ is invertible since $A$ is nonsingular, and thus the inverse $A^{-1}$ exists. Multiplying by $A^{−1}$ on the left, we have:

\[A=A^{−1}A^{2}=A^{−1}(AB+2A)=A^{−1}AB+2A^{−1}A=B+2I\]

where $I$ is the $3 \times 3$ identity matrix.

Therefore, if such a nonsingular matrix exists, it must be $A=B+2I$

Since, $B=\begin{bmatrix}-1 & 1 & -1\\0 & -1 & 0\\1 & 2 & -2\end{bmatrix}$,

Therefore, $A=B+2I$ = $\begin{bmatrix}1 & 1 & -1\\0 & 1 & 0\\1 & 2 & 0\end{bmatrix}$.

We still need to check that this matrix is in fact a nonsingular matrix.
To check the non-singularity and to find the inverse matrix at once, we consider the augmented matrix $[A∣I]$ and apply elementary row operations. We have:

\[[A|I]= \left[\begin{array}{ccc|ccc} 1 & 1 & -1 & 1 & 0 & 0\\0 & 1 & 0 & 0 & 1 & 0\\1 & 2 & 0 & 0 & 0 & 1\end{array}\right]\] \[\xrightarrow[]{R_{3}-R_{1}} \left[\begin{array}{ccc|ccc}1 & 1 & -1 & 1 & 0 & 0\\0 & 1 & 0 & 0 & 1 & 0\\0 & 1 & 1 & -1 & 0 & 1\end{array} \right]\] \[\xrightarrow[]{R_{1}-R_{2}; R_{3}-R_{2}} \left[ \begin{array}{ccc|ccc} 1 & 0 & -1 & 1 & -1 & 0\\0 & 1 & 0 & 0 & 1 & 0\\0 & 0 & 1 & -1 & -1 & 1 \end{array} \right]\] \[\xrightarrow[]{R_{1}+R_{3}} \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & -2 & 1\\0 & 1 & 0 & 0 & 1 & 0\\0 & 0 & 1 & -1 & -1 & 1 \end{array} \right]\]

The left part of the last matrix is the identity matrix, and thus the matrix $A$ is invertible and the inverse matrix is the right half:

\[A^{-1}=\begin{bmatrix}0 & -2 & 1\\0 & 1 & 0\\-1 & -1 & 1\end{bmatrix}\]

Example 2: Let $A$, $B$, $C$ be the following $3 \times 3$ matrices. $A=\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{bmatrix}$, $B=\begin{bmatrix}1 & 0 & 1\\0 & 3 & 0\\1 & 0 & 5\end{bmatrix}$, $C=\begin{bmatrix}-1 & 0 & 1\\0 & 5 & 6\\3 & 0 & 1\end{bmatrix}$.
Then compute and simplify the following expression. $(A^{T}−B)^{T}+C(B^{−1}C)^{−1}$.

Sol: We first use the properties of transpose matrices and inverse matrices and simplify the expression. Note that we have
$(A^{T}−B)^{T}=(A^{T})^{T}−B^{T}=A−B$ since the double transpose $(A^{T})^{T}=A$ and $B$ is a symmetric matrix. Also, note that we have
$(B^{−1}C)^{−1}=C^{−1}(B^{−1})^{−1}=C^{−1}B$ since $(B−1)−1=B$.

Care must be taken when you distribute the inverse sign in the first equality. We needed to switch the order of the product.
Then we have. $C(B^{−1}C)^{−1}=CC^{−1}B=IB=B$, where $I$ is the $3 \times 3$ identity matrix.

Therefore, the given expression can be simplified into $(A^{T}−B)^{T}+C(B^{−1}C)^{−1}$=$$A−B+B=A$$.

Hence we have:

$(A^T−B)^{T}+C(B^{−1}C)^{−1}$=$A$=$\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{bmatrix}$

Rank of a Matrix

The rank of a matrix is defined as the maximum number of linearly independent row or column vectors.

Example 1: Determine the values of a real number $a$ such that the matrix $A=\begin{bmatrix}3 & 0 & a\\2 & 3 & 0\\0 & 18a & a+1\end{bmatrix}$ is nonsingular.
Sol: We apply elementary row operations and obtain:

\[A=\begin{bmatrix}3 & 0 & a\\2 & 3 & 0\\0 & 18a & a+1\end{bmatrix}\] \[\xrightarrow[]{R_{1} - R_{2}} \begin{bmatrix}1 & -3 & a\\2 & 3 & 0\\0 & 18a & a+1\end{bmatrix}\] \[\xrightarrow[]{R_{2} - 2R_{1}} \begin{bmatrix}1 & -3 & a\\0 & 9 & -2a\\0 & 18a & a+1\end{bmatrix}\] \[\xrightarrow[]{R_{3} - (2a)R_{2}} \begin{bmatrix}1 & -3 & a\\0 & 9 & -2a\\0 & 0 & 4a^{2} + a + 1\end{bmatrix}\]

