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Free Matrix Worksheet with Answers PDF
Problem 1: If matrix A has 6 number elements, then determine the order of the matrix.
Solution:
Given that, the number of elements is 8.
Let’s write all the possible factors of the number 6.
i.e., 6 = 1 x 6
6 = 3 x 2
6 = 2 x 3
6 = 6 x 1
we can get the number 6 in four ways.
Therefore, there are four possible ways or orders of the 6 number matrix 6 elements are 1 x 6, 2 x 3, 3 x 2, and 6 x 1.
Problem 2: What is the order of a matrix given below? The matrix \( A = \left[
\begin{matrix}
1 & 3 & 7\cr
14 & 5 & 6\cr
\end{matrix}
\right]
\)
Solution:
Given that, the matrix \( A = \left[
\begin{matrix}
1 & 3 & 7\cr
14 & 5 & 6\cr
\end{matrix}
\right]
\)
The number of rows in a given matrix is A = 2.
The number of columns in a given matrix is A = 3
Thus, the order of matrix A is 2 x 3.
Problem 3:Â Prove that the additive identity of \( A = \left[
\begin{matrix}
2 & 3 & 4\cr
6 & 9& 7\cr
\end{matrix}
\right]
\) is a null matrix.
Solution:
Given that, the non zero matrix is \( A = \left[
\begin{matrix}
2 & 3 & 4\cr
6 & 9& 7\cr
\end{matrix}
\right]
\)
To prove that the additive identity of a given non-zero matrix is a null matrix, we need to add the given matrix with the null matrix.
\( A = \left[
\begin{matrix}
2 & 3 & 4\cr
6 & 9& 7\cr
\end{matrix}
\right]
\) + \( A = \left[
\begin{matrix}
0 & 0 & 0\cr
0 & 0& 0\cr
\end{matrix}
\right]
\)
The value of \( A = \left[
\begin{matrix}
2 & 3 & 4\cr
6 & 9& 7\cr
\end{matrix}
\right]
\)
Thus, the addition of a null matrix to the given matrix also gives the same matrix.
Therefore, the null matrix is the additive identity of the given matrix.
Problem 4:Â Determine if the given matrix is a unit matrix. The value of \( A = \left[
\begin{matrix}
1 & 1\cr
2 & 1\cr
\end{matrix}
\right]
\)
Solution:
The given matrix is \( A = \left[
\begin{matrix}
1 & 1\cr
2 & 1\cr
\end{matrix}
\right]
\)
The Identity Matrix consists of all the diagonal elements as 1 and other elements as zero. In the given matrix A, all the diagonal matrices are 1 but the remaining elements are not zero.
Therefore, the given matrix \( A = \left[
\begin{matrix}
1 & 1\cr
2 & 1\cr
\end{matrix}
\right]
\) is not an identity matrix.
Problem 5: Find the values of ‘m’ and ‘n’ in the given matrix A such that A is a strictly upper triangular matrix. The matrix \( A = \left[
\begin{matrix}
3m & 4\cr
n & 5\cr
\end{matrix}
\right]
\)
Solution:
Given matrix is \( A = \left[
\begin{matrix}
3m & 4\cr
n & 5\cr
\end{matrix}
\right]
\)
For the strictly upper triangular matrix, the elements below the diagonal are zero and the elements of the main diagonal are zero.
Therefore, we must have 3m = 0 that is m = 0 and n = 0.
Hence, the values of m and n are 0.
Problem 6: The matrix \( A = \left[
\begin{matrix}
12 & 5\cr
6 & 4\cr
\end{matrix}
\right]
\) and \( B = \left[
\begin{matrix}
4 & 3\cr
2 & 5\cr
\end{matrix}
\right]
\). What is the added value of Matrices A and B?
Solution:
As given in the question,
The matrix A is \( A = \left[
\begin{matrix}
12 & 5\cr
6 & 4\cr
\end{matrix}
\right]
\) and \( B = \left[
\begin{matrix}
4 & 3\cr
2 & 5\cr
\end{matrix}
\right]
\)
Now, we will find the addition value of A and B.
Here, both the matrices have the same order of 2 × 2. Now, we can add the elements of the first matrix with the respective elements of the second matrix.
