Class 10 Optional Mathematics Matrix and Determinant

Here, the given row matrix \(R=\begin{bmatrix}3 & 2 & 1\end{bmatrix}\) has order \(1\times 3\).

Interchange the rows and columns (Row \(i\) becomes Column \(i\)):

Here, the given column matrix \(W=\begin{bmatrix}-3 \\ 4 \\ 5\end{bmatrix}\) has order \(3\times 1\).

Interchange the rows and columns (Row \(i\) becomes Column \(i\)):

Here, the given square matrix \(X=\begin{bmatrix}-1 & 5 \\ 3 & 7\end{bmatrix}\) has order \(2\times 2\).

Interchange the rows and columns (Row \(i\) becomes Column \(i\)):

Here, the given matrix \(Y=\begin{bmatrix}3 & 4 & 7 \\ 2 & -1 & 10\end{bmatrix}\) has order \(2\times 3\).

Interchange the rows and columns (Row \(i\) becomes Column \(i\)):

Here, the given matrix \(Z=\begin{bmatrix}0 & 1 \\ 5 & -4 \\ 4 & 5\end{bmatrix}\) has order \(3\times 2\).

Interchange the rows and columns (Row \(i\) becomes Column \(i\)):

Here, the given square matrix \(M=\begin{bmatrix}6 & 7 & 1 \\ 0 & 5 & 2 \\ -1 & 3 & 4\end{bmatrix}\) has order \(3\times 3\).

Interchange the rows and columns (Row \(i\) becomes Column \(i\)):

Here, the given matrix is \(P=\begin{bmatrix}9 & 4 \\ 8 & 5\end{bmatrix}\).

Interchange the rows and columns of \(P\) (Row 1 \(\begin{bmatrix}9&4\end{bmatrix}\) becomes Column 1, Row 2 \(\begin{bmatrix}8&5\end{bmatrix}\) becomes Column 2):

Again, interchange the rows and columns of \(P^T\):

The matrix \((P^T)^T\) is identical to the original matrix \(P\).

Here, the given matrix is \(M=\begin{bmatrix}p & q & r \\ a & b & c \\ x & y & z\end{bmatrix}\).

a. Finding \(M^T\) and \((M^T)^T\)

Change the rows of \(M\) into columns:

Take the transpose of \(M^T\):

b. Relation of \(M\) and \((M^T)^T\)

All corresponding elements of \(M\) and \((M^T)^T\) are identical.

Here, \(A=\begin{bmatrix}1 & 3 \\ -2 & 4\end{bmatrix}\) and \(B=\begin{bmatrix}0 & -3 \\ 2 & 5\end{bmatrix}\).

a. Find \(A^T+B^T\)

First, \(A^T=\begin{bmatrix}1 & -2 \\ 3 & 4\end{bmatrix},\quad B^T=\begin{bmatrix}0 & 2 \\ -3 & 5\end{bmatrix}\).

b. Find \((A+B)^T\)

c. Comparison

Both results are equal to \(\begin{bmatrix}1 & 0 \\ 0 & 9\end{bmatrix}\), so \((A+B)^T=A^T+B^T\).

Reason: the transpose of the sum of two matrices is equal to the sum of their transposes.

Here, \(A=\begin{bmatrix}1 & 3 \\ 5 & 7\end{bmatrix}\) and \(B=\begin{bmatrix}2 & 4 \\ 6 & 8\end{bmatrix}\).

a. Proof

L.H.S.:

R.H.S.:

From equations (i) and (ii), L.H.S. = R.H.S., \((A+B)^T=A^T+B^T\) is proved.

b. Symmetric or not

A matrix \(M\) is symmetric if \(M^T=M\). Here \(A+B=\begin{bmatrix}3 & 7 \\ 11 & 15\end{bmatrix}\) but \((A+B)^T=\begin{bmatrix}3 & 11 \\ 7 & 15\end{bmatrix}\), and \(a_{12}=7\neq a_{21}=11\).