From this, we see that the matrix A is nonsingular if and only if the (3,3) entry $4a^{2}+a+1$ is not zero. By the quadratic formula, we see that $a=\frac{-1 \pm \sqrt{-15}}{8}$ are solutions of $4a^{2}+a+1=0$. Note that these are not real numbers. Thus, for any real number a, we have $4a^{2}+a+1 \neq 0$. Hence, we can divide the third row by this number, and eventually we can reduce it to the identity matrix.
So the rank of $A$ is 3, and $A$ is nonsingular for any real number $a$.

Example 2: Let $P$ be a square matrix and its characteristic polynomial is given by $p(t)=(t−1)^{3}(t−2)^{2}(t−3)^{4}(t−4)$. Find the rank of $A$.
Sol: Note that the degree of the characteristic polynomial $p(t)$ is the size of the matrix $P$. Since the degree of $p(t)$ is 3+2+4+1=10, the size of the matrix $P$ is 10 $\times$ 10. From the characteristic polynomial, we see that the eigenvalues of $P$ are 1,2,3,4. In particular, 0 is not an eigenvalue of $P$. Hence the null space of $P$ is zero dimensional, that is, the nullity of $P$ is 0. By the rank-nullity theorem, we have:
rank of $P$ + nullity of $P=10$. Hence the rank of $P$ is 10.

Solution of a System of Linear Equations

Let a system of linear equations be represented by the equation $AX=B$ and let $C=[A:B]$ be its augmented matrix. Then, the system is:

Consistent, if $rank(A)=rank(C)$ and has either unique or infinitely many solutions
(i) If $rank(A)=rank(C)=n$, the system has a unique solution
(ii) If $rank(A)=rank(C) < n$, the system has infinitely many solutions
Inconsistent, if $rank(A) \neq rank(C)$ and has no solutions

Example 1: Solve the following system of linear equations and give the vector form for the general solution.

\[x_{1}−x_{3}−2x_{5}=1\] \[x_{2}+3x_{3}−x_{5}=2\] \[x_{1}−2x_{3}+x_{4}−3x_{5}=0\]

Sol: We solve the system by Gauss-Jordan elimination.
The augmented matrix of the system is given by

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

We apply the elementary row operations as follows:

\[\left[ \begin{array}{ccccc|c} 1 & 0 & -1 & 0 & -2 & 1\\0 & 1 & 3 & 0 & -1 & 2\\2 & 0 & -2 & 1 & -3 & 0 \end{array} \right]\] \[\xrightarrow[]{R_{3}-2R_{1}} \left[ \begin{array}{ccccc|c} 1 & 0 & -1 & 0 & -2 & 1\\0 & 1 & 3 & 0 & -1 & 2\\0 & 0 & 0 & 1 & 1 & -2 \end{array} \right]\]

Then the last matrix is in reduced row echelon form.
The variables $x_{1},x_{2},x_{4}$ correspond to the leading 1’s of the last matrix, hence they are dependent variables and the rest $x_{3},x_{5}$ are free variables.
From the last matrix we obtain the general solution

\[x_{1}=x_{3}+2x_{5}+1; x_{2}=−3x_{3}+x_{5}+2; x_{4}=−x_{5}−2\]

The vector form for the general solution is obtained by substituting these into the vector x. We have

\[x=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{bmatrix}\] \[=\begin{bmatrix}x_{3} + 2x_{5} + 1\\-3x_{3} + x_{5} + 2\\x_{3}\\-x_{5}-2\\x_{5}\end{bmatrix}\] \[=x_{3}\begin{bmatrix}1\\-3\\1\\0\\0\end{bmatrix} + x_{5}\begin{bmatrix}2\\1\\0\\-1\\1\end{bmatrix} + \begin{bmatrix}1\\2\\0\\-2\\0\end{bmatrix}\]

Therefore, the vector form for the general solution is given by

\[x=x_{3}\begin{bmatrix}1\\-3\\1\\0\\0\end{bmatrix} + x_{5}\begin{bmatrix}2\\1\\0\\-1\\1\end{bmatrix} + \begin{bmatrix}1\\2\\0\\-2\\0\end{bmatrix}\]

where $x_{3},x_{5}$ are free variables.

Example 2: For what value(s) of $a$ does the system have nontrivial solutions?