After, addition the matrices is \( A +B= \left[
\begin{matrix}
12+4 & 5+\3cr
6+2 & 4+5\cr
\end{matrix}
\right]
\)
Therefore, the addition matrices value of A +B is \( \left[
\begin{matrix}
16 & 8\cr
8 & 9\cr
\end{matrix}
\right]
\)
Project 7: Subtract X and Y where
\( X = \left[
\begin{matrix}
12 & 8\cr
5 & -4\cr
\end{matrix}
\right]
\) and \( Y = \left[
\begin{matrix}
-5 & -4\cr
2 & 1\cr
\end{matrix}
\right]
\)
Solution:
Given matrices are \( X = \left[
\begin{matrix}
12 & 8\cr
5 & -4\cr
\end{matrix}
\right]
\) and \( Y = \left[
\begin{matrix}
-5 & -4\cr
2 & 1\cr
\end{matrix}
\right]
\)
Both the matrices have the same order of 2 × 2. So, we will subtract the elements of the first matrix with the respective elements of the second matrix.
So, the matrices X-Y is \( = \left[
\begin{matrix}
12 & 8\cr
5 & -4\cr
\end{matrix}
\right]
\) – \( \left[
\begin{matrix}
-5 & -4\cr
2 & 1\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
12-(-5) & 8-(-4) \cr
5-2Â & -4-1\cr
\end{matrix}
\right]
\)Â = \( \left[
\begin{matrix}
17 & 12\cr
3 & -5\cr
\end{matrix}
\right]
\)
Thus, the subtraction of X and Y is \( \left[
\begin{matrix}
17 & 12\cr
3 & -5\cr
\end{matrix}
\right]
\)
Problem 8: The matrix is,
\( A = \left[
\begin{matrix}
2 & 0\cr
3 & 5\cr
\end{matrix}
\right]
\)Â and \( B = \left[
\begin{matrix}
6 & 12\cr
4 & 5\cr
\end{matrix}
\right]
\). What is the final product value of two matrices?
Solution:
As given in the question,
The matrix A is \( A = \left[
\begin{matrix}
2 & 0\cr
3 & 5\cr
\end{matrix}
\right]
\) and \( B = \left[
\begin{matrix}
6 & 12\cr
4 & 5\cr
\end{matrix}
\right]
\)
Now, we need to find out the final product value of two matrices.
After the multiplication of two matrices is \( \left[
\begin{matrix}
(2*6) + (0*4) & (2*12) + (0*5) \cr
(3*6) + (5*4) & (3*12) + (5*5) \cr
\end{matrix}\right] \)
\( \left[
\begin{matrix}
12+0 & 24+0\cr
18+20 & 36+25\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
12 & 24\cr
38 & 61\cr
\end{matrix}
\right] \)
Therefore, the final product value of the two matrices is \( \left[
\begin{matrix}
12 & 24\cr
38 & 61\cr
\end{matrix}
\right] \).
Problem 9: Find the transpose of a column matrix \( A = \left[
\begin{matrix}
8\cr
26\cr
12\cr
\end{matrix}
\right]
\)
Solution:
The given matrix is \( A = \left[
\begin{matrix}
8\cr
26 \cr
12\cr
\end{matrix}
\right]
\)
To find the transpose of the given column matrix, the elements of the column matrix must be written as row elements.
A=[8 26 12]
Therefore, the transpose of the given column matrix is A=[8 26 12].
Problem 10: Find 3B where
\( B= \left[
\begin{matrix}
3 & 7\cr
2 & 13\cr
\end{matrix}
\right] \)?
Solution:
The given matrix is \( B = \left[
\begin{matrix}
3 & 7\cr
2 & 13\cr
\end{matrix}
\right] \)
To find the value of 3B, we need to multiply the value of 3 by all the elements of matrix B.
i.e., 3B= \( \left[
\begin{matrix}
3*3 & 3*7\cr
3*2 & 3*13\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
9 & 21\cr
6 & 39\cr
\end{matrix}
\right] \)
Therefore, the required matrix is 3C \( =\left[
\begin{matrix}
9 & 21\cr
6 & 39\cr
\end{matrix}
\right] \)
Problem 11: Let \( A =\left[
\begin{matrix}
8 & 5\cr
-4 & 3\cr
\end{matrix}
\right]
\) and \( B= \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right]
\). Prove that AB = BA = A?
Solution:
Given that, the A and B matrices values.
The matrix A is the order of 2 × 2 and the matrix B is the order of 2 × 2.
So, the number of columns in matrix A equals the number of rows in matrix B. Hence, AB can be found. Also, the number of columns in B equals the number of rows in A. So, BA can also be found.
Now, find the value of AB,
AB = \( \left[
\begin{matrix}
8 & 5\cr
-4 & 3\cr
\end{matrix}
\right]
\) \( \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
8*1 + 5*0 & 8*0+ 5*1\cr
-4*1 + 3*0 & -4*0+ 3*1\cr
\end{matrix}
\right]
\)Â = \( \left[
\begin{matrix}
8 & 5\cr
-4 & 3\cr
\end{matrix}
\right]
\) = A
Next, find the value of the BA.