Here, \(A^T=B\) and \(A^T=\begin{bmatrix}a&c\\b&d\end{bmatrix}\).

Equating the corresponding elements of equal matrices:

• Element \((1,1)\): \(a=4\)
• Element \((1,2)\): \(c=-3\)
• Element \((2,1)\): \(b=-2\)
• Element \((2,2)\): \(d=-5\)

Here, the order of the first matrix is \(1\times 2\) and the order of the second matrix is \(2\times 1\).

The number of columns of the first matrix (\(2\)) equals the number of rows of the second matrix (\(2\)), so the product is defined and its order is \(1\times 1\).

Here, the order of the first matrix is \(2\times 1\) and the order of the second matrix is \(1\times 2\).

The number of columns of the first matrix (\(1\)) equals the number of rows of the second matrix (\(1\)), so the product is defined and its order is \(2\times 2\).

Here, the order of the first matrix is \(1\times 3\) and the order of the second matrix is \(3\times 1\).

The number of columns of the first matrix (\(3\)) equals the number of rows of the second matrix (\(3\)), so the product is defined and its order is \(1\times 1\).

Here, the order of the first matrix is \(3\times 2\) and the order of the second matrix is \(2\times 3\).

The number of columns of the first matrix (\(2\)) equals the number of rows of the second matrix (\(2\)), so the product is defined and its order is \(3\times 3\).

Here, the order of the first matrix is \(2\times 2\) and the order of the second matrix is \(2\times 2\).

The number of columns of the first matrix (\(2\)) equals the number of rows of the second matrix (\(2\)), so the product is defined and its order is \(2\times 2\).

Here, the order of the first matrix is \(2\times 2\) and the order of the second matrix is \(3\times 2\).

The number of columns of the first matrix (\(2\)) is not equal to the number of rows of the second matrix (\(3\)).

Here, \(Q=\begin{bmatrix}3 & 6 \\ -2 & -4\end{bmatrix}\) and \(Q^2=Q\times Q\).

Here, \(M=\begin{bmatrix}-2 & 3 \\ -1 & 2\end{bmatrix}\) and \(N=\begin{bmatrix}4 & -3 \\ 5 & 0\end{bmatrix}\).

Calculate \(MN\)

Calculate \(NM\)

Here, \(C=\begin{bmatrix}3 & 5 \\ 0 & 1\end{bmatrix}\), \(D=\begin{bmatrix}-2 & 4 \\ -1 & 3\end{bmatrix}\) and \(E=\begin{bmatrix}2 & -3 \\ 6 & 4\end{bmatrix}\).

Calculate \(D+E\)

Calculate \(C(D+E)\)

Here, \(A=\begin{bmatrix}1 & -1 \\ -1 & 1\end{bmatrix}\).

L.H.S.:

R.H.S.:

From equations (i) and (ii), L.H.S. = R.H.S.,

Here, \(X=\begin{bmatrix}1 & -1 \\ 3 & -3\end{bmatrix}\) and \(Y=\begin{bmatrix}2 & -5 \\ 2 & -5\end{bmatrix}\).

A matrix whose elements are all zero is a zero matrix \(O\).

Here, \(R=\begin{bmatrix}2 & -3 \\ 1 & -2\end{bmatrix}\).

Note on the printed question

For this matrix, \(A^2-A+6I=\begin{bmatrix}12 & 0 \\ 0 & 12\end{bmatrix}\neq O\), so the identity as printed is not true. The identity that holds is \(A^2-A-6I=O\) (the sign of \(6I\) appears to be misprinted). The full verification is given below.

Here, \(A=\begin{bmatrix}2 & 1 \\ 4 & -1\end{bmatrix}\).

Calculate \(A^2\)

Calculate \(A^2-A-6I\)

Here, \(UV=U+V\).