$x_{1}+2x_{2}+x_{3}=0$ $−x_{1}−x_{2}+x_{3}=0$ $3x_{1}+4x_{2}+ax_{3}=0$

Sol: First note that the system is homogeneous and hence it is consistent. Thus if the system has a nontrivial solution, then it has infinitely many solutions.

This happens if and only if the system has at least one free variable. The number of free variables is $n−r$, where $n$ is the number of unknowns and $r$ is the rank of the augmented matrix.

To find the rank, we reduce the augmented matrix by elementary row operations.

\[\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 0\\-1 & -1 & 1 & 0\\3 & 4 & a & 0\end{array} \right]\] \[\xrightarrow[R_{3}-3R_{1}]{R_{2}+R_{1}} \left[ \begin{array}{ccc|c} 1 & 2 & 1 & 0\\0 & 1 & 2 & 0\\0 & -2 & a-3 & 0\end{array} \right]\] \[\xrightarrow[]{R_{3}+2R_{2}} \left[ \begin{array}{ccc|c} 1 & 2 & 1 & 0\\0 & 1 & 2 & 0\\0 & 0 & a+1 & 0\end{array} \right]\]

The last matrix is in row echelon form.

Thus if $a$+1=0, then the third row is a zero row, hence the rank is 2. In this case we have $n−r$=3−2=1 free variable. Thus there are infinitely many solutions. In particular, the system has nontrivial solutions.

On the other hand, if $a+1 \neq 0$, then the rank is 3 and there is no free variables since $n − r$ = 3 − 3 = 0.

To summarize, the system has nontrivial solutions exactly when $a$ = −1

PYQs

Echelon Form

1) Reduce the following matrix to row echelon form and hence find its rank: $\begin{bmatrix}{1} & {2} & {3} & {4} \\ {2} & {1} & {4} & {5} \\ {1} & {5} & {5} & {7} \\ {8} & {1} & {14} & {17}\end{bmatrix}$.

[2015, 10M]

</div>

Congruence and Similarity

1) Reduce the quadratic form $q(x, y, z) :=x^{2}+2 y^{2}-4 x z+4 y z+7 z^{2}$ to canonical form. Is $q$ positive definite?

[2007, 15M]

2) Find the quadratic form $q(x, y)$ corresponding to the symmetric matrix $A= \begin{bmatrix}{5} & {-3} \\ {-3} & {8}\end{bmatrix}$. Is this quadratic form positive definite? Justify your answer.

[2006, 15M]

3) Define a positive definite quadratic form. Reduce the quadratic form $x_1^2+x_3^2+2x_1x_2+2x_2x_3$ to canonical form. Is this quadratic form positive definite?

[2004, 15M]

4) Reduce the quadratic form given below to canonical form and find its rank and signature.

\[x^{2}+4 y^{2}+9 z^{2}+u^{2}-12 y z+6 z x-4 x y-2 x u-6 z u\]

[2003, 15M]

5) Show that the real quadratic form $\phi=n\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)-\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2}$ in $n$ variables is positive semi-definite.

[2001, 15M]

Inverse of a Matrix

1) Let $A$ be a $3\times 2$ matrix and $B$ a $2\times 3$ matrix. Show that $C=A\cdot B$ is a singular matrix.

[2018, 10M]

2) Using elementary row operations, find the inverse of $A= \begin{bmatrix}{1} & {2} & {1} \\ {1} & {3} & {2} \\ {1} & {0} & {1}\end{bmatrix}$.

[2016, 6M]

3) Find the inverse of the matrix: $A= \begin{bmatrix}{1} & {3} & {1} \\ {2} & {-1} & {7} \\ {3} & {2} & {-1}\end{bmatrix}$ by using elementary row operations. Hence solve the system of linear equations $x+3 y+z=10$, $2 x-y+7 z=12$, $3 x+2 y-z=4$.

[2013, 10M]

4) Let $A$ be a non-singular $n \times n$ square matrix. Show that $A \cdot (a d j A)= \vert A \vert \cdot I_{n}$. Hence show that $\vert adj(adj A) \vert =\vert A \vert^{(n-1)^{2}}$.

[2011, 10M]

5) Show that the matrix $A$ is invertible if and only if the $adj(A)$ is invertible. Hence find $\vert a d j(A) \vert$.

[2008, 12M]

6) Let $A$ be a non-singular matrix. Show that if $I+A+A^{2}+\ldots .+A^{n}=0$, then $A^{-1}=A^{n}$.

[2008, 15M]

7) Find the inverse of the matrix given below using elementary row operations only:

\[\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\]

[2005, 15M]

Rank of a Matrix

1) Let $A=\begin{bmatrix} 5& 7& 2& 1\\ 1& 1& -8& 1\\ 2& 3& 5& 0\\ 3& 4& -3& 1 \end{bmatrix}$.