\( BA= \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right]
\) \( \left[
\begin{matrix}
8 & 5\cr
-4 & 3\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
1*8 + 0*(-4) & 1*5+ 0*3\cr
0*8+ 1*(-4) & 0*5+ 1*3\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
8 & 5\cr
-4 & 3\cr
\end{matrix}
\right]
\) = A
Clearly, \( \left[
\begin{matrix}
8 & 5\cr
-4 & 3\cr
\end{matrix}
\right]
\) = \( \left[
\begin{matrix}
8 & 5\cr
-4 & 3\cr
\end{matrix}
\right]
\)
Therefore, the value of AB and BA is equal. Hence its proved AB = BA = A.
Problem 12: If \( A = \left[
\begin{matrix}
cosθ & 0\cr
0 & cosθ\cr
\end{matrix}
\right] \) and \( B = \left[
\begin{matrix}
sinθ & 0\cr
0 & sinθ\cr
\end{matrix}
\right] \) then show that A2 +B2 = I.
Solution:
In the given question, the matrices are \( A = \left[
\begin{matrix}
cosθ & 0\cr
0 & cosθ\cr
\end{matrix}
\right] \) and \( B = \left[
\begin{matrix}
sinθ & 0\cr
0 & sinθ\cr
\end{matrix}
\right] \)
Now, we will prove that the value of A2 +B2 = I.
First, find the value of A2
So, A2 = A x A = \( \left[
\begin{matrix}
cosθ & 0\cr
0 & cosθ\cr
\end{matrix}
\right] \) x \( \left[
\begin{matrix}
cosθ & 0\cr
0 & cosθ\cr
\end{matrix}
\right] \)
=Â \( \left[
\begin{matrix}
cos^2θ+0 &(0+0)\cr
(0+0) & cos^2θ+0\cr
\end{matrix}
\right] \)
The value of \( A^2 = \left[
\begin{matrix}
cos^2θ & 0\cr
0 & cos^2θ\cr
\end{matrix}
\right] \)
Now, find the value of B^2
So, the value is B x B = \( \left[
\begin{matrix}
Sinθ & 0\cr
0 & Sinθ\cr
\end{matrix}
\right] \) x \( \left[
\begin{matrix}
Sinθ & 0\cr
0 & Sinθ\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
sin^2θ+0 &(0+0)\cr
(0+0) & sin^2θ+0\cr
\end{matrix}
\right]\)
The value of \( B^2= \left[
\begin{matrix}
sin^2θ & 0\cr
0 & sin^2θ\cr
\end{matrix}
\right] \)
Now, adding the corresponding terms. We get,
A^2+B^2 = \( \left[
\begin{matrix}
cos^2θ+sin^2θ & 0 \cr
0 & cos^2θ+sin^2θ \cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right] \) = I.
Problem 13: Verify that X2 = I when \( X = \left[
\begin{matrix}
5 & -4\cr
6 & -5\cr
\end{matrix}
\right] \) ?
Solution:
As given in the question, the value of X is \( \left[
\begin{matrix}
5 & -4\cr
6 & -5\cr
\end{matrix}
\right] \)
Now, we will verify that X2 = I.
First, we will find the value of the X2.
The value of X2 is X x X = \( \left[
\begin{matrix}
5 & -4\cr
6 & -5\cr
\end{matrix}
\right] \) x \( \left[
\begin{matrix}
5 & -4\cr
6 & -5\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(25-24) & (20-20)\cr
(30-30) & (-24+25)\cr
\end{matrix}
\right] \)
So, the final value is \( \left[
\begin{matrix}
1 & 0\cr
0 & 1\cr
\end{matrix}
\right] \)
Hence, proved the value of X2 is I.
Problem 14:Â Prove that A(B + C) = AB + AC. The matrix of \( A=\left[
\begin{matrix}
1 & 3\cr
5 & -1\cr
\end{matrix}
\right] \), \( B= \left[
\begin{matrix}
1 & -1 & 2\cr
3 & 5 & 2\cr
\end{matrix}
\right] \), and \( C = \left[
\begin{matrix}
1 & 3 & 2\cr
-4 & 1 & 3\cr
\end{matrix}
\right] \) ?