Equating the corresponding elements of \(UV=U+V\):

Here, \(X=\begin{bmatrix}6 & 3 \\ -2 & 0\end{bmatrix}\) and \(Y=\begin{bmatrix}2 & 5 \\ 1 & -4\end{bmatrix}\).

L.H.S.: first find \(XY\).

R.H.S.: \(Y^T=\begin{bmatrix}2 & 1 \\ 5 & -4\end{bmatrix}\) and \(X^T=\begin{bmatrix}6 & -2 \\ 3 & 0\end{bmatrix}\).

From equations (i) and (ii), L.H.S. = R.H.S.,

Here, \(A=\begin{bmatrix}0 & 2 \\ 1 & 3\end{bmatrix}\) and \(B=\begin{bmatrix}3 & 5 \\ 4 & 6\end{bmatrix}\).

Here, \(A=\begin{bmatrix}0 & 2 \\ 1 & 3\end{bmatrix}\), \(B=\begin{bmatrix}3 & 5 \\ 4 & 6\end{bmatrix}\), \(C=\begin{bmatrix}6 & 8 \\ 7 & 9\end{bmatrix}\) and \(I=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\).

First, \(BC\):

L.H.S.:

R.H.S. (using \(AB=\begin{bmatrix}8 & 12 \\ 15 & 23\end{bmatrix}\)):

From equations (i) and (ii), L.H.S. = R.H.S.,

Here, \(A=\begin{bmatrix}0 & 2 \\ 1 & 3\end{bmatrix}\), \(B=\begin{bmatrix}3 & 5 \\ 4 & 6\end{bmatrix}\), \(C=\begin{bmatrix}6 & 8 \\ 7 & 9\end{bmatrix}\) and \(I=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\).

First, \(B+C\):

L.H.S.:

R.H.S.: \(AB=\begin{bmatrix}8 & 12 \\ 15 & 23\end{bmatrix}\) and

From equations (i) and (ii), L.H.S. = R.H.S.,

Here, \(A=\begin{bmatrix}0 & 2 \\ 1 & 3\end{bmatrix}\) and \(I=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\).

Here, \(A=\begin{bmatrix}0 & 2 \\ 1 & 3\end{bmatrix}\) and \(B=\begin{bmatrix}3 & 5 \\ 4 & 6\end{bmatrix}\).

L.H.S.: \(AB=\begin{bmatrix}8 & 12 \\ 15 & 23\end{bmatrix}\), so

R.H.S.: \(B^T=\begin{bmatrix}3 & 4 \\ 5 & 6\end{bmatrix}\) and \(A^T=\begin{bmatrix}0 & 1 \\ 2 & 3\end{bmatrix}\).

From equations (i) and (ii), L.H.S. = R.H.S.,

Here, the determinant is of order \(1\times 1\).

Here, the determinant is of order \(1\times 1\), so its value is the element itself (it is not an absolute value).

Here,

Here,

Here,

Here,

Here,

Here,

Using \(x^3+y^3=(x+y)(x^2-xy+y^2)\) and \(x^3-y^3=(x-y)(x^2+xy+y^2)\):

Note on the printed answer

The answer key of the textbook prints \(2y^2\). Expanding the determinant as shown gives \(2y^3\).

Here, the matrix is of order \(1\times 1\).

Here, the matrix is of order \(1\times 1\).

Here,

Here,

Here,

Here,

Here, \(\begin{vmatrix}3&x\\2&8\end{vmatrix}=20\).

Here, \(\begin{vmatrix}5&-2\\2x&-3\end{vmatrix}=1\).

Here, \(\begin{vmatrix}3x&2\\-4&3\end{vmatrix}=8\).

Here, \(\begin{vmatrix}x&-2x\\3x&4x\end{vmatrix}=100\).

Here, \(A=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}\) and \(B=\begin{bmatrix}1 & -5 \\ 6 & 2\end{bmatrix}\).