(i) Find the rank of matrix $A$.

[15M]

(ii) Find the dimension of the subspace

\[V= \left\{ (x_1,x_2,x_3,x_4) \in R^4 \vert A\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=0 \right\}\]

[5M]

2) Let $T:R^2 \to R^2$ be a linear map such that that $T(2,1)=(5,7)$ and $T(1,2)+(3,3)$. If $A$ is the matrix corresponding to $T$ with repect to the standard bases $e_1$, $e_2$, then find $Rank(A)$.

[2019, 10M]

3) Using elementary row or column operations, find the rank of the matrix $\left[ \begin{array}{cccc}{0} & {1} & {-3} & {-1} \\ {0} & {0} & {1} & {1} \\ {3} & {1} & {0} & {2} \\ {1} & {1} & {-2} & {0}\end{array}\right]$

[2014, 10M]

4) Find the rank of the matrix $A=\begin{bmatrix}{1} & {2} & {3} & {4} & {5} \\ {2} & {3} & {5} & {8} & {12} \\ {3} & {5} & {8} & {12} & {17} \\ {5} & {8} & {12} & {17} & {23} \\ {8} & {12} & {17} & {23} & {30}\end{bmatrix}$

[2013, 8M]

5) Using elementary row operations, find the rank of the matrix $\begin{bmatrix}{3} & {-2} & {0} & {-1} \\ {0} & {2} & {2} & {1} \\ {1} & {-2} & {-3} & {-2} \\ {0} & {1} & {2} & {1}\end{bmatrix}$

[2006, 15M]

Solution of a System of Linear Equations

1) If $A=\begin{bmatrix} 1& 2& 1\\ 1& -4& 1\\ 3& 0& -3 \end{bmatrix}$ and $B=\begin{bmatrix} 2& 1& 1\\1& -1& 0\\2& 1& -1 \end{bmatrix}$, then show that $AB=6I_3$. Use this result to solve the following system of equations:

$2x+y+z=5$
$x-y=0$
$2x+y-z=1$

[2019, 10M]

2) For the system of linear equations $x+3y-2z=-1$
$5y+3z=-8$
$x-2y-5z-7$
determine which of the following statements are true and which are false:

(i) The system has no solution.
(ii) The system has a unique solution.
(iii) The system has infinitely many solutions.

[2018, 13M]

3) Consider the following system of equation in $x$, $y$, $z$:
$\quad x+2 y+2 z=1$
$\quad x+a y+3 z=3$
$\quad x+11 y+a z=b$

i) For which values of $a$ does the system have a unique solution?
ii) For which of values $(a, b)$ does the system have more than one solution?

[2017, 15M]

4) Using elementary row operation find the condition that the linear equations have a solution: $\quad x-2 y+z=a$
$\quad 2 x+7 y-3 z=b$
$\quad 3 x+5 y-2 z=c$

[2016, 7M]

5) Investigate the values of $\lambda$ and $\mu$ so that the equations $x+y+z=6$, $x+2 y+3 z=10$, $x+2 y+\lambda z=\mu$ have:
i) no solution
ii) unique solution,
iii) an infinite number of solutions.

[2014, 10M]

6) Let $A= \begin{bmatrix}{1} & {0} & {-1} \\ {3} & {4} & {5} \\ {0} & {6} & {7}\end{bmatrix}$, $X= \begin{bmatrix}{x} \\ {y} \\ {z}\end{bmatrix}$, $B=\begin{bmatrix}{2} \\ {6} \\ {5}\end{bmatrix}$.

Solve the system of equations given by $A X=B$.

Using the above, also solve the system of equations $A^{T} X=B$, where $A^{T}$ denotes the transpose of matrix $A$.

[2011, 10M]

7) Investigate for what values of $\mu$ and $\lambda$ the equations $\quad x+y+z=6$
$\quad x+2 y+3 z=10$
$\quad x+2 y+\lambda z=\mu$ have:
i) no solution,
ii) a unique solution,
iii) infinitely many solutions

[2006, 15M]

8) Verify whether the following system of equation is consistent:

$\quad x+3 z=5$
$\quad -2 x+5 y-z=0$
$\quad -x+4 y+z=4$

[2004, 15M]

9) Solve the following system of linear equations:

$\quad x_{1}-2 x_{2}-3 x_{3}+4 x_{4}=-1$
$\quad -x_{1}+3 x_{2}+5 x_{3}-5 x_{4}-2 x_{5}=0$
$\quad 2 x_{1}+x_{2}-2 x_{3}+3 x_{4}-4 x_{5}=17$

[2002, 15M]

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