Solution:
In the given questions are, the matrices are \( A=\left[
\begin{matrix}
1 & 3\cr
5 & -1\cr
\end{matrix}
\right] \), \( B= \left[
\begin{matrix}
1 & -1 & 2\cr
3 & 5 & 2\cr
\end{matrix}
\right] \), and \( C = \left[
\begin{matrix}
1 & 3 & 2\cr
-4 & 1 & 3\cr
\end{matrix}
\right] \)
Now, we will prove that A(B + C) = AB + AC.
First, take the L.H.S part that is A(B + C).
(B+C) = \( \left[
\begin{matrix}
1 & -1 & 2\cr
3 & 5 & 2\cr
\end{matrix}
\right] \) + \( \left[
\begin{matrix}
1 & 3 & 2\cr
-4 & 1 & 3\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(1+1) & (-1+3) & (2+2)\cr
(3-4) & (5+1) & (2+3)\cr
\end{matrix}
\right] \) = \( \left[
\begin{matrix}
2 & 2 & 4\cr
-1 & 6 & 5\cr
\end{matrix}
\right] \)
Now, multiply the B+C value with A, We get,
A(B+C) = \( \left[
\begin{matrix}
1 & 3 \cr
5 & -1 \cr
\end{matrix}
\right] \). \( \left[
\begin{matrix}
2 & 2 & 4\cr
-1 & 6 & 5\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(2-3)Â & (2+18) & (4+15)\cr
(10+1) & (10-6) & (20-5)\cr
\end{matrix}
\right] \)
A(B+C) = \( \left[
\begin{matrix}
-1 & 20 & 19\cr
11 & 4 & 15\cr
\end{matrix}
\right] \)
Next, find the value of the R.H.S that is AB+AC
First take AB= \( \left[
\begin{matrix}
1 & 3\cr
5 & -1\cr
\end{matrix}
\right] \) x \(\left[
\begin{matrix}
1 & -1 & 2\cr
3 & 5 & 2\cr
\end{matrix}
\right] \)
= \(\left[
\begin{matrix}
(1+9) & (-1+15) & (2+6)\cr
(5-3) & (-5-5) & (10-2)\cr
\end{matrix}
\right] \)
\( AB=\left[
\begin{matrix}
10 & 14 & 8\cr
2 & -10 & 8\cr
\end{matrix}
\right] \)
Now, find the value of AC,
AC = \( \left[
\begin{matrix}
1 & 3\cr
5 & -1\cr
\end{matrix}
\right] \) x \( \left[
\begin{matrix}
1 & 3 & 2\cr
-4 & 1 & 3\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(1-12)Â & (3+3) & (2+19)\cr
(5+4) & (15-1) & (10-3)\cr
\end{matrix}
\right] \)
So, the value of AC is \( \left[
\begin{matrix}
-11Â & 6 & 11\cr
9 & 14 & 7\cr
\end{matrix}
\right] \)
Now, let us add AB+ AC.
AB+AC = \( \left[
\begin{matrix}
10Â & 14 & 8\cr
2 & -10 & 8\cr
\end{matrix}
\right] \) + \( \left[
\begin{matrix}
-11Â & 6 & 11\cr
9 & 14 & 7\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
(10-11)Â & (14+6) & (8+11)\cr
(2+9) & (-10+14) & (8+7)\cr
\end{matrix}
\right] \)
= \( \left[
\begin{matrix}
-1 & 20 & 19\cr
11 & 4 & 15\cr
\end{matrix}
\right] \)
The value of A(B+C) and AB+AC both are Equal.
Hence, prove that L.H.S = R.H.S.
Problem 15: The matrix \( X = \left[
\begin{matrix}
11 & 2\cr
3 & 4\cr
\end{matrix}
\right]
\) and \( Y = \left[
\begin{matrix}
7 & 8\cr
6 & 5\cr
\end{matrix}
\right]
\). What is the added value of Matrices X and Y?
Solution:
As given in the question,
The matrix A is \( X = \left[
\begin{matrix}
11 & 2\cr
3 & 4\cr
\end{matrix}
\right]
\) and \( Y = \left[
\begin{matrix}
7 & 8\cr
6 & 5\cr
\end{matrix}
\right]
\)
Now, we will find the addition value of A and B.
Here, both the matrices have the same order of 2 × 2. Now, we can add the elements of the first matrix with the respective elements of the second matrix.
After, addition the matrices is \( X+Y= \left[
\begin{matrix}
11+7 & 2+8\cr
3+6 & 4+5\cr
\end{matrix}
\right]
\)
Therefore, the addition matrices value of X+Y is \( \left[
\begin{matrix}
18 & 10\cr
9 & 9\cr
\end{matrix}
\right]
\)