Here, \(A=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}\) and \(B=\begin{bmatrix}1 & -5 \\ 6 & 2\end{bmatrix}\).

Here, \(A=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}\) and \(B=\begin{bmatrix}1 & -5 \\ 6 & 2\end{bmatrix}\).

Here, \(A=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}\) and \(B=\begin{bmatrix}1 & -5 \\ 6 & 2\end{bmatrix}\).

Here, \(M=\begin{bmatrix}5 & 2 \\ 4 & 3\end{bmatrix}\) and \(N=\begin{bmatrix}1 & -3 \\ 6 & 2\end{bmatrix}\).

R.H.S.

L.H.S.

Here, \(A=\begin{bmatrix}5 & 1 \\ 2 & 1\end{bmatrix}\) and \(B=\begin{bmatrix}1 & -1 \\ -2 & 3\end{bmatrix}\).

As printed, \(AB=\begin{bmatrix}3 & -2 \\ 0 & 1\end{bmatrix}\neq I\).

Note on the printed matrix

As printed, \(AB\neq I\), so these two matrices are not inverses of each other. The matrix \(B\) is the inverse of \(A=\begin{bmatrix}3 & 1 \\ 2 & 1\end{bmatrix}\) (the entry \(5\) looks like a misprint of \(3\)). With this matrix:

Here, \(A=\begin{bmatrix}5 & -3 \\ -3 & 2\end{bmatrix}\) and \(B=\begin{bmatrix}2 & 3 \\ 3 & 5\end{bmatrix}\).

Here, \(A=\begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix}\) and \(B=\begin{bmatrix}3 & -1 \\ -5 & 2\end{bmatrix}\).

Here, \(A=\begin{bmatrix}4 & 7 \\ 5 & 9\end{bmatrix}\) and \(B=\begin{bmatrix}9 & -7 \\ -5 & 4\end{bmatrix}\).

Here, \(A=\begin{bmatrix}2 & 1 \\ 3 & 4\end{bmatrix}\).

Since \(|A|\neq 0\), the inverse exists. Interchange the diagonal elements and change the signs of the other two elements:

Here, \(B=\begin{bmatrix}7 & 4 \\ 3 & 2\end{bmatrix}\).

Since \(|B|\neq 0\), the inverse exists. Interchange the diagonal elements and change the signs of the other two elements:

Here, \(P=\begin{bmatrix}2 & 0 \\ 3 & 1\end{bmatrix}\).

Since \(|P|\neq 0\), the inverse exists. Interchange the diagonal elements and change the signs of the other two elements:

Here, \(R=\begin{bmatrix}\cos A&-\sin A\\\sin A&\cos A\end{bmatrix}\).

Since the matrices are inverses of each other, their product is the identity matrix.

Equating the corresponding elements:

Check: \(-a+b=-3+3=0\). ✓

Since the matrices are inverses of each other, their product is the identity matrix.

From the top-right element:

From the bottom-right element: \(-5q+3p=1\).

• If \(p=2\): \(-5q+6=1\), so \(q=1\). Check the top-left element: \(3(2)+1(4-9)=6-5=1\) ✓.
• If \(p=0\): \(q=-\frac{1}{5}\), and the top-left element gives \(-9q=\frac{9}{5}\neq 1\), so \(p=0\) is rejected.

Here, \(A^{-1}=\begin{bmatrix}3 & 2 \\ 5 & 4\end{bmatrix}\). We know that \(A=(A^{-1})^{-1}\).

Here, \(A=\begin{bmatrix}2 & 0 \\ 5 & 3\end{bmatrix}\) and \(B=\begin{bmatrix}4 & 1 \\ 7 & 2\end{bmatrix}\).

a. Determinants

b. Inverse matrices

c. \((AB)^{-1}\)

d. Verification

R.H.S.:

From equations (i) and (ii), L.H.S. = R.H.S.,